We now consider the parity binomial edge ideal \({\mathcal {I}}_{K_{n}}\) of a complete graph \(K_{n}\) on \(n\ge 3\) vertices. For \(1\le i<j\le n\), let
$$\begin{aligned} f_{ij} := x_iy_j-x_jy_i \quad \text { and } \quad g_{ij} := x_ix_j - y_jy_i. \end{aligned}$$
The parity binomial edge ideal of the complete graph is \({\mathcal {I}}_{K_n} = \left( g_{ij}\mid 1\le i<j\le n\right)\).
We need some further notation. For any \(I\subseteq [n]\) we denote \({\mathfrak {m}}_I := (x_i,y_i\mid i\in I)\). Let \({\mathfrak {p}}^{+} := (x_i+y_i\mid i\in [n])\) and \({\mathfrak {p}}^{-} :=(x_i-y_i\mid i\in [n])\). Denote \(P_{ij} := (g_{ij})+ {\mathfrak {m}}_{[n]\backslash \{i,j\}}\). By [11, Theorem 5.9], there is a decomposition of \({\mathcal {I}}_{K_n}\) as follows.
Proposition 3.1
For \(n\ge 3\), we have
$$\begin{aligned} {\mathcal {I}}_{K_n} = {\mathcal {J}}_{K_n} \cap \bigcap _{1\le i<j\le n} P_{ij}. \end{aligned}$$
In particular, \(\dim (R/{\mathcal {I}}_{K_n}) = n\).
We analyze \({\mathcal {I}}_{K_{n}}\) by regular sequences arising from successively adding the polynomials \(f_{kn}\) or saturating with respect to them. Let \(I_0 := {\mathcal {I}}_{K_n}\) and, inductively for \(1\le k\le n-1\), \(I_k := I_{k-1} + (f_{kn})\).
Lemma 3.2
For \(1\le k\le n-1\), we have
$$\begin{aligned} I_{k-1} \subseteq \bigcap _{1\le i<j\le n-1} P_{ij} \cap {\mathcal {J}}_{K_n} \cap \bigcap _{t=k}^{n-1}P_{tn}. \end{aligned}$$
Proof
By Proposition 2.2, \(f_{1n}, \ldots , f_{(k-1)n}\in {\mathcal {J}}_{K_n}\). Moreover, for all \((\ell ,n)\ne (i,j)\) we have \(f_{\ell n}\in P_{ij}\). Thus
$$\begin{aligned} (f_{1n}, \ldots , f_{(k-1)n}) \subseteq \bigcap _{1\le i<j\le n-1} P_{ij} \cap {\mathcal {J}}_{K_n} \cap \bigcap _{t=k}^{n-1}P_{tn}. \end{aligned}$$
Together with Proposition 3.1 the lemma is proven. \(\square\)
Lemma 3.3
For \(1\le k\le n-1\), we have
$$\begin{aligned} I_{k-1}: f_{kn} = P_{kn}. \end{aligned}$$
In particular, \({{\mathrm{depth}}}(R/(I_{k-1}: f_{kn}))= 3\), \({{\mathrm{reg}}}(R/(I_{k-1}: f_{kn})) =1\) and \(P_{R/(I_{k-1}: f_{kn})}(t) = (1-t)^{2n-3}(1+t)\).
Proof
One can check that \(I_{k-1}: f_{kn} \supseteq P_{kn}\) (in fact \({\mathcal {I}}_{K_{n}} : f_{kn} \supseteq P_{kn}\)) by simple calculations like \(x_{1}f_{kn} \equiv -y_{k}g_{1n} \mod {\mathcal {I}}_{K_{n}}\). Now, for all \((k,n)\ne (i,j)\), one can see that \(f_{kn}\) is contained in both \(P_{ij}\) and \({\mathcal {J}}_{K_n}\). By [1, Lemma 4.4], \(P_{ij}: f_{kn} = {\mathcal {J}}_{K_n}: f_{kn} = R\) and \(P_{kn}: f_{kn} = P_{kn}\) because \(P_{kn}\) is a prime that does not contain \(f_{kn}\). Hence by Lemma 3.2, we have \(I_{k-1}: f_{kn} \subseteq P_{kn}\) and thus \(I_{k-1}: f_{kn} = P_{kn}\).
