$K,L_{i}\subset\mathcal{T}$ discrete with $i\in I$. Then $\mathcal{T}$
satisfies _location symmetry_ if

$\begin{split}&f(K)\cap f(L_{i})=f(K)\cap f(L_{k})\quad\forall i,k\in I\\\ \
\implies\ &f(K)\subseteq\bigcap_{i\in
I}f(L_{i})\quad\lor\quad\exists!M\subset\mathcal{T}:\
\operatorname{span}(KL_{i})\setminus K=M\neq\emptyset\quad\forall i\in
I\end{split}$ (59)

where

$f(L):=\bigcap_{x\in L}J^{+}(x)$

If $K$ or $L_{i}$ is empty, we understand their intersection as being equal to
the full poset $\mathcal{T}$.

Alternatively, we can also phrase location symmetry in terms of a set of ORVs
$\mathcal{S}$ on this poset.

######  Definition G.2 (Location Symmetry for ORVs)

Let $\mathcal{T}$ be a poset and $I$ be an index set. Let
$\mathcal{X},\mathcal{Y}_{i}\subset\mathcal{S}$ with $i\in I$, not necessarily
disjoint or non-empty. Then $\mathcal{T}$ satisfies _location symmetry_

$\begin{split}&\bar{\mathcal{F}}_{s}(\mathcal{X}\mathcal{Y}_{i})=\bar{\mathcal{F}}_{s}(\mathcal{X}\mathcal{Y}_{k})\quad\forall
i,k\in I\\\ \ \implies\
&\bar{\mathcal{F}}_{s}(\mathcal{X})\subseteq\bar{\mathcal{F}}_{s}\left(\textstyle{\bigcup_{i\in
I}}\mathcal{Y}_{i}\right)\quad\lor\quad\exists!M\subset\mathcal{T}:\
O(\operatorname{span}(\mathcal{X}\mathcal{Y}_{k}))\setminus
O(\mathcal{X})=M\neq\emptyset\quad\forall k\in I\end{split}$ (60)

Especially in the respective second alternative, these definitions takes care
to play well with the respective sets / their locations potentially not being
disjoint, as this will be of importance for considering degenerate embeddings
of ORVs.

###### Lemma G.3

$\mathcal{T}$ is a conical poset if and only if it satisfies location
symmetry.

##### Proof:

For this proof, we will refer to ORVs directly, as that is more compact and
matches the notation we will use for the proofs going forward. All notation
matches Definition G.2.

“$\Rightarrow$”: Here we assume location symmetry and prove conicality. We
only require the former for the special case where $\mathcal{X}=\emptyset$
while $\mathcal{Y}_{1}$ and $\mathcal{Y}_{2}$ are any disjoint sets of ORVs
with
$\bar{\mathcal{F}}_{s}(\mathcal{Y}_{1})=\bar{\mathcal{F}}_{s}(\mathcal{Y}_{2})$.
Then, applying Definition 5.9 to empty sets of ORVs, we recover
$\bar{\mathcal{F}}_{s}(\mathcal{X})=\mathcal{T}$. Moreover, the pre-condition
$\bar{\mathcal{F}}_{s}(\mathcal{X}\mathcal{Y}_{1})=\bar{\mathcal{F}}_{s}(\mathcal{X}\mathcal{Y}_{2})$
for applying location symmetry is satisfied as it is equivalent to
$\bar{\mathcal{F}}_{s}(\mathcal{Y}_{1})=\bar{\mathcal{F}}_{s}(\mathcal{Y}_{2})$.
Therefore, location symmetry implies that

$\mathcal{T}=\bar{\mathcal{F}}_{s}(\mathcal{Y}_{1}\mathcal{Y}_{2})\ \lor\
O(\operatorname{span}(\mathcal{X}\mathcal{Y}_{1}))\setminus
O(\mathcal{X})=O(\operatorname{span}(\mathcal{X}\mathcal{Y}_{2}))\setminus
O(\mathcal{X})\,.$ (61)

Here, the second alternative immediately reduces to
$O(\operatorname{span}(\mathcal{Y}_{1}))=O(\operatorname{span}(\mathcal{Y}_{2}))$,
due to $\mathcal{X}$ being empty. The first is an equality as
$\bar{\mathcal{F}}_{s}(\mathcal{Y}_{1}\mathcal{Y}_{2})\subseteq\mathcal{T}$
trivially holds. Moreover, notice that the first case
$\mathcal{T}=\bar{\mathcal{F}}_{s}(\mathcal{Y}_{1}\mathcal{Y}_{2})$ is
equivalent to
$\mathcal{T}=\bar{\mathcal{F}}_{s}(\mathcal{Y}_{1})=\bar{\mathcal{F}}_{s}(\mathcal{Y}_{2})$,
and this can only be satisfied if both sets of ORVs are embedded into the
minimal element of the poset (which may not exist). If the minimal element
exists, we therefore have $O(\mathcal{Y}_{1})=O(\mathcal{Y}_{2})$, which in
particular implies
$O(\operatorname{span}(\mathcal{Y}_{1}))=O(\operatorname{span}(\mathcal{Y}_{2}))$
In other words, for both alternatives of location symmetry we get that for any
disjoint sets of ORVs $\mathcal{Y}_{1}$ and $\mathcal{Y}_{2}$,
$\bar{\mathcal{F}}(\mathcal{Y}_{1})=\bar{\mathcal{F}}(\mathcal{Y}_{2})\implies
O(\operatorname{span}(\mathcal{Y}_{1}))=O(\operatorname{span}(\mathcal{Y}_{2}))$,
reproducing the definition of conicality (Equation 15).

“$\Leftarrow$”: Let $\mathcal{X},\mathcal{Y}_{i}\subset\mathcal{S}$ with $i\in
I$ such that

$\bar{\mathcal{F}}_{s}(\mathcal{X}\mathcal{Y}_{i})=\bar{\mathcal{F}}_{s}(\mathcal{X}\mathcal{Y}_{k})\quad\forall
i,k\in I\,.$ (62)

We now show that in any conical space-time, this implies one of the two
conditions in the second line of Equation 60, thus proving location symmetry.

Applying the property Equation 15 of conical space-times to Equation 62, we
see that for any choice of $i\in I$, we can deduce the value of
$O(\mathcal{A})$ for all
$\mathcal{A}\in\operatorname{span}(\mathcal{X}\mathcal{Y}_{i})$ for ORVs
embedded in any conical poset. This implies for the set of respective
locations to be identical.

$O(\operatorname{span}(\mathcal{X}\mathcal{Y}_{i}))=O(\operatorname{span}(\mathcal{X}\mathcal{Y}_{k}))\quad\forall
i,k\in I\,.$ (63)

We now consider two cases and show that these yield respectively the two
alternatives for location symmetry, thus concluding the proof.

  1. 1.

Case 1 ($\exists
i:O(\mathcal{X})=O(\operatorname{span}(\mathcal{X}\mathcal{Y}_{i}))$) : In
this case, we can deduce that the same holds for all $\mathcal{Y}_{k}$ with
$k\in I$, due to Equation 62. Hence,

$\displaystyle\bar{\mathcal{F}}_{s}(\mathcal{X})$
$\displaystyle=\bar{\mathcal{F}}_{s}(\mathcal{X}\mathcal{Y}_{i})$ (64)
$\displaystyle\implies\bar{\mathcal{F}}_{s}(\mathcal{X})$
$\displaystyle\subseteq\bar{\mathcal{F}}_{s}(\mathcal{Y}_{i})\quad\forall i\in
I$ (65) $\displaystyle\iff\bar{\mathcal{F}}_{s}(\mathcal{X})$
$\displaystyle\subseteq\bigcap_{i\in
I}\bar{\mathcal{F}}_{s}(\mathcal{Y}_{i})=\bar{\mathcal{F}}_{s}\left(\bigcup_{i\in
I}\mathcal{Y}_{i}\right)\,,$ (66)

yielding the first alternative of Equation 60.

  2. 2.

Case 2 ($\forall i:O(\mathcal{X})\neq
O(\operatorname{span}(\mathcal{X}\mathcal{Y}_{i}))$) : This implies that
$O(\operatorname{span}(\mathcal{X}\mathcal{Y}_{i}))\setminus
O(\mathcal{X})\neq\emptyset$. We have already established in Equation 63 that
in conical space-times where Equation 62 is satisfied,
$O(\operatorname{span}(\mathcal{X}\mathcal{Y}_{k}))$ must be identical for all
$i\in I$, this implies the same holds for
$O(\operatorname{span}(\mathcal{X}\mathcal{Y}_{i}))\setminus O(\mathcal{X})$,
and yields the second alternative of Equation 60.

$\square$

We conclude by providing a weaker, but significantly shorter version of the
second alternative that will be useful for clarity in some later proofs.

###### Lemma G.4

Let $M\subseteq\mathcal{T}$ non-empty. Let $I,\mathcal{X},\mathcal{Y}_{k}$ as
in Definition G.2. Then

$O(\operatorname{span}(\mathcal{X}\mathcal{Y}_{k}))\setminus
O(\operatorname{span}(\mathcal{X}))=M\quad\forall k\in
I\quad\implies\quad\bigcap_{k}O(\mathcal{Y}_{k})\neq\emptyset\,.$ (67)

##### Proof:

First, we see that
$\operatorname{span}(\mathcal{X}\mathcal{Y}_{i})\setminus\mathcal{X}\subseteq\mathcal{Y}_{i}$.
As this carries over to their respective locations, we obtain

$\displaystyle\emptyset\neq
M=O(\operatorname{span}(\mathcal{X}\mathcal{Y}_{k}))\setminus
O(\operatorname{span}(\mathcal{X}))\quad\forall k\in I$ (68)
$\displaystyle\iff\ $ $\displaystyle\emptyset\neq
M=O(\operatorname{span}(\mathcal{X}\mathcal{Y}_{k}))\cap
O(\mathcal{Y}_{k})\setminus O(\operatorname{span}(\mathcal{X}))\quad\forall
k\in I$ (69) $\displaystyle\implies\ $ $\displaystyle\emptyset\neq M\subseteq
O(\operatorname{span}(\mathcal{X}\mathcal{Y}_{k}))\cap
O(\mathcal{Y}_{k})\quad\forall k\in I$ (70) $\displaystyle\implies\ $
$\displaystyle\emptyset\neq
M\subseteq\bigcap_{k}O(\operatorname{span}(\mathcal{X}\mathcal{Y}_{k}))\cap
O(\mathcal{Y}_{k})$ (71) $\displaystyle\implies\ $
$\displaystyle\emptyset\neq\bigcap_{k}O(\mathcal{Y}_{k})\,,$ (72)

where the last implication holds as intersecting larger sets always yields a
(potentially improper) superset. $\square$

###  G.3 Proofs for Section 5

To prove Theorem 6.3, we require some auxiliary lemmas, which we first prove
before stating and proving the main theorem.

###### Lemma G.5

Let $\mathcal{A},\mathcal{B}\subset\mathcal{S}$. Then

$\bar{\mathcal{F}}_{s}(\mathcal{A})\cap\bar{\mathcal{F}}_{s}(\mathcal{B}\setminus
e_{\mathcal{B}})\subseteq\bar{\mathcal{F}}(e_{\mathcal{B}})\ \forall
e_{\mathcal{B}}\in\mathcal{B}\quad\implies\quad\bar{\mathcal{F}}_{s}(\mathcal{A}\mathcal{B})=\bar{\mathcal{F}}_{s}(\mathcal{A}\mathcal{B}\setminus
e_{\mathcal{B}})\ \forall e_{\mathcal{B}}\in\mathcal{B}\,.$ (73)

##### Proof:

Let $e^{i}_{\mathcal{B}}:=e_{\mathcal{B}}$. We can transform

$\displaystyle\bar{\mathcal{F}}_{s}(\mathcal{A})\cap\bar{\mathcal{F}}_{s}(\mathcal{B}\setminus
e^{i}_{\mathcal{B}})$
$\displaystyle\subseteq\bar{\mathcal{F}}(e^{i}_{\mathcal{B}})$ (74)
$\displaystyle\implies\bar{\mathcal{F}}_{s}(\mathcal{A})\cap\bar{\mathcal{F}}_{s}(\mathcal{B}\setminus
e^{i}_{\mathcal{B}})$
$\displaystyle\subseteq\bar{\mathcal{F}}_{s}(\mathcal{A})\cap\bar{\mathcal{F}}(e^{i}_{\mathcal{B}})$
$\displaystyle\forall e^{i}_{\mathcal{B}}\in\mathcal{B}$ (75)
$\displaystyle\implies\bar{\mathcal{F}}_{s}(\mathcal{A})\cap\bar{\mathcal{F}}_{s}(\mathcal{B}\setminus
e_{\mathcal{B}}^{i})$
$\displaystyle\subseteq\bar{\mathcal{F}}_{s}(\mathcal{A})\cap\bar{\mathcal{F}}(e_{\mathcal{B}}^{i})\cap\bar{\mathcal{F}}_{s}(\mathcal{B}\setminus\\{e^{i}_{\mathcal{B}},e^{j}_{\mathcal{B}}\\})$
$\displaystyle\forall e_{\mathcal{B}}^{i},e_{\mathcal{B}}^{j}\in\mathcal{B}$
(76)
$\displaystyle\iff\bar{\mathcal{F}}_{s}(\mathcal{A})\cap\bar{\mathcal{F}}_{s}(\mathcal{B}\setminus
e_{\mathcal{B}}^{i})$
$\displaystyle\subseteq\bar{\mathcal{F}}_{s}(\mathcal{A})\cap\bar{\mathcal{F}}_{s}(\mathcal{B}\setminus
e_{\mathcal{B}}^{j})$ $\displaystyle\forall
e_{\mathcal{B}}^{i},e_{\mathcal{B}}^{j}\in\mathcal{B}\,.\\!\\!\\!$ (77)

Combining these for all $e_{\mathcal{B}}$, we get

$\displaystyle\bar{\mathcal{F}}_{s}(\mathcal{A})\cap\bar{\mathcal{F}}_{s}(\mathcal{B}\setminus
e_{\mathcal{B}}^{i})$
$\displaystyle=\bar{\mathcal{F}}_{s}(\mathcal{A})\cap\bar{\mathcal{F}}_{s}(\mathcal{B}\setminus
e_{\mathcal{B}}^{j})\quad\forall
e_{\mathcal{B}}^{i},e_{\mathcal{B}}^{j}\in\mathcal{B}$ (78)
$\displaystyle\implies\bar{\mathcal{F}}_{s}(\mathcal{A}\mathcal{B})$
$\displaystyle=\bar{\mathcal{F}}_{s}(\mathcal{A}\mathcal{B}\setminus
e_{\mathcal{B}})\quad\forall e_{\mathcal{B}}\in\mathcal{B}\,.$ (79)