Using this result, the invariants can be computed for the prime \(P_{kn}\) as follows: \({{\mathrm{depth}}}(R/(I_{k-1}:f_{kn})) = {{\mathrm{depth}}}(R/P_{kn}) = 3\), \({{\mathrm{reg}}}(R/(I_{k-1}: f_{kn})) = {{\mathrm{reg}}}(R/P_{kn}) =1\), and \(P_{R/(I_{k-1}: f_{kn})}(t) = P_{R/P_{kn}}(t) = (1-t)^{2n-3}(1+t)\). \(\square\)
Lemma 3.4
$$\begin{aligned} I_{n-2}: (x_n +y_n) = {\mathfrak {p}}^{-}\cap P_{n-1,n}. \end{aligned}$$
In particular, \({{\mathrm{depth}}}(R/(I_{n-2}: (x_n+y_n)))\ge 3\), \({{\mathrm{reg}}}(R/(I_{n-2}: (x_{n}+y_n)))\le 1\) and \(P_{R/(I_{n-2}: (x_{n}+y_n))}(t) = (1-t)^n +2t(1-t)^{2n-3}\).
Proof
For the lexicographic ordering on \(\Bbbk [x_1,\ldots , x_n, y_1,\ldots , y_n,t]\) induced by \(x_1>\ldots>x_n>y_1>\ldots>y_n>t\), the Gröbner basis for \(J=t{\mathfrak {p}}^{-} + (1-t)P_{n-1,n}\) is
$$\begin{aligned} {\mathcal {G}}&= \{(x_{n-1}-y_{n-1})t,\, (x_{n}-y_{n})t),\, x_{n-1}x_n-y_{n-1}y_n,\\&\quad x_i-y_i,\, (x_{n-1}-y_{n-1})y_i,\, (x_{n}-y_{n})y_i,\, (t-1)y_{i} \mid 1\le i\le n-2 \}. \end{aligned}$$
Thus,
$$\begin{aligned}&{\mathfrak {p}}^{-}\cap P_{n-1,n} \\&\quad = \left( x_{n-1}x_n-y_{n-1}y_n,\, x_i-y_i,\, (x_{n-1}-y_{n-1})y_i,\, (x_{n}-y_{n})y_i \mid 1\le i\le n-2\}\right) . \end{aligned}$$
This implies the containment \({\mathfrak {p}}^{-}\cap P_{n-1,n} \subseteq I_{n-2}: (x_{n}+y_n)\). Conversely, by Lemma 3.2
$$\begin{aligned} I_{n-2} \subseteq \bigcap _{1\le i<j\le n-1} P_{ij} \cap {\mathcal {J}}_{K_n} \cap P_{n-1,n}. \end{aligned}$$
For all \(1\le i<j\le n-1\), it is clear that \(x_n+y_n\in P_{ij}\) and so \(P_{ij}:(x_n+y_n)=R\). By [1, Lemma 4.4], \(P_{n-1,n}: (x_n+y_n)=P_{n-1,n}\). Moreover, by Proposition 2.2, we obtain that \({\mathcal {J}}_{K_n}: (x_n+y_n) = {\mathfrak {p}}^{-}\). This implies that \(I_{n-2}: (x_n+y_n) \subseteq {\mathfrak {p}}^{-} \cap P_{n-1,n}\) and thus the conclusion \(I_{n-2}: (x_{n}+y_n) = {\mathfrak {p}}^{-}\cap P_{n-1,n}\).