This is precisely the claim. $\square$

###### Lemma G.6

Let $\mathscr{A}$ be a set of unconditional affects relations and
$\mathscr{A}^{\prime}\subseteq\mathscr{A}$ consist of all affects relations in
$\mathscr{A}$ that are Irred3. Then for any non-degenerate embedding
$\mathcal{E}$ into an arbitrary space-time $\mathcal{T}$ satisfying compat we
have

$(X\operatorname{\vDash}Y\operatorname{\,|\,}\operatorname{do}(Z))\in\mathscr{A}^{\prime}\quad\implies\quad\bar{\mathcal{F}}_{s}(\mathcal{Y}\mathcal{X})\cap\bar{\mathcal{F}}_{s}(\mathcal{Z}\setminus
e_{\mathcal{Z}})\subseteq\bar{\mathcal{F}}(e_{\mathcal{Z}})\ \forall
e_{\mathcal{Z}}\in\mathcal{Z}\,.$ (80)

##### Proof:

As $X\operatorname{\vDash}Y\operatorname{\,|\,}\operatorname{do}(Z)$ satisfies
Irred3, we have for $\tilde{Z}:=Z\backslash e_{Z}$

$\forall e_{Z}\in Z:\quad
e_{Z}\operatorname{\vDash}Y\operatorname{\,|\,}\operatorname{do}(\tilde{Z})\
\lor\
e_{Z}\operatorname{\vDash}Y\operatorname{\,|\,}\operatorname{do}(\tilde{Z}X)\,,$
(81)

either of which is Irred1, if it holds. Therefore, by compat, we get

$\displaystyle\bar{\mathcal{F}}_{s}(\mathcal{Y})\cap\bar{\mathcal{F}}_{s}(\tilde{\mathcal{Z}})\subseteq\bar{\mathcal{F}}(e_{\mathcal{Z}})\
\lor\
\bar{\mathcal{F}}_{s}(\mathcal{Y})\cap\bar{\mathcal{F}}_{s}(\mathcal{X}\tilde{\mathcal{Z}})\subseteq\bar{\mathcal{F}}(e_{\mathcal{Z}})$
(82) $\displaystyle\implies\
\bar{\mathcal{F}}_{s}(\mathcal{Y}\mathcal{X})\cap\bar{\mathcal{F}}_{s}(\tilde{\mathcal{Z}})\subseteq\bar{\mathcal{F}}(e_{\mathcal{Z}})$
(83)

for any $e_{Z}$. $\square$

With these three auxiliary lemmas in place, we can continue to prove the
theorem.

###### Lemma G.7

Let $\mathscr{A}$ be a set of unconditional affects relations and
$\mathscr{A}^{\prime}\subseteq\mathscr{A}$ consist of all affects relations in
$\mathscr{A}$ that are Irred3. Then for any non-degenerate embedding
$\mathcal{E}$ into a conical space-time $\mathcal{T}$ satisfying compat, we
have

$(X\operatorname{\vDash}Y\operatorname{\,|\,}\operatorname{do}(Z))\in\mathscr{A}^{\prime}\quad\implies\quad\bar{\mathcal{F}}_{s}(\mathcal{Y})\cap\bar{\mathcal{F}}_{s}(\mathcal{X})\subseteq\bar{\mathcal{F}}_{s}(\mathcal{Z})\,.$
(84)

##### Proof:

By conjoining Lemma G.6 with Lemma G.5 for
$\mathcal{A}=\mathcal{Y}\mathcal{X}$ and $\mathcal{B}=\mathcal{Z}$, for any
affects relation
$X\operatorname{\vDash}Y\operatorname{\,|\,}\operatorname{do}(Z)$ irreducible
in the third argument, we get

$\bar{\mathcal{F}}_{s}(\mathcal{X}\mathcal{Y}\mathcal{Z})=\bar{\mathcal{F}}_{s}(\mathcal{X}\mathcal{Y}\mathcal{Z}\setminus
e_{\mathcal{Z}})\quad\forall e_{\mathcal{Z}}\in\mathcal{Z}\,.$ (85)

Due to conicality, Lemma G.3 implies that $\mathcal{T}$ satisfies location
symmetry. By using this property on the chain of equalities (setting
$\mathcal{X}$ in Equation 60 to $\mathcal{X}\mathcal{Y}$, and
$\mathcal{Y}_{i}$ to $\mathcal{Z}_{i}=\mathcal{Z}\setminus
e_{\mathcal{Z}}^{i}$) as well as the transformation of Lemma G.4, we arrive at

$\bar{\mathcal{F}}_{s}(\mathcal{X}\mathcal{Y})\subseteq\bar{\mathcal{F}}_{s}(\mathcal{Z})\
\lor\ \bigcap_{i}O(\mathcal{Z}_{i})\neq\emptyset\,,$ (86)

where $Z_{i}:=Z\setminus e_{i}$.

For the second alternative, beware that the different $\mathcal{Z}_{i}$ are
not disjoint. Nonetheless, $\bigcap_{i}\mathcal{Z}_{i}=\emptyset$. Therefore,
multiple ORVs must share the same location. As this corresponds to a
degenerate embedding, we arrive at the claim. $\square$

See 6.3

##### Proof:

Due to Lemma G.7, compat implies the following for an Irred1 and Irred3
affects relation

$(X\operatorname{\vDash}Y\operatorname{\,|\,}\operatorname{do}(Z))\in\mathscr{A}^{\prime}\quad\implies\quad\bar{\mathcal{F}}_{s}(\mathcal{Y})\cap\bar{\mathcal{F}}_{s}(\mathcal{X})\subseteq\bar{\mathcal{F}}_{s}(\mathcal{Z})\,.$
(87)

Conjoining this with compat for this affects relation (which implies
$\bar{\mathcal{F}}_{s}(\mathcal{Y})\cap\bar{\mathcal{F}}_{s}(\mathcal{Z})\subseteq\bar{\mathcal{F}}_{s}(\mathcal{X})$)
and using elementary set theory yields

$\displaystyle\bar{\mathcal{F}}_{s}(\mathcal{Y})\cap\bar{\mathcal{F}}_{s}(\mathcal{Z})\subseteq\bar{\mathcal{F}}_{s}(\mathcal{X})\cap\bar{\mathcal{F}}_{s}(\mathcal{Y})\
\land\
\bar{\mathcal{F}}_{s}(\mathcal{Y})\cap\bar{\mathcal{F}}_{s}(\mathcal{X})\subseteq\bar{\mathcal{F}}_{s}(\mathcal{Z})\cap\bar{\mathcal{F}}_{s}(\mathcal{Y})$
(88) $\displaystyle\implies\ $
$\displaystyle\bar{\mathcal{F}}_{s}(\mathcal{Y})\cap\bar{\mathcal{F}}_{s}(\mathcal{X})=\bar{\mathcal{F}}_{s}(\mathcal{Y})\cap\bar{\mathcal{F}}_{s}(\mathcal{Z})$
(89)

Due to conicality, Lemma G.3 implies that $\mathcal{T}$ satisfies location
symmetry. Using location symmetry, we arrive at

$\bar{\mathcal{F}}_{s}(\mathcal{Y})\subseteq\bar{\mathcal{F}}_{s}(\mathcal{X}\mathcal{Z})\
\lor\ O(\mathcal{X})\cap O(\mathcal{Z})\neq\emptyset\,.$ (90)

As the second option would imply the embedding to be degenerate, the claim,
corresponding to the first option, follows. $\square$

See 6.4

##### Proof:

Every affects relation $X\vDash Y|\mathrm{do}(Z)$ in $\mathscr{A}^{\prime}$ is
Irred1 and Irred3. By Lemma D.6, due to the absence of affects relations with
Clus3, there exists $e^{1}_{X}\in X$ such that
$e^{1}_{X}\operatorname{\vDash}Y$. Therefore, for any compatible embedding
$\mathcal{E}$, we have
$\bar{\mathcal{F}}_{s}(\mathcal{Y})\subseteq\bar{\mathcal{F}}(e^{1}_{\mathcal{X}})$.
However, due to $X\vDash Y|\mathrm{do}(Z)$ being Irred1, we also have
$X\setminus
e^{1}_{X}\operatorname{\vDash}Y\operatorname{\,|\,}\operatorname{do}(Ze^{1}_{X})$,
which again by Lemma D.6 implies that there is $e^{2}_{X}\in X\setminus
e^{1}_{X}$ such that $e^{2}_{X}\operatorname{\vDash}Y$. Repeating this
argument recursively, we find that any compatible embedding $\mathcal{E}$ of
such affects relations in a space-time will satisfy
$\bar{\mathcal{F}}_{s}(\mathcal{Y})\subseteq\bar{\mathcal{F}}(e_{\mathcal{X}})\
\forall e_{\mathcal{X}}\in\mathcal{X}$, or in short,
$\bar{\mathcal{F}}_{s}(\mathcal{Y})\subseteq\bar{\mathcal{F}}_{s}(\mathcal{X})$.

Next we use the fact that $X\vDash Y|\mathrm{do}(Z)$ is Irred3, which gives
$s_{Z}\operatorname{\vDash}Y\operatorname{\,|\,}\operatorname{do}(\tilde{s}_{Z}X)$
or
$s_{Z}\operatorname{\vDash}Y\operatorname{\,|\,}\operatorname{do}(\tilde{s}_{Z})$
for all choices of $s_{Z}\subsetneq Z$. Applying Lemma D.6, both these
alternatives yield $\exists e^{1}_{Z}\in Z$ s.t.
$e^{1}_{Z}\operatorname{\vDash}Y$ and
$\bar{\mathcal{F}}_{s}(\mathcal{Y})\subseteq\bar{\mathcal{F}}(e^{1}_{\mathcal{Z}})$.
Moreover, $X\vDash Y|\mathrm{do}(Z)$ being Irred3 implies in particular that
$Z\setminus
e^{1}_{Z}\operatorname{\vDash}Y\operatorname{\,|\,}\operatorname{do}(e_{Z}^{1}X)$
or $Z\setminus
e^{1}_{Z}\operatorname{\vDash}Y\operatorname{\,|\,}\operatorname{do}(e_{Z}^{1})$,
both of which imply by Lemma D.6 that $\exists e^{2}_{Z}\in Z\setminus
e^{1}_{Z}$ s.t. $e^{2}_{Z}\operatorname{\vDash}Y$. Repeating this procedure
recursively, we obtain that for any compatible embedding $\mathcal{E}$ of
these affects relations, we have
$\bar{\mathcal{F}}_{s}(\mathcal{Y})\subseteq\bar{\mathcal{F}}(e_{\mathcal{Z}})\
\forall e_{\mathcal{Z}}\in\mathcal{Z}$, or in short,
$\bar{\mathcal{F}}_{s}(\mathcal{Y})\subseteq\bar{\mathcal{F}}_{s}(\mathcal{Z})$.

Combining both results, we recover the claim. $\square$

###  G.4 Proofs for Appendix E

See E.3

##### Proof:

Suppose by contradiction that $X\not\vDash Y|\\{\mathrm{do}(Z),W\\}$ (which
amounts to
$P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|XZW)=P_{\mathcal{G}_{\mathrm{do}(Z)}}(Y|ZW)$)
and
$(X\not\mathrel{\perp\mkern-11.0mu\perp}Y|ZW)_{\mathcal{G}_{\mathrm{do}}(XZ)}$.
The latter is equivalent to saying that there exist values $z$ of $Z$, $w$ of
$W$ and distinct values $x$ and $x^{\prime}$ of $X$ such that
$P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|X=x,Z=z,W=w)\neq
P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|X=x^{\prime},Z=z,W=w)$. However, this
implies that $P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|XZ)$ does have a non-trivial
dependence on $X$ and contradicts
$P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|XZW)=P_{\mathcal{G}_{\mathrm{do}(Z)}}(Y|ZW)$
(which would imply that it is equal to an $X$-independent quantity). This
proves the result. $\square$

See E.12

##### Proof:

Suppose $X\vDash Y|\mathrm{do}(Z),W$ holds and is a Red4 affects relation.
Then writing out this affects relation along with the two conditions implied
by the reducibility (while recalling that $s_{W}\cup\tilde{s}_{W}=W$), we have

$\displaystyle\begin{split}P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|XZW)&\neq
P_{\mathcal{G}_{\mathrm{do}(Z)}}(Y|ZW)\\\
P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Ys_{W}|XZ\tilde{s}_{W})&=P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|XZ\tilde{s}_{W})P_{\mathcal{G}_{\mathrm{do}(XZ)}}(s_{W}|XZ\tilde{s}_{W})\\\
P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Ys_{W}|Z\tilde{s}_{W})&=P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|Z\tilde{s}_{W})P_{\mathcal{G}_{\mathrm{do}(XZ)}}(s_{W}|Z\tilde{s}_{W}).\end{split}$
(91)

The last two equalities are equivalent to the following.

$\displaystyle\begin{split}P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|XZW)&=P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|XZ\tilde{s}_{W})\\\
P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|ZW)&=P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|Z\tilde{s}_{W})\end{split}$
(92)

Plugging these back into the first expression of Equation 91 we obtain
$P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|XZ\tilde{s}_{W})\neq
P_{\mathcal{G}_{\mathrm{do}(Z)}}(Y|Z\tilde{s}_{W})$, which is equivalent to
$X\vDash Y|\mathrm{do}(Z),\tilde{s}_{W}$. $\square$

See E.17

##### Proof:

For $i\in\\{1,2,3\\}$, the proofs are entirely analogous to that of Theorem
4.11 and will not be repeated. We carry out the proof for $i=4$ below.

Again, start by assuming that $X\vDash Y|\mathrm{do}(Z),W$ holds and satisfies
Clus4 and Red4. From Definition E.16 and Definition E.8 this implies the
following conditions (see also Equation 92).

$\displaystyle\begin{split}P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|XZ\tilde{s}_{W})&=P_{\mathcal{G}_{\mathrm{do}(Z)}}(Y|Z\tilde{s}_{W}),\quad\forall\tilde{s}_{W}\subsetneq
W,\\\
P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|XZW)&=P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|XZ\tilde{s}_{W})\quad\exists\tilde{s}_{W}\subseteq
W,\\\
P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|ZW)&=P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|Z\tilde{s}_{W})\quad\exists\tilde{s}_{W}\subseteq
W.\end{split}$ (93)

These imply that
$P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|XZW)=P_{\mathcal{G}_{\mathrm{do}(Z)}}(Y|ZW)$,
which is equivalent to $X\not\vDash Y|\mathrm{do}(Z),W$. This contradicts our
initial assumption that $X\vDash Y|\mathrm{do}(Z),W$ and establishes the
result. $\square$

See E.18

##### Proof:

The proofs for $i\in\\{1,2,3\\}$ are entirely analogous as those of Theorem
4.12 whenever we are given that $X$ is not a cause of $W$. This is because the
main step of the proof is to infer from $X\vDash Y|\mathrm{do}(Z)$ that
$(X\not\perp^{d}Y|Z)_{\mathcal{G}_{\mathrm{do}(XZ)}}$, and the analogous
inference holds for conditional affects relations when $X$ is not a cause of
$W$, due to Lemma E.4 i.e., $X\vDash Y|\\{\mathrm{do}(Z),W\\}$ implies
$(X\not\perp^{d}Y|ZW)_{\mathcal{G}_{\mathrm{do}(XZ)}}$ whenever $X$ is not a
cause of $W$. We therefore do not repeat the proof.

The proof for $i=4$ is also analogous, but we repeat it for completeness.
Suppose that $X\vDash Y|\mathrm{do}(Z),W$ holds and is a Clus4 affects
relation. This implies in particular that for all $e_{W}\in W$, $X\not\vDash
Y|\mathrm{do}(Z),e_{W}$. Writing this out, we have

$\displaystyle\begin{split}P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|XZW)&\neq
P_{\mathcal{G}_{\mathrm{do}(Z)}}(Y|ZW)\\\
P_{\mathcal{G}_{\mathrm{do}(XZ)}}(Y|XZe_{W})&=P_{\mathcal{G}_{\mathrm{do}(Z)}}(Y|Ze_{W}),\qquad\forall
e_{W}\in W.\end{split}$ (94)

Further, using Lemma E.3, the non-affects relations $X\not\vDash
Y|\mathrm{do}(Z),e_{W}$ imply the conditional independences
$(X\mathrel{\perp\mkern-11.0mu\perp}Y|Z,e_{W})_{\mathcal{G}_{\mathrm{do}(XZ)}}$.
Now consider the corresponding d-separation for each $e_{W}\in W$, we can
either have $(X\perp^{d}Y|Z,e_{W})_{\mathcal{G}_{\mathrm{do}(XZ)}}$ or
$(X\not\perp^{d}Y|Z,e_{W})_{\mathcal{G}_{\mathrm{do}(XZ)}}$. In the latter
case we could have a d-connection with a corresponding conditional
independence, which would make the model fine-tuned. So we consider the case
where $(X\perp^{d}Y|Z,e_{W})_{\mathcal{G}_{\mathrm{do}(XZ)}}$ holds for all
$e_{W}\in W$ (for all other cases the above argument establishes the result
about fine-tuning). However, by assumption that $X$ is not a cause of $W$, we
have $X(W)=W$ and we know from Lemma E.4 that
$(X\perp^{d}Y|Z,e_{W})_{\mathcal{G}_{\mathrm{do}(XZ)}}$ implies $X\not\vDash
Y|\\{\mathrm{do}(Z),W\\}$, which contradicts our initial assumption and
established the result. $\square$

See E.19

##### Proof:

Any $X\vDash Y|\mathrm{do}(Z)$ which is Clus2, by Lemma E.2.3 is equivalent to
$X\vDash s_{Y}|\mathrm{do}(Z),\tilde{s}_{Y}$ for each partition
$s_{Y}\tilde{s}_{Y}=Y$.

If for the case of $\tilde{s}_{Y}\geq 2$ – which is only possible for $Y\geq
2$ – this affects relation were not Clus4, there would exist
$X\operatorname{\vDash}s_{Y}|\mathrm{do}(Z),\tilde{s}^{\prime}_{Y}$ with
$\tilde{s}^{\prime}_{Y}\subsetneq\tilde{s}_{Y}$, implying
$X\operatorname{\vDash}s_{Y}\tilde{s}^{\prime}_{Y}|\mathrm{do}(Z)$ by Lemma
E.2.1. This is in contradiction to the original affects relation being Clus2
and proves the claim. $\square$


# Zeta-functions of Curves over Finite Fields

Kin Wai (Edisy) Chan
Mentor: Dr. Eric Wambach

###### Abstract

Curves over finite fields are of great importance in cryptography and coding
theory. Through studying their zeta-functions, we would be able to find out
vital arithmetic and geometric information about them and their Jacobians,
including the number of rational points on this kind of curves. In this paper,
I investigate if it is possible to construct a curve over finite fields of a
given genus $g$ whose zeta-function is given as a product of zeta-functions of
$g$ elliptic curves, and find out alternative methods if it is not possible.
Basically, I look for conditions which those $g$ elliptic curves should
satisfy such that their product (of their Jacobians) is isogenous to the
Jacobian of a curve of a given genus $g$. Then from this isogenous
relationship I can determine the characteristic polynomial of the Frobenius
endomorphism of the Jacobian of the new curve and by this characteristic
polynomial I can thus determine the zeta-function of this new curve. By using
the zeta-functions of curves in the form as generating functions, the number
of rational points on curves can even be found out, which may lead to further
researches relating to some applications in cryptography, coding theory and
even information theory.

##  1 Introduction

It is known that to determine all possible zeta-functions of the curves over
finite fields of a given genus g is a very difficult task. In some sense, this
is the same task of determining whether a given abelian variety is isogenous
to a Jacabian. For $g=1$, the problem is completely solved by Deuring[3]. For
g=2, Serre[4] and R$\mathrm{\ddot{u}}$ck[5] give the partial answer.

In this paper, I investigate for which $g$ it is possible to apply an
elementary construction method using elliptic curves to find some of the zeta-
functions of curves of genus $g$ whose Jacobian is isogenous to the product of
$g$ elliptic curves. That is to say, I find out when and how the product of
$g$ elliptic curves satisfying certain conditions is isogenous to the Jacobian
of a curve of genus $g$. This is possible for $g=3$ (Corollary 3.4), but it is
impossible for $g\geq 4$ (Remark 4.3). A direct result of Corollary 3.4 is
Theorem 4.2 where the conditions are more numerical than those in Corollary
3.4.

There is a fact that the reciprocal of the numerator of the zeta-function of a
curve is the characteristic polynomial of the Frobenius endomorphism of the
Jacobian of the curve. Hence, it suffices to determine which polynomial can
become the characteristic polynomial of the Frobenius endomorphism of the
Jacobian of a curve of genus 3 in order to determine some of the zeta-
functions of curves of genus 3.

In Section 2, I introduce some basic facts about elliptic curves and study
2-torsion points. An elementary construction method using elliptic curves is
presented in Section 3. I determine part of the characteristic polynomials of
the Frobenius endomorphisms of the Jacobians of curves of genus 3 and explain
briefly why this cannot be done for $g\geq 4$ in Section 4.

I adopt the following notations throughout the paper:

  * •

$q=p^{r}$: $p$ is a prime and $r\in\mathbb{N}$;

  * •

$\chi/\mathbb{F}_{q}$: a (smooth projective) algebraic curve over
$\mathbb{F}_{q}$;

  * •

$\mathcal{E}/\mathbb{F}_{q}$: an elliptic curve over $\mathbb{F}_{q}$;

  * •

$g=g(\chi)$: the genus of $\chi$;

  * •

$\mathbb{F}_{q}(\chi)$: the function field of $\chi$;

  * •

$\chi(\mathbb{F}_{q})$: the set of the $\mathbb{F}_{q}$–rational points of
$\chi$;

  * •

$\mathcal{J}_{\chi}$: the Jacobian variety of $\chi$;

  * •

$\pi_{\mathcal{A}}$: the Frobenius morphism of a variety
$\mathcal{A}/\mathbb{F}_{q}$;

  * •

$\pi$: the Frobenius morphism of $\mathbb{P}^{1}/\mathbb{F}_{q}$.

##  2 Elliptic curve

An elliptic curve is a 1–dimensional smooth projective variety of genus 1 with
a specified base point, i.e. an algebraic curve of genus 1 having a specified
rational point. By considering the map
$\mathcal{E}/\mathbb{F}_{q}\rightarrow\mathbb{P}^{1}/\mathbb{F}_{q}$ such that
$[x:y:1]\mapsto[x:1]$ and $[0:1:0]\mapsto[1:0]$ , we know that $\mathcal{E}$
is a covering of degree 2 over the projective line $\mathbb{P}^{1}$ with set
of ramified points being the set $\mathcal{E}[2]$ of the 2-torsion points. It
is clear that $|\mathcal{E}[2]|=4$ for odd $p$ and $|\mathcal{E}[2]|=1$ or $2$
for $p=2$.

Since I attempt to apply a similar method in [1] to do the generalization, I
need to use the following lemma from [9] as well.

######  Lemma 2.1.

[9, Theorem 4.1] Let $q=p^{r}$ a prime $p$ and an integer $r\geq 1$. Let $t$
be an integer with $|t|\leq 2\sqrt{q}$. Then there is an elliptic curve over
$\mathbb{F}_{q}$ with $q+1+t$ rational points if and only if $t$ satisfies one
of the following conditions:

  * (i)

$\gcd(p,t)=1$;

  * (ii)

$r$ is even: $t=\pm 2\sqrt{q}$;

  * (iii)

$r$ is even and $p\not\equiv 1\pmod{3}$: $t=\pm\sqrt{q}$;

  * (iv)

$r$ is odd and $p=2$ or 3: $t=\pm\sqrt{pq}$;

  * (v)

$r$ is odd: $t=0$;

  * (vi)

$r$ is even and $p\not\equiv 1\pmod{4}$: $t=0$.

Let $R_{\mathcal{E}}$ be the set of ramified points of $\mathbb{P}^{1}$ for
the covering
$\mathcal{E}/\mathbb{F}_{q}\rightarrow\mathbb{P}^{1}/\mathbb{F}_{q}$. By
looking at the form of the covering map, we can easily know that
$|R_{\mathcal{E}}|=|\mathcal{E}[2]|$. Conversely, if we have
$R\subseteq\mathbb{P}^{1}$ such that $\pi(R)=R$ and a $Q\in R$ such that
$\pi(Q)=Q$ for the Frobenius map $\pi$, and if $|R|=4$ for odd $p$ and $|R|=1$
or 2 for $p=2$ as well, then there exists an automorphism $g$ in the
projective general linear group $\mathrm{PGL}(2,q)$ such that $g(Q)=[1:0]$ and
an elliptic curve $\mathcal{E}/\mathbb{F}_{q}$ such that $R_{\mathcal{E}}=R$.

|
---|---
$\textstyle{\mathcal{E}[2]\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{E}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{R_{\mathcal{E}}}$$\textstyle{\mathbb{P}^{1}}$

Now if we suppose that $p$ is odd, then an elliptic curve $\mathcal{E}$ is
totally determined by $R_{\mathcal{E}}$. That is to say if there exists an
automorphism $\sigma\in\mathrm{PGL}(2,q)$ such that
$\sigma(R_{\mathcal{E}_{1}})=R_{\mathcal{E}_{2}}$, then
$\mathcal{E}_{1}\cong\mathcal{E}_{2}$. Throughout this paper, p is assumed to
be odd, so we have $|R_{\mathcal{E}}|=|\mathcal{E}[2]|=4$ for any elliptic
curve $\mathcal{E}/\mathbb{F}_{q}$.

From now on, I fix the map
$\mathcal{E}/\mathbb{F}_{q}\rightarrow\mathbb{P}^{1}/\mathbb{F}_{q}$ so that
the notation $R_{\mathcal{E}}$ makes sense.

In studying elliptic curves, we sometimes use another more convenient form of
Weierstrass equation for an elliptic curve. This is called the Legendre form.

######  Definition 2.2.

An elliptic curve $\mathcal{E}$ is in Legendre form if its Weierstrass
equation can be written as

$y^{2}=x(x-1)(x-\lambda).$

In this paper, $\lambda$ is called a Legendre coefficient of $\mathcal{E}$. If
an elliptic curve $\mathcal{E}$ is in Legendre form, then $\mathcal{E}$ is
denoted as $\mathcal{E}_{\lambda}$ instead. In fact, we have the following
proposition illustrating the usefulness of Legendre form.

######  Proposition 2.3.

[6, Proposition III.1.7 (a)(b)] Let $p$ be an odd prime. Then every elliptic
curve $\mathcal{E}/\mathbb{F}_{q}$ is isomorphic (over $\bar{\mathbb{F}}_{q}$)
to an elliptic curve in Legendre form

$\mathcal{E}_{\lambda}:y^{2}=x(x-1)(x-\lambda)$

for some $\lambda\in\bar{\mathbb{F}}_{q}$, $\lambda\neq 0,1$ (Note that
$\mathcal{E}_{\lambda}$ is defined over $\bar{\mathbb{F}}_{q}$). Moreover, the
$j$-invariant of $\mathcal{E}_{\lambda}$ is

$j(\mathcal{E}_{\lambda})=2^{8}\frac{(\lambda^{2}-\lambda+1)^{3}}{\lambda^{2}(\lambda-1)^{2}}.$

In order to facilitate our discussion on elliptic curves, I define a new term
regarding the relation between $\lambda_{i}$ of elliptic curves
$\mathcal{E}_{\lambda_{i}}$.

######  Definition 2.4.

Let $\lambda_{1}$ and $\lambda_{2}$ be Legendre coefficients of
$\mathcal{E}_{\lambda_{1}}$ and $\mathcal{E}_{\lambda_{2}}$ respectively. Then
$\lambda_{1}$ is Legendre equivalent to $\lambda_{2}$ if

$\lambda_{2}\in\\{\lambda_{1},1-\lambda_{1},\frac{1}{\lambda_{1}},\frac{1}{1-\lambda_{1}},\frac{\lambda_{1}-1}{\lambda_{1}},\frac{\lambda_{1}}{\lambda_{1}-1}\\}.$

It is not difficult to check that it is indeed an equivalence relation. Then
we have an important proposition telling us when two elliptic curves in
Legendre forms is isomorphic to each other.