In order to prove the second part, note that
$$\begin{aligned} {\mathfrak {p}}^{-}+P_{n-1,n} = (x_{n-1}+y_{n-1}, x_n+y_n) + {\mathfrak {m}}_{[n-2]}. \end{aligned}$$
Therefore one reads off \({{\mathrm{depth}}}(R/({\mathfrak {p}}^{-}+P_{n-1,n})) = 2\) and \({{\mathrm{reg}}}(R/({\mathfrak {p}}^{-}+P_{n-1,n}))=0\). It is clear that \({{\mathrm{depth}}}(R/{\mathfrak {p}}^{-}) =n\) and \({{\mathrm{reg}}}(R/{\mathfrak {p}}^{-})=0\). From the exact sequence
$$\begin{aligned} 0\longrightarrow R/ ({\mathfrak {p}}^{-}\cap P_{n-1,n}) \longrightarrow R/ {\mathfrak {p}}^{-}\oplus R/P_{n-1,n} \longrightarrow R/({\mathfrak {p}}^{-}+P_{n-1,n})\longrightarrow 0, \end{aligned}$$
we obtain, using Lemma 2.1, that
$$\begin{aligned} {{\mathrm{depth}}}(R/I_{n-2}: (x_{n}+y_n))&= {{\mathrm{depth}}}(R/ ({\mathfrak {p}}^{-}\cap P_{n-1,n})) \ge \min \{n,3,2+1\}=3,\\ {{\mathrm{reg}}}(R/I_{n-2}: (x_{n}+y_n))&= {{\mathrm{reg}}}(R/ ({\mathfrak {p}}^{-}\cap P_{n-1,n})) \le \max \{0,1,0+1\} =1, \end{aligned}$$
and furthermore,
$$\begin{aligned} P_{R/I_{n-2}: (x_{n}+y_n)}(t)&= P_{R/ {\mathfrak {p}}^{-}}(t) + P_{R/P_{n-1,n}}(t) - P_{R/({\mathfrak {p}}^{-}+P_{n-1,n})}(t) \\&= (1-t)^n +(1-t)^{2n-3}(1+t) - (1-t)^{2n-2}\\&= (1-t)^n + 2t(1-t)^{2n-3}. \end{aligned}$$
\(\square\)
Lemma 3.5
Let \(J :=(x_n+y_n, I_{n-2})\). Then
$$\begin{aligned} {{\mathrm{depth}}}(R/J)&\ge \min \{n,{{\mathrm{depth}}}(S/{\mathcal {I}}_{K_{n-1}})\},\\ {{\mathrm{reg}}}(R/J)&\le \max \{1, {{\mathrm{reg}}}(S/{\mathcal {I}}_{K_{n-1}})\}, \end{aligned}$$
and \(P_{R/J}(t) = t(1-t)^n + (1-t)^2 P_{S/{\mathcal {I}}_{K_{n-1}}}(t),\) where \(S = \Bbbk [x_i,y_i\mid 1\le i \le n-1]\).
Proof
In order to prove the lemma, we first check two following claims:
Claim 1
\((J,x_n) = (x_n,y_n, {\mathcal {I}}_{K_{n-1}}).\)
Since \(y_n = (x_n+y_n) - x_n\in (x_n,J)\) and \({\mathcal {I}}_{K_{n-1}}\subseteq I_{n-2}\), we have \((x_n,y_n, {\mathcal {I}}_{K_{n-1}}) \subseteq (J,x_n)\). Conversely, \(x_n+y_n, g_{in}, f_{in} \in (x_n,y_n)\) for \(1\le i\le n-1\) and thus \((J,x_n)\subseteq (x_n,y_n, {\mathcal {I}}_{K_{n-1}})\).
Claim 2
\(J: x_n = {\mathfrak {p}}^+.\)
One can compute \(x_n(x_i+y_i) = (x_ix_n-y_iy_n) + y_i(x_n+y_n)\in J\) for \(1\le i\le n\), so that \(x_n{\mathfrak {p}}^{+} \subseteq J\) which implies that \({\mathfrak {p}}^{+} \subseteq J: x_n\). Conversely, for \(1\le i<j\le n\), we have
$$\begin{aligned} g_{ij}&= x_ix_j - y_iy_j = (x_i-y_i)x_j + y_i(x_j-y_j)= (x_i+y_i)x_j - y_i(x_j+y_j),\\ f_{ij}&= x_iy_j-x_jy_i = (x_i+y_i)y_j - y_i(x_j+y_j)= (x_i-y_i)y_j - y_i(x_j-y_j). \end{aligned}$$
Thus, by Proposition 2.2, \({\mathcal {J}}_{K_n} \subseteq {\mathfrak {p}}^{+} \cap (x_1-y_1,\ldots ,x_{n-1}-y_{n-1}, x_n,y_n)\) and \(f_{kn}\in {\mathfrak {p}}^{+} \cap (x_1-y_1,\ldots ,x_{n-1}-y_{n-1}, x_n,y_n)\) for all \(1\le k\le n-2\). Together with Proposition 3.1,
$$\begin{aligned} J \subseteq \bigcap _{1\le i<j\le n-1} P_{ij} \cap {\mathfrak {p}}^{+} \cap (x_1-y_1,\ldots ,x_{n-1}-y_{n-1}, x_n,y_n). \end{aligned}$$
By [1, Lemma 4.4], \(J: x_n \subseteq {\mathfrak {p}}^+\) and thus the claim holds.