######  Proposition 2.5.

[7, Proposition 4(1.3)] Let $\mathcal{E}_{\lambda_{1}}$ and
$\mathcal{E}_{\lambda_{2}}$ be two elliptic curves in Legendre forms. Then
$\mathcal{E}_{\lambda_{1}}$ is isomorphic to $\mathcal{E}_{\lambda_{2}}$ (over
$\bar{\mathbb{F}}_{q}$)111In fact, to be more precise,
$\mathcal{E}_{\lambda_{1}}$ is isomorphic to $\mathcal{E}_{\lambda_{2}}$ over
$\mathbb{F}_{q}(j(\mathcal{E}_{\lambda_{1}}))=\mathbb{F}_{q}(j(\mathcal{E}_{\lambda_{2}}))$.
if and only if $\lambda_{1}$ is Legendre equivalent to $\lambda_{2}$.

The following lemma is also used in an essential place in the Lemma 2.7, which
is a significant lemma in this paper.

######  Lemma 2.6.

[1, Lemma 2.3] Let $\mathcal{E}_{i}/\mathbb{F}_{q}$ $(i=1,2)$ be elliptic
curves such that $\mathcal{E}_{1}$ is not isomorphic to $\mathcal{E}_{2}$ and
$|\mathcal{E}_{1}(\mathbb{F}_{q})\cap\mathcal{E}_{1}[2]|=|\mathcal{E}_{2}(\mathbb{F}_{q})\cap\mathcal{E}_{2}[2]|$,
then there exists an elliptic curve $\mathcal{E}_{3}/\mathbb{F}_{q}$ such that
$\mathcal{E}_{3}$ is isomorphic to $\mathcal{E}_{2}$ and
$|R_{\mathcal{E}_{1}}\cap R_{\mathcal{E}_{3}}|=3$.

Now we have sufficient machinery to prove the Lemma 2.7, which will be used
later in an important place. Note that $j_{i}$ represents the $j$-invariant of
$\mathcal{E}_{i}$.

######  Lemma 2.7.

Let $\mathcal{E}_{i}/\mathbb{F}_{q}$ $(i=1,2,3)$ be pairwise non-isomorphic
elliptic curves. Suppose $j_{3}\in\mathbb{F}_{q}$ and
$\mathcal{E}_{i}/\mathbb{F}_{q}$ $(i=1,2,3)$ satisfy any one of the following
sets of conditions:

  * (i)

$|\mathcal{E}_{1}(\mathbb{F}_{q})\cap\mathcal{E}_{1}[2]|=|\mathcal{E}_{2}(\mathbb{F}_{q})\cap\mathcal{E}_{2}[2]|=|\mathcal{E}_{3}(\mathbb{F}_{q})\cap\mathcal{E}_{3}[2]|=2$;

  * (ii)

$\displaystyle\frac{\lambda_{2}(1-\lambda_{1})}{\lambda_{2}-\lambda_{1}}$ is
Legendre equivalent to $\lambda_{3}$,

or

  * (i)

$|\mathcal{E}_{1}(\mathbb{F}_{q})\cap\mathcal{E}_{1}[2]|=|\mathcal{E}_{2}(\mathbb{F}_{q})\cap\mathcal{E}_{2}[2]|=|\mathcal{E}_{3}(\mathbb{F}_{q})\cap\mathcal{E}_{3}[2]|=4$;

  * (ii)

$\displaystyle\frac{\lambda_{2}}{\lambda_{1}}$ is Legendre equivalent to
$\lambda_{3}$,

where $\lambda_{1}$, $\lambda_{2}$ and $\lambda_{3}$ are Legendre coefficients
of $\mathcal{E}_{1}$, $\mathcal{E}_{2}$ and $\mathcal{E}_{3}$ respectively.
Then there exists elliptic curves $\mathcal{E}^{\prime}_{i}/\mathbb{F}_{q}$
$(i=1,2,3)$ such that $\mathcal{E}^{\prime}_{i}$ is isomorphic to
$\mathcal{E}_{i}$ over $\mathbb{F}_{q}$ ($\forall i=1,2,3$) and:

  * (i)

$|R_{\mathcal{E}^{\prime}_{i}}\cap R_{\mathcal{E}^{\prime}_{j}}|=3$ $\forall
i\neq j$;

  * (ii)

$|R_{\mathcal{E}^{\prime}_{1}}\cap R_{\mathcal{E}^{\prime}_{2}}\cap
R_{\mathcal{E}^{\prime}_{3}}|=2$.

###### Proof.

By Lemma 2.6, we may assume without loss of generality that $\mathcal{E}_{1}$
and $\mathcal{E}_{2}$ possess the property $|R_{\mathcal{E}_{1}}\cap
R_{\mathcal{E}_{2}}|=3$. We may also assume:

$\mathcal{E}_{1}:y^{2}=(x-p_{1})(x-p_{2})(x-p_{3})\quad\mathrm{and}\quad\mathcal{E}_{2}:y^{2}=(x-p_{1})(x-p_{2})(x-p_{4}).$

Then we can consider
$R_{\mathcal{E}_{1}}=\\{\mathcal{O},[p_{1}:1],[p_{2}:1],[p_{3}:1]\\}$ and
$R_{\mathcal{E}_{2}}=\\{\mathcal{O},[p_{1}:1],[p_{2}:1],[p_{4}:1]\\}$, where
$\mathcal{O}$ is the point of infinity. If we change $\mathcal{E}_{1}$ and
$\mathcal{E}_{2}$ into Legendre forms, we may assume
$\lambda_{1}=\displaystyle\frac{p_{3}-p_{1}}{p_{2}-p_{1}}$ and
$\lambda_{2}=\displaystyle\frac{p_{4}-p_{1}}{p_{2}-p_{1}}$. Now, it is
sufficient to find out an elliptic curve
$\mathcal{E}^{\prime}_{3}/\mathbb{F}_{q}$ which is isomorphic to
$\mathcal{E}_{3}$ over $\mathbb{F}_{q}$ and satisfies the conditions:

  * (i)

$|R_{\mathcal{E}^{\prime}_{3}}\cap R_{\mathcal{E}_{j}}|=3$ for $j=1,2$;

  * (ii)

$|R_{\mathcal{E}_{1}}\cap R_{\mathcal{E}_{2}}\cap
R_{\mathcal{E}^{\prime}_{3}}|=2$.

Let $\phi_{i}$ be the covering $\mathcal{E}_{i}\rightarrow\mathbb{P}^{1}$.
Then $\phi_{i}$ maps $\mathcal{E}_{i}[2]$ onto $R_{\mathcal{E}_{i}}$
bijectively, and thus
$\phi_{i}(\mathcal{E}_{i}(\mathbb{F}_{q})\cap\mathcal{E}_{i}[2])=\mathbb{P}^{1}(\mathbb{F}_{q})\cap
R_{\mathcal{E}_{i}}$. This implies $|\mathbb{P}^{1}(\mathbb{F}_{q})\cap
R_{\mathcal{E}_{1}}|=|\mathbb{P}^{1}(\mathbb{F}_{q})\cap
R_{\mathcal{E}_{2}}|=|\mathbb{P}^{1}(\mathbb{F}_{q})\cap
R_{\mathcal{E}_{3}}|$.

######  Case 1.

$|\mathbb{P}^{1}(\mathbb{F}_{q})\cap R_{\mathcal{E}_{i}}|=2$

In this case, $R_{\mathcal{E}_{i}}$ contains two distinct rational points and
two other conjugate points. Without loss of generality, let $[p_{1}:1]$ and
$[p_{2}:1]$ be the conjugate points. Now, we consider
$R_{\mathcal{E}^{\prime}_{3}}=\\{[p_{1}:1],[p_{2}:1],[p_{3}:1],[p_{4}:1]\\}$.
Then we have an elliptic curve $\mathcal{E}^{\prime}_{3}$, which is defined
over $\mathbb{F}_{q}$ (since $p_{1}$ and $p_{2}$ are conjugate pairs and
$p_{3},p_{4}\in\mathbb{F}_{q}$), with the desired conditions. It suffices to
show that $\mathcal{E}^{\prime}_{3}$ is isomorphic to $\mathcal{E}_{3}$ over
$\mathbb{F}_{q}$.

Let $\sigma=\bigg{[}\bigg{(}\begin{array}[]{cc}0&1\\\
1&-p_{3}\end{array}\bigg{)}\bigg{]}\in\mathrm{PGL}(2,q)$, then
$\displaystyle\sigma(R_{\mathcal{E}^{\prime}_{3}})=\\{\mathcal{O},[\frac{1}{p_{1}-p_{3}}:1],[\frac{1}{p_{2}-p_{3}}:1],[\frac{1}{p_{4}-p_{3}}:1]\\}$.
The new elliptic curve $\sigma(\mathcal{E}^{\prime}_{3})$ determined by
$\sigma(R_{\mathcal{E}^{\prime}_{3}})$ is therefore isomorphic to
$\mathcal{E}^{\prime}_{3}$. Since a Legendre coefficient of
$\sigma(\mathcal{E}^{\prime}_{3})$ is

$\displaystyle\lambda_{\sigma}$ $\displaystyle=$
$\displaystyle\frac{\frac{1}{p_{4}-p_{3}}-\frac{1}{p_{1}-p_{3}}}{\frac{1}{p_{2}-p_{3}}-\frac{1}{p_{1}-p_{3}}}$
$\displaystyle=$
$\displaystyle\frac{\frac{p_{1}-p_{4}}{(p_{4}-p_{3})(p_{1}-p_{3})}}{\frac{p_{1}-p_{2}}{(p_{2}-p_{3})(p_{1}-p_{3})}}$
$\displaystyle=$
$\displaystyle\frac{\frac{p_{4}-p_{1}}{p_{2}-p_{1}}(1-\frac{p_{3}-p_{1}}{p_{2}-p_{1}})}{\frac{p_{4}-p_{1}}{p_{2}-p_{1}}-\frac{p_{3}-p_{1}}{p_{2}-p_{1}}}$
$\displaystyle=$
$\displaystyle\frac{\lambda_{2}(1-\lambda_{1})}{\lambda_{2}-\lambda_{1}}$

which is Legendre equivalent to $\lambda_{3}$. By Proposition 2.5, this
implies $\sigma(\mathcal{E}^{\prime}_{3})\cong\mathcal{E}_{3}$ (over
$\mathbb{F}_{q}(j_{3})$) and in turn
$\mathcal{E}^{\prime}_{3}\cong\mathcal{E}_{3}$ (over $\mathbb{F}_{q}(j_{3})$).
Since, however, $j_{3}\in\mathbb{F}_{q}$, $\mathcal{E}^{\prime}_{3}$ is in
fact isomorphic to $\mathcal{E}_{3}$ over $\mathbb{F}_{q}$.

######  Case 2.

$|\mathbb{P}^{1}(\mathbb{F}_{q})\cap R_{\mathcal{E}_{i}}|=4$

In this case, all points in $R_{\mathcal{E}_{i}}$ are rational points. Now, we
consider
$R_{\mathcal{E}^{\prime}_{3}}=\\{\mathcal{O},[p_{1}:1],[p_{3}:1],[p_{4}:1]\\}$.
Again, we get an elliptic curve $\mathcal{E}^{\prime}_{3}$, which is trivially
defined over $\mathbb{F}_{q}$, satisfying the required conditions. Clearly, if
we change $\mathcal{E}^{\prime}_{3}$ into Legendre form, we may have
$\displaystyle\lambda^{\prime}_{3}=\frac{p_{4}-p_{1}}{p_{3}-p_{1}}=\frac{\frac{p_{4}-p_{1}}{p_{2}-p_{1}}}{\frac{p_{3}-p_{1}}{p_{2}-p_{1}}}=\frac{\lambda_{2}}{\lambda_{1}}$,
which is Legendre equivalent to $\lambda_{3}$. We immediately get the
desirable result $\mathcal{E}^{\prime}_{3}\cong\mathcal{E}_{3}$ (over
$\mathbb{F}_{q}$) similarly as in case 1.

∎

##  3 Construction of a curve of genus 3

######  Proposition 3.1.

Let $p$ be an odd prime. Let $\mathcal{E}_{i}/\mathbb{F}_{q}$ $(i=1,2,3)$ be
pairwise non-isomorphic elliptic curves and $R_{\mathcal{E}_{i}}$ $(i=1,2,3)$
be the sets of ramified points of $\mathbb{P}^{1}$ for
$\mathcal{E}_{i}/\mathbb{F}_{q}\rightarrow\mathbb{P}^{1}/\mathbb{F}_{q}$
satisfy the following conditions:

  * (i)

$|R_{\mathcal{E}^{\prime}_{i}}\cap R_{\mathcal{E}^{\prime}_{j}}|=3$ $\forall
i\neq j$;

  * (ii)

$|R_{\mathcal{E}_{1}}\cap R_{\mathcal{E}_{2}}\cap R_{\mathcal{E}_{3}}|=2$;

Then there exists a curve $\chi/\mathbb{F}_{q}$ of genus 3 such that the
Jacobian $\mathcal{J}_{\chi}$ is isogenous to
$\mathcal{E}_{1}\times\mathcal{E}_{2}\times\mathcal{E}_{3}$.

###### Proof.

$\mathbb{F}_{q}(\mathbb{P}^{1})=\mathbb{F}_{q}(t)$ can be seen as the subfield
of $\mathbb{F}_{q}(\mathcal{E}_{i})$ $(i=1,2,3)$. Clearly, for any $i\neq j$,
$\mathbb{F}_{q}(\mathcal{E}_{i})\cap\mathbb{F}_{q}(\mathcal{E}_{j})=\mathbb{F}_{q}(\mathbb{P}^{1})$
as $\mathbb{F}_{q}(\mathcal{E}_{i})(i=1,2,3)$ is a quadratic extension of
$\mathbb{F}_{q}(t)$. Then we have a composite field
$\mathbb{F}_{q}(\mathcal{E}_{1})\cdot\mathbb{F}_{q}(\mathcal{E}_{2})$ which is
an abelian extension of $\mathbb{F}_{q}(t)$ of degree 4. Now if
$\mathbb{F}_{q}(\mathcal{E}_{3})$ is not contained in
$\mathbb{F}_{q}(\mathcal{E}_{1})\cdot\mathbb{F}_{q}(\mathcal{E}_{2})$, we will
get a new composite field
$\mathbb{F}_{q}(\mathcal{E}_{1})\cdot\mathbb{F}_{q}(\mathcal{E}_{2})\cdot\mathbb{F}_{q}(\mathcal{E}_{3})$
which is an abelian extension of $\mathbb{F}_{q}(t)$ of degree 8. Let
$\mathcal{Y}/\mathbb{F}_{q}$ and $\chi/\mathbb{F}_{q}$ be (smooth) algebraic
curves corresponding to the composite fields
$\mathbb{F}_{q}(\mathcal{E}_{1})\cdot\mathbb{F}_{q}(\mathcal{E}_{2})$ and
$\mathbb{F}_{q}(\mathcal{E}_{1})\cdot\mathbb{F}_{q}(\mathcal{E}_{2})\cdot\mathbb{F}_{q}(\mathcal{E}_{3})$
respectively.

$\textstyle{\chi\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{Y}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{E}_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{E}_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathcal{E}_{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathbb{P}^{1}}$

$\textstyle{\mathbb{F}_{q}(\chi)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathbb{F}_{q}(\mathcal{Y})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathbb{F}_{q}(\mathcal{E}_{1})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathbb{F}_{q}(\mathcal{E}_{2})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathbb{F}_{q}(\mathcal{E}_{3})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathbb{F}_{q}(\mathbb{P}^{1})}$

The set of ramified points of $\mathbb{P}^{1}$ for the covering
$\chi\rightarrow\mathbb{P}^{1}$ is the union $R_{\mathcal{E}_{1}}\cup
R_{\mathcal{E}_{2}}\cup R_{\mathcal{E}_{3}}$. By Abhyankar’s Lemma [8,
Proposition III.8.9], we know that every point of this set has ramification
index 2. In addition,

$\displaystyle|R_{\mathcal{E}_{1}}\cup R_{\mathcal{E}_{2}}\cup
R_{\mathcal{E}_{3}}|$ $\displaystyle=$
$\displaystyle|R_{\mathcal{E}_{1}}|+|R_{\mathcal{E}_{2}}|+|R_{\mathcal{E}_{3}}|-|R_{\mathcal{E}_{1}}\cap
R_{\mathcal{E}_{2}}|-|R_{\mathcal{E}_{2}}\cap
R_{\mathcal{E}_{3}}|-|R_{\mathcal{E}_{3}}\cap R_{\mathcal{E}_{1}}|$
$\displaystyle+|R_{\mathcal{E}_{1}}\cap R_{\mathcal{E}_{2}}\cap
R_{\mathcal{E}_{3}}|$ $\displaystyle=$ $\displaystyle 4+4+4-3-3-3+2$
$\displaystyle=$ $\displaystyle 5.$

Therefore by the Hurwitz formula [6, Theorem II.5.9], we get

$\displaystyle 2\textrm{genus}(\chi)-2$ $\displaystyle=$
$\displaystyle(2\textrm{genus}(\mathbb{P}^{1})-2)(\textrm{deg}(\phi))+\sum_{P\in\chi}(e_{\phi}(P)-1)$
$\displaystyle=$ $\displaystyle(2\times 0-2)\times 8+4\times 5\times(2-1)$
$\displaystyle=$ $\displaystyle 4.$

(Note that each ramified point on $\mathbb{P}^{1}$ has four pre-images in
$\chi$). We thus have $\textrm{genus}(\chi)=3$. Now it only remains to prove
that
$\mathbb{F}_{q}(\mathcal{E}_{3})\not\leq\mathbb{F}_{q}(\mathcal{E}_{1})\cdot\mathbb{F}_{q}(\mathcal{E}_{2})$
and the Jacobian of $\chi$ is isogenous to
$\mathcal{E}_{1}\times\mathcal{E}_{2}\times\mathcal{E}_{3}$. The first
statement can be verified by Lemma 3.2 and the second one can be proved by a
special case of Theorem 3.3. ∎

######  Lemma 3.2.

Let $p$ be an odd prime and $\mathcal{E}_{i}/\mathbb{F}_{q}$ $(i=1,2,3)$ be
pairwise non-isomorphic elliptic curves. Then
$\mathbb{F}_{q}(\mathcal{E}_{3})\not\leq\mathbb{F}_{q}(\mathcal{E}_{1})\cdot\mathbb{F}_{q}(\mathcal{E}_{2})$.
Here $\mathbb{F}_{q}(\mathcal{E}_{i})$ $(i=1,2,3)$ is viewed as a fixed degree
$2$ extension of $\mathbb{F}_{q}(\mathbb{P}^{1})$.

######  Remark.

An elementary proof of lemma 3.2 is included here for convenience. However it
should be noted that this lemma indeed follows from theorem 3.3.

###### Proof.

Since $\mathrm{Char}(\mathbb{F}_{q})\neq 2$, we can assume
$\mathcal{E}_{i}/\mathbb{F}_{q}$ is given by the equation $y^{2}=f_{i}(x)$,
where $f_{i}(x)$ is a cubic polynomial in $\mathbb{F}_{q}[x]$. Let’s
standardize $f_{i}$ by $f_{i}(x)=4x^{3}+b_{i,2}x^{2}+2b_{i,4}x+b_{i,6}$. Now
let $K=\mathbb{F}_{q}(x)$. Then by considering $y^{2}-f_{i}(x)\in K[y]$, we
may have

$\displaystyle\mathbb{F}_{q}(\mathcal{E}_{i})$ $\displaystyle=$
$\displaystyle\mathrm{Frac}(\mathbb{F}_{q}[x,y]/(y^{2}-f_{i}(x)))$
$\displaystyle=$ $\displaystyle\mathbb{F}_{q}(x)(y_{i})$ $\displaystyle=$
$\displaystyle K(y_{i}),$

where $y_{i}$ is a root of the polynomial $y^{2}-f_{i}(x)$ in $K[y]$. It
suffices to prove $K(y_{3})\not\leq K(y_{1})\cdot K(y_{2})$ instead.

Suppose the contrary is true, then $y_{3}$ is a linear combination of the
basis $\\{1,y_{1},y_{2},y_{1}y_{2}\\}$ of $K(y_{1})\cdot
K(y_{2})=K(y_{1},y_{2})$ over $K$. Let
$y_{3}=c_{0}+c_{1}y_{1}+c_{2}y_{2}+c_{3}y_{1}y_{2}$, where $c_{j}\in K$
$(j=0,1,2,3)$. Then

$\displaystyle y_{3}^{2}$ $\displaystyle=$ $\displaystyle
c_{0}^{2}+2c_{0}c_{1}y_{1}+2c_{0}c_{2}y_{2}$ (1)
$\displaystyle+y_{1}^{2}(c_{1}^{2}+2c_{1}c_{2}y_{2})+y_{2}^{2}(c_{2}^{2}+2c_{2}c_{3}y_{1})$
$\displaystyle+2c_{0}c_{3}y_{1}y_{2}+c_{3}^{2}y_{1}^{2}y_{2}^{2}$
$\displaystyle=$ $\displaystyle
C_{0}+f_{1}(x)C_{1}+f_{2}(x)C_{2}+C_{3}f_{1}(x)f_{2}(x),$

$\mathrm{where}\left\\{\begin{array}[]{ccc}C_{0}&=&c_{0}^{2}+2c_{0}c_{1}y_{1}+2c_{0}c_{2}y_{2}+2c_{0}c_{3}y_{1}y_{2}\\\
C_{1}&=&c_{1}^{2}+2c_{1}c_{2}y_{2}\\\ C_{2}&=&c_{2}^{2}+2c_{2}c_{3}y_{1}\\\
C_{3}&=&c_{3}^{2}\end{array}\right.\in K(y_{1},y_{2}).$

However, since $y_{3}^{2}=f_{3}(x)$ which is a cubic polynomial in
$\mathbb{F}_{q}[x]$ and $f_{1}(x)f_{2}(x)$ is a polynomial of degree 6 in
$\mathbb{F}_{q}[x]$, by comparing coefficients in (1) we know that
$c_{3}^{2}=C_{3}=0$. Therefore,
$C_{0}+f_{1}(x)C_{1}+f_{2}(x)C_{2}=y_{3}^{2}\in K$. It implies that
$\forall\sigma\in\mathrm{Gal}(K(y_{1},y_{2})/K),$

$\displaystyle\sigma(C_{0}+f_{1}(x)C_{1}+f_{2}(x)C_{2})=C_{0}+f_{1}(x)C_{1}+f_{2}(x)C_{2}$
$\displaystyle\Rightarrow$
$\displaystyle(\sigma-\iota)C_{0}+f_{1}(x)(\sigma-\iota)C_{1}+f_{2}(x)(\sigma-\iota)C_{2}=0,$
(2)

where $\iota\in\mathrm{Gal}(K(y_{1},y_{2})/K)$ is the identity. Since 222Note:
$(\sigma-\iota)C_{i}$ is only a notation used for convenience to represent
$\sigma C_{i}-C_{i}$$(\sigma-\iota)C_{2}=2c_{2}c_{3}(\sigma-\iota)y_{1}=0$ for
$c_{3}=0$, instead of (2) we should have

$\displaystyle(\sigma-\iota)C_{0}+f_{1}(x)(\sigma-\iota)C_{1}=0.$ (3)

Now we know that

$\left\\{\begin{array}[]{ccc}(\sigma-\iota)C_{0}&=&2c_{0}c_{1}(\sigma-\iota)y_{1}+2c_{0}c_{2}(\sigma-\iota)y_{2}\\\
(\sigma-\iota)C_{1}&=&2c_{1}c_{2}(\sigma-\iota)y_{2}\end{array}\right..$

Consider the automorphism $\sigma$ in $\mathrm{Gal}(K(y_{1},y_{2})/K)$ defined
by

$\left\\{\begin{array}[]{ccccc}y_{1}&\mapsto&\bar{y_{1}}&=&-y_{1}\\\
y_{2}&\mapsto&\bar{y_{2}}&=&-y_{2}\\\
y_{1}y_{2}&\mapsto&\bar{y_{1}}\bar{y_{2}}&=&y_{1}y_{2}\end{array}\right.$

(Since $y_{i}$ and $\bar{y_{i}}$ both satisfy the equation $y^{2}-f_{i}(x)=0$,
$y_{i}\bar{y_{i}}=-f_{i}(x)=-y_{i}^{2}\Rightarrow\bar{y_{i}}=-y_{i}$ as
$y_{i}\neq 0$). Then we obtain

$\left\\{\begin{array}[]{ccc}(\sigma-\iota)C_{0}&=&-4c_{0}c_{1}y_{1}-4c_{0}c_{2}y_{2}\\\
(\sigma-\iota)C_{1}&=&-4c_{1}c_{2}y_{2}\end{array}\right.$

which are linearly independent over $K$ if both $(\sigma-\iota)C_{0}$ and
$(\sigma-\iota)C_{1}$ are non-vanishing. This is a contradiction to (3) and
hence the lemma follows. If $(\sigma-\iota)C_{0}=0$ or
$(\sigma-\iota)C_{1}=0$, then we get $f_{1}(x)(\sigma-\iota)C_{1}=0$ or
$(\sigma-\iota)C_{0}=0$ respectively, which also implies
$(\sigma-\iota)C_{0}=(\sigma-\iota)C_{1}=0$. Then we can merely consider a few
cases:

  * (i)

$c_{0}=0$, $c_{1}=0$ and $c_{2}\neq 0$;

  * (ii)

$c_{0}=0$, $c_{1}\neq 0$ and $c_{2}=0$;

  * (iii)

$c_{0}=0$, $c_{1}=0$ and $c_{2}=0$;

  * (iv)

$c_{0}\neq 0$, $c_{1}=0$ and $c_{2}=0$.

For cases (i), $y_{3}=c_{2}y_{2}\Rightarrow
f_{3}(x)=y_{3}^{2}=c_{2}^{2}y_{2}^{2}=c_{2}^{2}f_{2}(x)$. Again, by comparing
coefficients we may have $c_{2}^{2}=1$. This implies that $y_{3}=\pm y_{2}$,
which is not possible. Similarly, case (ii) is impossible as well. Cases (iii)
and (iv) are trivially excluded, so the proof is completed. ∎

######  Theorem 3.3.

[2, Theorem C] Let $\mathcal{C}$ be a (smooth, projective) algebraic curve
defined over an arbitrary field. Let
$H_{1},...H_{k}\leq\mathrm{Aut}(\mathcal{C})$ be (finite) subgroups with
$H_{i}\cdot H_{j}=H_{j}\cdot H_{i}\quad i\neq j$, and let $g_{ij}$ denote the
genus of the quotient curve $\mathcal{C}/H_{i}\cdot H_{j}$. Then, for
$n_{1},...,n_{k}\in\mathbb{Z}$, the conditions

$\sum n_{i}n_{j}g_{ij}=0,\\\ \sum_{j}n_{j}g_{ij}=0,\quad 1\leq i\leq n,$

are each equivalent to $\sum n_{i}\varepsilon_{n_{i}}=0$, where
$\varepsilon\in\mathrm{End}^{0}(\mathcal{J}_{\mathcal{C}})$ are idempotents,
and hence both imply the isogeny relation

$\prod_{n_{i}>0}\mathcal{J}^{n_{i}}_{\mathcal{C}/H_{i}}\sim\prod_{n_{i}<0}\mathcal{J}^{|n_{i}|}_{\mathcal{C}/H_{i}}.$

In particular, if $g_{ij}=0$ for $2\leq i<j\leq k$ and if

$g_{\mathcal{C}}=g_{\mathcal{C}/H_{2}}+...+g_{\mathcal{C}/H_{k}}$

then we have (by taking $H_{1}=\\{1\\}$ above):

$\mathcal{J}_{\mathcal{C}}\sim\mathcal{J}_{\mathcal{C}/H_{2}}\times...\times\mathcal{J}_{\mathcal{C}/H_{k}}.$

By considering $\mathcal{C}=\chi$, and taking $H_{i}$ as the following
subgroups of
$\mathrm{Aut}(\chi)=\mathrm{Aut}_{\mathbb{F}_{q}(t)}(\mathbb{F}_{q}(\chi))$:

$\left\\{\begin{array}[]{ccc}H_{1}&=&\\{1\\}\\\
H_{2}&=&<\sigma_{2},\sigma_{3}>\\\ H_{3}&=&<\sigma_{1},\sigma_{3}>\\\
H_{4}&=&<\sigma_{1},\sigma_{2}>\\\ \end{array}\right.$

where

$\sigma_{1}:\left\\{\begin{array}[]{ccc}y_{1}&\mapsto&-y_{1}\\\
y_{2}&\mapsto&y_{2}\\\ y_{3}&\mapsto&y_{3}\\\ \end{array}\right.$
$\sigma_{2}:\left\\{\begin{array}[]{ccc}y_{1}&\mapsto&y_{1}\\\
y_{2}&\mapsto&-y_{2}\\\ y_{3}&\mapsto&y_{3}\\\ \end{array}\right.$
$\sigma_{3}:\left\\{\begin{array}[]{ccc}y_{1}&\mapsto&y_{1}\\\
y_{2}&\mapsto&y_{2}\\\ y_{3}&\mapsto&-y_{3}\\\ \end{array}\right.,$

we can easily check that the condition $H_{i}\cdot H_{j}=H_{j}\cdot H_{i}$
$\forall i,j=1,2,3,4$ is satisfied. Since
$\mathbb{F}_{q}(\chi/H_{2})=\mathbb{F}_{q}(\chi)_{H_{2}}=\mathbb{F}_{q}(t)(y_{1})=\mathbb{F}_{q}(\mathcal{E}_{1})$,
we have $\chi/H_{2}\cong\mathcal{E}_{1}$. Similarly, we can also derive
$\chi/H_{3}\cong\mathcal{E}_{2}$ and $\chi/H_{4}\cong\mathcal{E}_{3}$, and
thus we know $g_{12}=g_{13}=g_{14}=1$. Moreover, as
$\mathbb{F}_{q}(\chi/H_{i}\cdot H_{j})=\mathbb{F}_{q}(\chi)_{H_{i}\cdot
H_{j}}=\mathbb{F}_{q}(t)=\mathbb{F}_{q}(\mathbb{P}^{1})$ for $2\leq i<j\leq
4$, $\chi/{H_{i}\cdot H_{j}}\cong\mathbb{P}^{1}$ $(2\leq i<j\leq 4)$, which
implies $g_{ij}=0$ for $(2\leq i<j\leq 4)$. Therefore,
$g_{\chi}=g_{\chi/H_{2}}+g_{\chi/H_{3}}+g_{\chi/H_{4}}$. Then using the
particular case in the Theorem 3.3, we have

$\displaystyle\mathcal{J}_{\chi}$ $\displaystyle\sim$
$\displaystyle\mathcal{J}_{\chi/H_{2}}\times\mathcal{J}_{\chi/H_{3}}\times\mathcal{J}_{\chi/H_{4}}$
$\displaystyle=$
$\displaystyle\mathcal{J}_{\mathcal{E}_{1}}\times\mathcal{J}_{\mathcal{E}_{2}}\times\mathcal{J}_{\mathcal{E}_{3}}.$

This completes the proof of Proposition 3.1.

######  Corollary 3.4.

Let $p$ be odd. Let $\mathcal{E}_{i}/\mathbb{F}_{q}$ $(i=1,2,3)$ be pairwise
non-isomorphic elliptic curves. Suppose $j_{3}\in\mathbb{F}_{q}$. Then there
exists a curve $\chi/\mathbb{F}_{q}$of genus 3 such that the Jacobian
$\mathcal{J}_{\chi}$ is isogenous to
$\mathcal{E}_{1}\times\mathcal{E}_{2}\times\mathcal{E}_{3}$ if one of the
following is satisfied:

  * (i)

$|\mathcal{E}_{1}(\mathbb{F}_{q})\cap\mathcal{E}_{1}[2]|=|\mathcal{E}_{2}(\mathbb{F}_{q})\cap\mathcal{E}_{2}[2]|=|\mathcal{E}_{3}(\mathbb{F}_{q})\cap\mathcal{E}_{3}[2]|=2$
and $\displaystyle\frac{\lambda_{2}(1-\lambda_{1})}{\lambda_{2}-\lambda_{1}}$
is Legendre equivalent to $\lambda_{3}$;

  * (ii)

$|\mathcal{E}_{1}(\mathbb{F}_{q})\cap\mathcal{E}_{1}[2]|=|\mathcal{E}_{2}(\mathbb{F}_{q})\cap\mathcal{E}_{2}[2]|=|\mathcal{E}_{3}(\mathbb{F}_{q})\cap\mathcal{E}_{3}[2]|=4$
and $\displaystyle\frac{\lambda_{2}}{\lambda_{1}}$ is Legendre equivalent to
$\lambda_{3}$.

###### Proof.

This is the direct consequence of Lemma 2.7 and Proposition 3.1. ∎

##  4 Main Result

Before we go through the main result, we should first know some facts. If
there are three distinct integers $t_{i}$ $(i=1,2,3)$ satisfying one of the
conditions in Lemma 2.1, then there exist three pairwise non-isomorphic
elliptic curves $\mathcal{E}_{i}/\mathbb{F}_{q}$ $(i=1,2,3)$ such that
$|\mathcal{E}_{i}(\mathbb{F}_{q})|=q+1+t_{i}$. $\mathcal{E}_{i}$ is called an
elliptic curve corresponding to $t_{i}$ in this paper.

######  Definition 4.1.

Let $t_{i}$ $(i=1,2,3)$ be three distinct integers satisfying one of the
conditions in Lemma 2.1. If there exists $\mathcal{E}_{i}/\mathbb{F}_{q}$
$(i=1,2,3)$ which are elliptic curves corresponding to $t_{i}$ with Legendre
coefficients $\lambda_{i}$ such that:

  * (i)

$\displaystyle\frac{\lambda_{i_{2}}(1-\lambda_{i_{1}})}{\lambda_{i_{2}}-\lambda_{i_{1}}}$
is Legendre equivalent to $\lambda_{i_{3}}$, where
$\\{i_{1},i_{2},i_{3}\\}=\\{1,2,3\\}$, and $j_{i_{3}}\in\mathbb{F}_{q}$; Then
$t_{i}$ $(i=1,2,3)$ are said to be weakly Legendre consistent and
$\mathcal{E}_{i}$ are called elliptic curves corresponding to $t_{i}$ weakly.
Or else

  * (ii)

$\displaystyle\frac{\lambda_{i_{2}}}{\lambda_{i_{1}}}$ is Legendre equivalent
to $\lambda_{i_{3}}$, where $\\{i_{1},i_{2},i_{3}\\}=\\{1,2,3\\}$,
$j_{i_{3}}\in\mathbb{F}_{q}$ and no $\mathcal{E}_{i}$ has the property
$|\mathcal{E}_{i}(\mathbb{F}_{q})\cap\mathcal{E}_{i}[2]|=2$; Then $t_{i}$
$(i=1,2,3)$ are said to be strongly Legendre consistent and $\mathcal{E}_{i}$
are called elliptic curves corresponding to $t_{i}$ strongly.

With this new definition, the main theorem in this paper can thus be
formulated.

######  Theorem 4.2.

Let $q=p^{r}$ for a prime $p>2$ and an integer $r\geq 1$. Let $t_{i}$
$(i=1,2,3)$ be three distinct integers satisfying one of the conditions in
Lemma 2.1. Then the function $\prod_{i=1}^{3}(T^{2}+t_{i}T+q)$ is the
characteristic polynomial of the Frobenius endomorphism of the Jacobian of a
curve of genus 3 over $\mathbb{F}_{q}$ if one of the following conditions is
satisfied:

  * (i)

$q+1+t_{i}\equiv 2\pmod{4}$ $\forall i=1,2,3$ and $t_{i}$ $(i=1,2,3)$ are
weakly Legendre consistent;

  * (ii)

$q+1+t_{i}\equiv 0\pmod{4}$ $\forall i=1,2,3$ and $t_{i}$ $(i=1,2,3)$ are
strongly Legendre consistent.

###### Proof.

In the case (i), let $\mathcal{E}_{i}/\mathbb{F}_{q}$ be the elliptic curves
corresponding to $t_{i}$ weakly. Then it is trivial that
$|\mathcal{E}_{1}(\mathbb{F}_{q})\cap\mathcal{E}_{1}[2]|=|\mathcal{E}_{2}(\mathbb{F}_{q})\cap\mathcal{E}_{2}[2]|=|\mathcal{E}_{3}(\mathbb{F}_{q})\cap\mathcal{E}_{3}[2]|=2$
and by definiton (without loss of generality)
$\displaystyle\frac{\lambda_{2}(1-\lambda_{1})}{\lambda_{2}-\lambda_{1}}$ is
Legendre equivalent to $\lambda_{3}$ and $j_{3}\in\mathbb{F}_{q}$. Therefore
by Corollary 3.4, there exists a curve $\chi/\mathbb{F}_{q}$of genus 3 such
that the Jacobian $\mathcal{J}_{\chi}$ is isogenous to
$\mathcal{E}_{1}\times\mathcal{E}_{2}\times\mathcal{E}_{3}$. In other word,
the characteristic polynomial of the Frobenius endomorphism of the Jacobian of
this curve is $\prod_{i=1}^{3}(T^{2}+t_{i}T+q)$.

In the case (ii), let $\mathcal{E}_{i}/\mathbb{F}_{q}$ be the elliptic curves
corresponding to $t_{i}$ strongly. Then similarly by definiton,
$|\mathcal{E}_{1}(\mathbb{F}_{q})\cap\mathcal{E}_{1}[2]|=|\mathcal{E}_{2}(\mathbb{F}_{q})\cap\mathcal{E}_{2}[2]|=|\mathcal{E}_{3}(\mathbb{F}_{q})\cap\mathcal{E}_{3}[2]|=4$,
$\displaystyle\frac{\lambda_{2}}{\lambda_{1}}$ is Legendre equivalent to
$\lambda_{3}$ and $j_{3}\in\mathbb{F}_{q}$. Thus by applying Corollary 3.4, we
get the same result as in the previous case. ∎

######  Remark 4.3.

By adopting similar method in [1], we are only able to do the case up to
$g=3$. For the cases $g\geq 4$, the method is invalid since the Proposition
3.1 breaks down. Strictly speaking, we cannot construct a curve of genus $g$
by just considering the corresponding (smooth) algebraic curve to the
composite field of the function fields of $g$ pairwise non-isomorphic elliptic
curves, and then applying the Hurwitz formula to prove it is of genus g.

##  5 Further Research

  1. 1.

Find out all of the zeta-functions of curves of genus 2 over finite fields,
including those not being determined in the paper [1].

  2. 2.

Find out the intrinsic reason(s) why curves of genus 4 or above cannot be
constructed by similar method in the case $g=2$ or $3$.

  3. 3.

Improve the method used in this paper such that the conditions used for the
construction are weaker.

##  6 Acknowledgement

I hereby give my sincere appreciation to my mentor, Dr. Eric Wambach. He gave
me a lot of support and valuable advice about both my project and my future
study plan during the ten-week SURF program. I give my special thanks to my
sponsor, the Caltech Alumni Association (Hong Kong Chapter), for letting me
gain this meaningful research experience. I am truly grateful for the great
support from Mr. Chris Lyons, who helped me to tackle many problems
encountered in the project while I was in Caltech. I would like to thank Mr.
Gary Tsang as well for teaching me how to use LaTeX and helping me deal with
the problems in compiling this paper. Lastly, I offer my heartfelt thanks to
all people who have given me a hand in any aspect.

## References

  * [1] C. XING, _Zeta-functions of curves of genus 2 over finite fields_ , Journal of Algebra 308 (2007) 734-741.
  * [2] E. KANI AND M. ROSEN, _Idempotent relations and factors of Jacobians_ , Mathematische Annalen 284 (1989) 307-327.
  * [3] M. DEURING, _Die Typen der Multiplikatorenringe elliptischer Functionenk $\ddot{o}$rper_, Abh. Math. Sem. Univ. Hamburg 14 (1941) 197-272.
  * [4] J. P. SERRE, _Rational points on curves over finite fields_ , notes by F. GOUVEA of lectures at Harvard University, 1985.
  * [5] H. G. R$\mathrm{\ddot{U}}$CK, _Abelian surfaces and Jacobian varieties over finite fields_ , Compos. Math. 76 (1990) 351-366.
  * [6] J. H. SILVERMAN, _The Arithmetic of Elliptic Curves_ , Springer-Verlag, New York, 1986.
  * [7] D. HUSEMLLER, _Elliptic Curves (2nd ed.)_ , Springer, New York, 2004.
  * [8] H. STICHTENOTH, _Algebraic Function Fields and Codes_ , Springer, Berlin, 1993
  * [9] W. C. WATERHOUSE, _Abelian varieties over finite fields_ , Ann. Sci. $\mathrm{\acute{E}}$cole Norm. Sup. (4) 2 (1963) 521-560.


# The sequentially Cohen-Macaulay property of edge ideals of edge-weighted
graphs

Ly Thi Kieu Diem  Department of Mathematics and Informatics, Hanoi National
University of Education, 136 Xuan Thuy, Hanoi, Vietnam
<EMAIL_ADDRESS>,  Nguyên Công Minh School of Applied Mathematics and
Informatics
Hanoi University of Science and Technology, 1 Dai Co Viet, Hanoi, Vietnam
<EMAIL_ADDRESS>and  Thanh Vu  Hanoi University of Science and
Technology, 1 Dai Co Viet, Hanoi, Vietnam<EMAIL_ADDRESS>Dedicated to
Professor Ngo Viet Trung on the occasion of his 70th birthday

###### Abstract.

Let $I(G_{\mathbf{w}})$ be the edge ideal of an edge-weighted graph
$(G,\mathbf{w})$. We prove that $I(G_{\mathbf{w}})$ is sequentially Cohen-
Macaulay for all weight functions $\mathbf{w}$ if and only if $G$ is a
Woodroofe graph.

###### Key words and phrases:

sequentially Cohen-Macaulay; edge-weighted graph; monomial ideal

###### 2010 Mathematics Subject Classification:

05E40, 13F55, 13D02

##  1\. Introduction

Let $S=K[x_{1},\ldots,x_{n}]$ be a standard graded polynomial ring over an
arbitrary field $K$. Let $G$ be a simple graph with vertex set
$V=\\{x_{1},\ldots,x_{n}\\}$ and edge set $E(G)$. By abuse of notation, we
also use $x_{i}x_{j}$ to denote an edge $\\{x_{i},x_{j}\\}$ of $G$. Assume
that $\mathbf{w}:E(G)\rightarrow\operatorname{\mathbb{Z}}_{>0}$ is a weight
function on edges of $G$. The edge ideal of the edge-weighted graph
$(G,\mathbf{w})$ is defined by

$I(G_{\mathbf{w}})=\big{(}(x_{i}x_{j})^{\mathbf{w}(x_{i}x_{j})}\mid\\{i,j\\}\in
E(G)\big{)}\subseteq S.$

In particular, if every edge of $G$ has weight one then $I(G_{\mathbf{w}})$
becomes the usual edge ideal $I(G)$.

Edge ideals of edge-weighted graphs were introduced by Paulsen and Sather-
Wagstaff [PS]. In this work, the authors described a primary decomposition of
$I(G_{\mathbf{w}})$ and studied the Cohen-Macaulay property of
$I(G_{\mathbf{w}})$ when the underlying graph $G$ is a cycle, a tree, or a
complete graph. In particular, they proved that $I(G_{\mathbf{w}})$ is Cohen-
Macaulay for all weight functions $\mathbf{w}$ when $G$ is a complete graph.
In our first main result, we prove the converse of this result.

######  Theorem 1.1.

Let $G$ be a simple graph. The following statements are equivalent:

  1. (1)

$I(G_{\mathbf{w}})$ is Cohen-Macaulay for all weight functions $\mathbf{w}$;

  2. (2)

$I(G_{\mathbf{w}})$ is Cohen-Macaulay for all weight functions $\mathbf{w}$
such that $\mathbf{w}(x_{i}x_{j})\in\\{1,2\\}$ for all edges $x_{i}x_{j}\in
E(G)$;

  3. (3)

$G$ is a disjoint union of finitely many complete graphs.

In [FSTY], Fakhari, Shibata, Terai, and S. Yassemi characterized the unmixed
property of $I(G_{\mathbf{w}})$ when $G$ is a very well-covered graph and
proved that this is equivalent to the Cohen-Macaulay property of
$I(G_{\mathbf{w}})$. In this context, Terai [T] proposed the following
conjecture

######  Conjecture (Terai).

Let $G$ be a Cohen-Macaulay very well-covered graph. Then $I(G_{\mathbf{w}})$
is sequentially Cohen-Macaulay for all weight functions $\mathbf{w}$.

We first recall the definition of sequentially Cohen-Macaulay modules over
$S$.

######  Definition 1.

Let $M$ be a graded module over $S$. We say that $M$ is sequentially Cohen-
Macaulay if there exists a filtration

$0=M_{0}\subset M_{1}\subset\cdots\subset M_{r}=M$

of $M$ by graded $S$-modules such that
$\dim(M_{i}/M_{i-1})<\dim(M_{i+1}/M_{i})$ for all $i$, where $\dim$ denotes
Krull dimension, and $M_{i}/M_{i-1}$ is Cohen-Macaulay for all $i$. An ideal
$J$ is said to be sequentially Cohen-Macaulay if $S/J$ is a sequentially
Cohen-Macaulay $S$-module. A graph $G$ (resp. $(G,\mathbf{w})$) is said to be
sequentially Cohen-Macaulay if $I(G)$ (resp. $I(G_{\mathbf{w}})$) is.

The notion of sequentially Cohen-Macaulay was introduced by Stanley [S] as a
generalization of the Cohen-Macaulay property in connection with the work of
Björner and Wachs on nonpure shellability [BW1, BW2]. When $J$ is a
sequentially Cohen-Macaulay ideal, it is well-known that $J$ is Cohen-Macaulay
if and only if $J$ is unmixed.

In motivation to study the conjecture of Terai, we can classify graphs for
which $G_{\mathbf{w}}$ are sequentially Cohen-Macaulay for all weight
functions $\mathbf{w}$. To introduce our result, we first define a special
class of simple graphs which contain $5$-cycles and chordal graphs. A
chordless cycle $C_{t}$ of length $t$ is a cycle with no chord $\\{i,j\\}$ for
$j\neq i+1$. Equivalently, the induced graph of $G$ on $\\{1,\ldots,t\\}$ is
the cycle on $t$ vertices.

######  Definition 2.

A simple graph $G$ is said to be a Woodroofe graph if $G$ has no chordless
cycles of length other than $3$ or $5$.

In [Wo, Theorem 1], Woodroofe proved that if $G$ is a Woodroofe graph, then it
is vertex-decomposable. So, it is sequentially Cohen-Macaulay. Our second main
result of this paper states that Woodroofe graphs are precisely graphs for
which $G_{\mathbf{w}}$ are sequentially Cohen-Macaulay for all weight
functions $\mathbf{w}$.

######  Theorem 1.2.

Let $G$ be a simple graph. The following statements are equivalent:

  1. (1)

$(G,\mathbf{w})$ is sequentially Cohen-Macaulay for all weight functions
$\mathbf{w}$;

  2. (2)

$(G,\mathbf{w})$ is sequentially Cohen-Macaulay for all weight functions
$\mathbf{w}$ such that $\mathbf{w}(x_{i}x_{j})\in\\{1,2\\}$ for all edges
$x_{i}x_{j}\in E(G)$;

  3. (3)

$G$ is a Woodroofe graph.

Now we explain the organization of the paper. In Section 2, we prove Theorem
1.1 and Theorem 1.2. In Section 3, we give some applications of our main
results. In particular, we provide counterexamples to Terai’s conjecture.

##  2\. Proof of the main results

Throughout the paper, we denote $S=K[x_{1},\ldots,x_{n}]$ a standard graded
polynomial ring over a field $K$. Let $\mathfrak{m}=(x_{1},\ldots,x_{n})$ be
the maximal homogeneous ideal of $S$. We first recall some notation and
results.

For a finitely generated graded $S$-module $L$, the depth of $L$ is defined to
be

$\operatorname{depth}(L)=\min\\{i\mid H_{\mathfrak{m}}^{i}(L)\neq 0\\},$

where $H^{i}_{\mathfrak{m}}(L)$ denotes the $i$-th local cohomology module of
$L$ with respect to $\mathfrak{m}$.

######  Definition 3.

A finitely generated graded $S$-module $L$ is called Cohen-Macaulay if
$\operatorname{depth}L=\dim L$. A homogeneous ideal $I\subseteq S$ is said to
be Cohen-Macaulay if $S/I$ is a Cohen-Macaulay $S$-module.

Let $I$ be a monomial ideal in $S$. In [H], Hochster introduced the set of
associated radical ideals of $I$, namely $\sqrt{I:u}$ for monomials $u\notin
I$ and proved that the Cohen-Macaulay property of $I$ can be characterized in
terms of its associated radicals. In [JS], Jafari and Sabzrou showed that the
sequentially Cohen-Macaulay property of $I$ can also be characterized in terms
of its associated radicals.

######  Lemma 2.1.

A monomial ideal $I$ is (sequentially) Cohen-Macaulay if and only if
$\sqrt{I:u}$ is (sequentially) Cohen-Macaulay for all monomials $u\notin I$.

###### Proof.

The statement for Cohen-Macaulay property follows from [H, Theorem 7.1] and
the fact that $\dim S/\sqrt{I:u}\leq\dim S/I$ for all monomials $u\notin I$.
See also [JS, Proposition 2.8]. The statement for sequentially Cohen-Macaulay
is [JS, Proposition 2.23]. ∎

The associated radicals also play an important role in studying the regularity
of monomial ideals (see [MNPTV]). The associated radicals can be computed as
follows. For a monomial $f$ in $S$, the support of $f$, denoted
$\operatorname{supp}(f)$, is the set of $x_{i}$ such that $x_{i}|f$. The
radical of $f$ is defined by
$\sqrt{f}=\prod_{x_{i}\in\operatorname{supp}f}x_{i}$. For an exponent
$\mathbf{a}\in\operatorname{\mathbb{N}}^{n}$, denote $x^{\mathbf{a}}$ the
monomial $x_{1}^{a_{1}}\cdots x_{n}^{a_{n}}$.

######  Lemma 2.2.

Let $J$ be a monomial ideal in $S$ generated by the monomials
$f_{1},\ldots,f_{r}$ and $x^{\mathbf{a}}$ a monomial in $S$. Then
$\sqrt{I:x^{\mathbf{a}}}$ is generated by

$\sqrt{f_{1}/\gcd(f_{1},x^{\mathbf{a}})},\ldots,\sqrt{f_{r}/\gcd(f_{r},x^{\mathbf{a}})}.$

###### Proof.

See [MNPTV, Lemma 2.24]. ∎

Let $G$ denote a finite simple graph over the vertex set
$V(G)=\\{x_{1},x_{2},\ldots,x_{n}\\}$ and the edge set $E(G)$. A subgraph
$H=G[W]$ is called an induced subgraph of $G$ on $W\subset V(G)$ if for any
vertices $u,v\in W$ then $uv\in E(H)$ if and only if $uv\in E(G)$. Let
$\mathbf{w}:E(G)\to\operatorname{\mathbb{Z}}_{>0}$ be a weight function on the
edges of $G$. We first show that the property that $I(G_{\mathbf{w}})$ are
(sequentially) Cohen-Macaulay for all weight functions $\mathbf{w}$ is
equivalent to the property that all induced subgraphs of $G$ are
(sequentially) Cohen-Macaulay.

######  Lemma 2.3.

Let $G$ be a simple graph. The following statements are equivalent.

  1. (1)

$(G,\mathbf{w})$ is (sequentially) Cohen-Macaulay for all weight functions
$\mathbf{w}$;

  2. (2)

$(G,\mathbf{w})$ is (sequentially) Cohen-Macaulay for all weight functions
$\mathbf{w}$ such that $\mathbf{w}(x_{i}x_{j})\in\\{1,2\\}$ for all edges
$x_{i}x_{j}\in E(G)$;

  3. (3)

$G[W]$ is (sequentially) Cohen-Macaulay for all subsets $W\subseteq V(G)$.

###### Proof.

It is obvious that $(1)\Rightarrow(2)$. Now, we prove $(2)\Rightarrow(3)$. Let
$W$ be any subset of $V(G)$. Let $\mathbf{w}$ be the weight function defined
as follows:

$\mathbf{w}(e)=\begin{cases}2\text{ if }e\in G[W],\\\ 1\text{
otherwise}.\end{cases}$

Let $x^{\mathbf{a}}=\prod_{x_{j}\in W}x_{j}$. We first show that

(1) $\sqrt{I(G_{\mathbf{w}}):x^{\mathbf{a}}}=I(G[W])+(\text{some variables not
in W})+I(G[W^{\prime}]),$

where $W^{\prime}$ is a subset of $V(G)\setminus W$. By Lemma 2.2
$\sqrt{I(G_{\mathbf{w}}):x^{\mathbf{a}}}$ is generated by
$\sqrt{e^{\mathbf{w}(e)}/\gcd(e^{\mathbf{w}(e)},x^{\mathbf{a}})}$ for all edge
$e$ of $G$. We have three cases:

Case 1. $e\in G[W]$. In this case, $\mathbf{w}(e)=2$ and
$e^{2}/\gcd(e^{2},x^{\mathbf{a}})=e$.

Case 2. $|\operatorname{supp}e\cap W|=1$. Assume that $e=xy$ with $x\in W$ and
$y\notin W$. Then $e/\gcd(e,x^{\mathbf{a}})=y\notin W$.

Case 3. $|\operatorname{supp}e\cap W|=0$. Then $e/\gcd(e,x^{\mathbf{a}})=e$.

Eq. (1) follows. By Lemma 2.1, $\sqrt{I(G_{\mathbf{w}}):x^{\mathbf{a}}}$ is
(sequentially) Cohen-Macaulay. In particular, $I(G[W])+I(G[W^{\prime}])$ is
(sequentially) Cohen-Macaulay. Since $W\cap W^{\prime}=\emptyset$, by [V,
Lemma 4.1] and [Wo, Lemma 20], we deduce that $I(G[W])$ is (sequentially)
Cohen-Macaulay.

$(3)\Rightarrow(1).$ By Lemma 2.2, any minimal generator of
$\sqrt{I(G_{\mathbf{w}}):x^{\mathbf{a}}}$ for any weight function $\mathbf{w}$
and any monomial $x^{\mathbf{a}}$ such that $x^{\mathbf{a}}\notin
I(G_{\mathbf{w}})$ is either $xy$ where $xy$ is an edge of $G$ or a variable.
Hence, $\sqrt{I(G_{\mathbf{w}}):x^{\mathbf{a}}}=I(G[W])+(\text{some
variables})$ for some subset $W$ of $V(G)$. By assumption, they are
(sequentially) Cohen-Macaulay. By Lemma 2.1, $I(G_{\mathbf{w}})$ is
(sequentially) Cohen-Macaulay. ∎

By Lemma 2.3, we see that studying the class of graphs for which
$I(G_{\mathbf{w}})$ is (sequentially) Cohen-Macaulay is equivalent to the
problem of finding obstructions to (nonpure) shellability of flag complexes.
This observation leads us to the proof of the main results.

###### Proof of Theorem 1.1.

By Lemma 2.3, the Theorem follows from the following facts.

  1. (1)

Disjoint unions of complete graphs are Cohen-Macaulay [V].

  2. (2)

Induced subgraphs of a disjoint union of complete graphs are disjoint unions
of complete graphs.

  3. (3)

$P_{3}$ a path of length $2$ is not Cohen-Macaulay.

  4. (4)

$P_{3}$ is not an induced subgraph of $G$ then $G$ is a disjoint union of
complete graphs.

∎

###### Proof of Theorem 1.2.

By Lemma 2.3, the Theorem follows from the definition of Woodroofe graphs and
the following facts.

  1. (1)

Woodroofe graphs are sequentially Cohen-Macaulay [Wo, Theorem 1].

  2. (2)

Induced subgraphs of a Woodroofe graph are Woodroofe graphs.

  3. (3)

The cycles $C_{t}$ are not sequentially Cohen-Macaulay for $t\neq 3,5$ (see
[FT, Proposition 4.1] and [Wo, Theorem 10]).

∎

######  Remark 2.4.

  1. (1)

The simplicial complexes corresponding to edge ideals of disjoint unions of
complete graphs are matroid of rank $2$.

  2. (2)

One can show that the classification of hypergraphs $\mathcal{H}$ whose edge
ideals of edge-weighted hypergraphs $(\mathcal{H},\mathbf{w})$,
$I(\mathcal{H}_{\mathbf{w}})$ is (sequentially) Cohen-Macaulay for all weight
functions $\mathbf{w}$ also reduces to study the obstructions to (nonpure)
shellablity of simplicial complexes.

  3. (3)

We can show that $I(\mathcal{H}_{\mathbf{w}})$ is Cohen-Macaulay for all
weight functions $\mathbf{w}$ if and only if $\Delta(I_{\mathcal{H}})$ is a
matroid.

  4. (4)

The obstruction to nonpure shellability is much more subtle (see [Wa] for some
partial results) and we will leave that for future work.

##  3\. Applications

In this section, we will give some applications of our results. Firstly, one
knows that when $J$ is sequentially Cohen-Macaulay, then $J$ is Cohen-Macaulay
if and only if it is unmixed. Therefore, we obtain

######  Corollary 3.1.

Let $G$ be a Woodroofe graph and
$\mathbf{w}:E(G)\to\operatorname{\mathbb{Z}}_{>0}$ a weight function. Then
$I(G_{\mathbf{w}})$ is Cohen-Macaulay if and only if $I(G_{\mathbf{w}})$ is
unmixed.

When $G$ is a Cohen-Macaulay graph, a weight function $\mathbf{w}$ on edges of
$G$ is called Cohen-Macaulay if $(G,\mathbf{w})$ is Cohen-Macaulay. Before
giving our next application, we recall the result of Paulsen and Sather-
Wagstaff [PS, Theorem 4.4] on edge-weighted graph $(C_{5},\mathbf{w})$. They
proved that $\mathbf{w}$ is Cohen-Macaulay if and only if there exists a
vertex $v$ so that the weights on edges of $C_{5}$ starting from $v$ in
clockwise order are of the form $a,b,c,d,a$ and that $a\leq b\geq c\leq d\geq
a$. We call such a vertex $v$ a balancing vertex of $\mathbf{w}$.

Let $H$ be a graph formed by connecting two $5$-cycles by a path. By [HMT,
Theorem 2.4], $H$ is Cohen-Macaulay if and only if this path is of length $1$.
We may assume that the vertices of $H$ are
$\\{x_{1},\ldots,x_{5},y_{1},\ldots,y_{5}\\}$ and edges of $H$ are
$\\{x_{1}x_{2},\ldots,x_{4}x_{5},x_{1}x_{5},y_{1}y_{2},\ldots,y_{4}y_{5},y_{1}y_{5},x_{1}y_{1}\\}$.
Note that $I(H)+(x_{i})$ and $I(H)+(y_{i})$ are not Cohen-Macaulay for
$i\in\\{2,5\\}$. With this assumption, we have

######  Proposition 3.2.

The edge-weighted graph $(H,\mathbf{w})$ is Cohen-Macaulay if and only if
$\mathbf{w}$ satisfies the following conditions:

  1. (1)

$\mathbf{w}(x_{1}y_{1})\leq\min\\{\mathbf{w}(x_{1}x_{2}),\mathbf{w}(x_{1}x_{5}),\mathbf{w}(y_{1}y_{2}),\mathbf{w}(y_{1}y_{5})\\},$

  2. (2)

The induced edge-weighted graphs of $(H,\mathbf{w})$ on
$\\{x_{1},\ldots,x_{5}\\}$ and $\\{y_{1},\ldots,y_{5}\\}$ are Cohen-Macaulay.

  3. (3)

Balancing vertices of $\mathbf{w}$ on $\\{x_{1},\ldots,x_{5}\\}$ and
$\\{y_{1},\ldots,y_{5}\\}$ can be chosen among $\\{x_{1},x_{3},x_{4}\\}$ and
$\\{y_{1},y_{3},y_{4}\\}$ respectively.

###### Proof.

Denote $I=I(H_{\mathbf{w}})$. Let $(H_{1},\mathbf{w}_{1})$ and
$(H_{2},\mathbf{w}_{2})$ be the induced edge-weighted graphs of
$(H,\mathbf{w})$ on $\\{x_{1},\ldots,x_{5}\\}$ and $\\{y_{1},\ldots,y_{5}\\}$
respectively.

First, we prove that if $\mathbf{w}$ is Cohen-Macaulay, it must satisfy the
above conditions. For (1), assume by contradiction that
$\mathbf{w}(x_{1}y_{1})=a>\mathbf{w}(y_{1}y_{2})=b$. Let
$c=\max(\mathbf{w}(y_{3}y_{4}),\mathbf{w}(y_{4}y_{5}))$. Then

$\sqrt{I:y_{1}^{a-1}y_{4}^{c}}=I(H[x_{1},\ldots,x_{5},y_{1}])+(y_{2},y_{3},y_{5}).$

In particular, it is not Cohen-Macaulay. By Lemma 2.1, $I(H_{\mathbf{w}})$ is
not Cohen-Macaulay, a contradiction. By symmetry, $\mathbf{w}$ must satisfy
condition (1).

We now prove that $(H_{2},\mathbf{w}_{2})$ must be Cohen-Macaulay. Assume by
contradiction that, $(H_{2},\mathbf{w}_{2})$ is not Cohen-Macaulay. By Lemma
2.1, there exists an exponent $y^{\mathbf{b}}$ such that
$\sqrt{I(H_{2},\mathbf{w}_{2}):y^{\mathbf{b}}}$ is not Cohen-Macaulay. Then we
have

$\sqrt{I(H_{\mathbf{w}}):x_{2}^{a_{2}}x_{4}^{a_{4}}y^{\mathbf{b}}}=\sqrt{I(H_{2},\mathbf{w}_{2}):y^{\mathbf{b}}}+(x_{1},x_{3},x_{5}),$

where $a_{2}=\max(\mathbf{w}(x_{2}x_{1}),\mathbf{w}(x_{3}x_{2}))$ and
$a_{4}=\max(\mathbf{w}(x_{3}x_{4}),\mathbf{w}(x_{4}x_{5}))$. In particular, it
is not Cohen-Macaulay. By Lemma 2.1, $I(H_{\mathbf{w}})$ is not Cohen-
Macaulay, a contradiction. By symmetry, $\mathbf{w}$ must satisfy condition
(2).

Now note that if $\mathbf{w}(x_{2}x_{3})<\mathbf{w}(x_{3}x_{4})$ then
$\sqrt{I:x_{3}^{b}}=I+(x_{2})$ where $b=\mathbf{w}(x_{3}x_{4})-1$. Since
$I+(x_{2})$ is not Cohen-Macaulay, this implies a contradiction. Hence,
$\mathbf{w}(x_{2}x_{3})\geq\mathbf{w}(x_{3}x_{4})$. By symmetry, we deduce
that $\mathbf{w}(x_{4}x_{5})\geq\mathbf{w}(x_{3}x_{4})$. By [PS, Theorem 4.4]
and the previous claim that $(H_{1},\mathbf{w}_{1})$ is Cohen-Macaulay, we
deduce that a balancing vertex of $\mathbf{w}$ on $\\{x_{1},\ldots,x_{5}\\}$
can be chosen among $\\{x_{1},x_{3},x_{4}\\}$. By symmetry, $\mathbf{w}$ must
satisfy condition (3).

It remains to prove that if $\mathbf{w}$ satisfies conditions (1), (2), (3),
then $I=I(H_{\mathbf{w}})$ is Cohen-Macaulay. By Lemma 2.1, it suffices to
prove that $\sqrt{I:x^{\mathbf{a}}y^{\mathbf{b}}}$ is Cohen-Macaulay for all
exponents $\mathbf{a},\mathbf{b}$ such that
$x^{\mathbf{a}}y^{\mathbf{b}}\notin I$. Denote
$I_{\mathbf{a},\mathbf{b}}=\sqrt{I:x^{\mathbf{a}}y^{\mathbf{b}}}$. First, we
have

Claim A. Assume that $(G,\mathbf{w})$ is an edge-weighted graph and
$x^{\mathbf{a}}$ is an exponent such that $a_{i}<\mathbf{w}(e)$ for all edges
$e$ adjacent to $i$ then

(2)
$\sqrt{I(G_{\mathbf{w}}):x^{\mathbf{a}}}=\sqrt{I(G_{\mathbf{w}}):x^{\mathbf{b}}},$

with $x^{\mathbf{b}}=x_{1}^{a_{1}}\cdots
x_{i-1}^{a_{i-1}}x_{i+1}^{a_{i+1}}\cdots x_{n}^{a_{n}}$. In other words, we
may assume that $a_{i}=0$.

By symmetry, we may assume that $a_{1}\geq b_{1}$. Since
$x^{\mathbf{a}}y^{\mathbf{b}}\notin I(H_{\mathbf{w}})$, we must have
$b_{1}<\mathbf{w}(x_{1}y_{1})\leq\min(\mathbf{w}(y_{1}y_{2}),\mathbf{w}(y_{1}y_{5}))$.
By Claim A, we may assume that $b_{1}=0$. There are two cases as follows.

Case 1. $a_{1}\geq\mathbf{w}(x_{1}y_{1})$. Then

(3)
$I_{\mathbf{a},\mathbf{b}}=\sqrt{I:x^{\mathbf{a}}y^{\mathbf{b}}}=(y_{1})+\sqrt{I(H_{1},\mathbf{w}_{1}):x^{\mathbf{a}}}+\sqrt{I(H_{2},\mathbf{w}_{2}):y^{\mathbf{b}}}$

Since $(H_{1},\mathbf{w}_{1})$ is Cohen-Macaulay by [PS, Theorem 4.4],
$I_{\mathbf{a},\mathbf{b}}$ is not Cohen-Macaulay if and only if
$(y_{1})+\sqrt{I(H_{2},\mathbf{w}_{2}):y^{\mathbf{b}}}=(y_{1},y_{2})+I(H_{2})$
or $I(H_{2})+(y_{1},y_{5})$. Assume by contradiction that
$(y_{1})+\sqrt{I(H_{2},\mathbf{w}_{2}):y^{\mathbf{b}}}=(y_{1},y_{2})+I(H_{2})$.
Then we must have $b_{3}\geq\mathbf{w}(y_{2}y_{3})\geq\mathbf{w}(y_{3}y_{4})$.
But then $y_{4}\in\sqrt{I(H_{2},\mathbf{w}_{2}):y^{\mathbf{b}}}$, a
contradiction. Hence, $I_{\mathbf{a},\mathbf{b}}$ is Cohen-Macaulay.

Case 2. $a_{1}<\mathbf{w}(x_{1}y_{1})$. By Claim A, we may assume that
$a_{1}=0$. In particular,

(4)
$I_{\mathbf{a},\mathbf{b}}=\sqrt{I(H_{1},\mathbf{w}_{1}):x^{\mathbf{a}}}+\sqrt{I(H_{2},\mathbf{w}_{2}):y^{\mathbf{b}}}+(x_{1}y_{1}).$

If $x_{1}$ or $y_{1}$ appears in $I_{\mathbf{a},\mathbf{b}}$, with an argument
similar to Case 1, we deduce that $I_{\mathbf{a},\mathbf{b}}$ is the sum of
two Cohen-Macaulay ideals on different sets of variables and an ideal
generated by some other variables. Hence, it is Cohen-Macaulay. Thus, we may
assume that $x_{1},y_{1}$ does not appear in $I_{\mathbf{a},\mathbf{b}}$. By
Lemma 2.2, $\sqrt{I(H_{1},\mathbf{w}_{1}):x^{\mathbf{a}}}=I(H_{1})+(x_{i}\mid
i\in W_{1})$, where $W_{1}\subseteq\\{1,\ldots,5\\}$. By the following facts,
$W_{1}$ must belong to
$P=\\{\\{2,4\\},\\{2,3,4\\},\\{3,5\\},\\{3,4,5\\},\\{3\\},\\{4\\},\emptyset\\}$.

  1. (1)

By assumption, $1\notin W_{1}$.

  2. (2)

Since the balancing vertex of $(H_{1},\mathbf{w}_{1})$ can be chosen in the
set $\\{3,4,1\\}$, $\mathbf{w}(x_{2}x_{3})\geq\mathbf{w}(x_{3}x_{4})$. Hence,
if $2\in W_{1}$ then $4\in W_{1}$. Similarly, if $5\in W_{1}$ then $3\in
W_{1}$.

  3. (3)

By [PS, Theorem 4.4] and Lemma 2.1, $I(H_{1})+(x_{i}\mid i\in W_{1})$ is a
Cohen-Macaulay ideal, $W_{1}$ cannot be $\\{3,4\\}$.

  4. (4)

$W_{1}$ cannot contain $\\{2,5\\}$.

Proof of (4). Assume by contradiction that $2,5\in W_{1}$. Since $a_{1}=0$, we
must have $a_{3}\geq\mathbf{w}(x_{2}x_{3})$. Similarly,
$a_{4}\geq\mathbf{w}(x_{4}x_{5})$. Since the balancing vertex of
$(H_{1},\mathbf{w}_{1})$ can be chosen in the set $\\{3,4,1\\}$, we have
$\mathbf{w}(x_{3}x_{4})\leq\min(\mathbf{w}(x_{2}x_{3}),\mathbf{w}(x_{4}x_{5}))$.
Hence, $\mathbf{w}(x_{3}x_{4})\leq a_{3},a_{4}$. But this implies that
$x^{\mathbf{a}}\in I$, a contradiction.

Now, it is easy to check that if $W_{1},W_{2}$ belong to $P$, we have
$I(H)+(x_{i}\mid i\in W_{1})+(y_{j}\mid j\in W_{2})$ is Cohen-Macaulay. The
Proposition follows. ∎

Finally, by Theorem 1.2, any Cohen-Macaulay very well-covered graph that is
not Woodroofe is a counterexample to Terai’s conjecture. We provide some
concrete examples below. Recall that a simple graph is called very well-
covered if the size of every minimal vertex cover is half the number of
vertices. In particular, it is unmixed.

######  Example 3.3.

Let $H$ be a suspension of a cycle $C_{t}$ for $t\neq 3,5$; i.e., the set of
edges and the set of vertices are

$E(H)=\\{x_{1}x_{2},x_{2}x_{3},\ldots,x_{t-1}x_{t},x_{t}x_{1},x_{1}y_{1},\ldots,x_{t}y_{t}\\}\text{
and }V(H)=\\{x_{1},y_{1},\ldots,x_{t},y_{t}\\}.$

Let $\mathbf{w}$ be a weight function on $E(H)$ taking value $w\geq 2$ for the
edges $x_{i}x_{i+1}$ and value $1$ otherwise. Then, $H$ is a Cohen-Macaulay
very well-covered graph, but $(H,\mathbf{w})$ is not sequentially Cohen-
Macaulay.

###### Proof.

The graph $H$ is Cohen-Macaulay by [SVV, Theorem 2.1] (also see [V]). By
definition, $H$ is very well-covered. Since

$\sqrt{I(H_{\mathbf{w}}):\prod_{i=1}^{t}x_{i}^{w-1}}=I(C_{t})+(y_{1},\ldots,y_{t})$

and $I(C_{t})$ is not sequentially Cohen-Macaulay by [FT, Proposition 4.1]. By
Lemma 2.1, $I(H_{\mathbf{w}})$ is not sequentially Cohen-Macaulay. ∎

### Acknowledgments

This paper was done while the first author was visiting the Vietnam Institute
for Advanced Study in Mathematics (VIASM). He would like to thank the VIASM
for the hospitality and financial support, and he also thanks the Vietnam
National Foundation for Science and Technology Development (NAFOSTED) for its
support under grant number 101.04-2021.19.

### Conflict of interest

The authors declare no potential conflict of interests.

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