Now, we turn to the proof of the lemma. By Claim 1,
$$\begin{aligned} {{\mathrm{depth}}}(R/(J,x_n)) = {{\mathrm{depth}}}(S/{\mathcal {I}}_{K_{n-1}}) \text { and } {{\mathrm{reg}}}(R/(J,x_n)) = {{\mathrm{reg}}}(S/{\mathcal {I}}_{K_{n-1}}). \end{aligned}$$
Moreover, by Claim 2, we have
$$\begin{aligned} {{\mathrm{depth}}}(R/J: x_n)={{\mathrm{depth}}}(R/{\mathfrak {p}}^+)=n \text { and } {{\mathrm{reg}}}(R/J:x_n) ={{\mathrm{reg}}}(R/{\mathfrak {p}}^+)=0. \end{aligned}$$
From the exact sequence
$$\begin{aligned} 0 \longrightarrow R/(J: x_n)(-1) \longrightarrow R/J \longrightarrow R/(J,x_n)\longrightarrow 0 \end{aligned}$$
we obtain
$$\begin{aligned} {{\mathrm{depth}}}(R/J) \ge \min \{n,{{\mathrm{depth}}}(S/{\mathcal {I}}_{K_{n-1}})\} \text { and } {{\mathrm{reg}}}(R/J)\le \max \{1, {{\mathrm{reg}}}(S/{\mathcal {I}}_{K_{n-1}})\}. \end{aligned}$$
Moreover,
$$\begin{aligned} P_{R/J}(t)&= tP_{R/J: x_n}(t) + P_{R/(J, x_n)}(t) = tP_{R/{\mathfrak {p}}^+}(t) + P_{R/(x_n,y_n, {\mathcal {I}}_{K_{n-1}})}(t)\\&= t(1-t)^{n} + (1-t)^2 P_{S/{\mathcal {I}}_{K_{n-1}}}(t), \end{aligned}$$
as required. \(\square\)
Theorem 3.6
The Hilbert–Poincaré polynomial of \(R/{\mathcal {I}}_{K_{n}}\) is
$$\begin{aligned} P_{R/{\mathcal {I}}_{K_n}}(t) =2(1-t)^n + \left[ -1+3t + \left( \frac{n^2+n-6}{2}\right) t^2 + \left( \frac{n^2-3n+2}{2}\right) t^3 \right] (1-t)^{2n-3}. \end{aligned}$$
In particular, \({{\mathrm{depth}}}(R/{\mathcal {I}}_{K_n}) \ge 3\) and \({{\mathrm{reg}}}(R/{\mathcal {I}}_{K_n})\le 3\).
Proof
The proof is by induction on n. If \(n=3\), then a simple calculation (e.g. in Macaulay2) gives the result. Now assume \(n\ge 4\). For any \(1\le k\le n-1\) there is an exact sequence
$$\begin{aligned} 0\longrightarrow R/(I_{k-1}: f_{kn})(-2) \xrightarrow {\cdot f_{kn}} R/I_{k-1} \longrightarrow R/I_{k} \longrightarrow 0. \end{aligned}$$
By Lemmas 2.1 and 3.3, \({{\mathrm{depth}}}(R/I_{k-1}) \ge \min \{3, {{\mathrm{depth}}}(R/I_{k})\}\), \({{\mathrm{reg}}}(R/I_{k-1}) \le \max \{3, {{\mathrm{reg}}}(R/I_{k})\}\) and \(P_{R/I_{k-1}}(t) = t^2(1-t)^{2n-3}(1+t)+ P_{R/I_{k}}(t)\). This implies that \({{\mathrm{depth}}}(R/I_0) \ge \min \{3,{{\mathrm{depth}}}(R/I_{n-2})\}\), \({{\mathrm{reg}}}(R/I_{0}) \le \max \{3,{{\mathrm{reg}}}(R/I_{n-2})\}\) and
$$\begin{aligned} P_{R/I_{0}}(t) = (n-2)t^2(1-t)^{2n-3}(1+t)+ P_{R/I_{n-2}}(t). \end{aligned}$$
Now consider the following exact sequence
$$\begin{aligned} 0\longrightarrow R/(I_{n-2}: (x_{n}+y_n))(-1) \longrightarrow R/I_{n-2} \longrightarrow R/(x_n+y_n, I_{n-2}) \longrightarrow 0. \end{aligned}$$
Let \(S := \Bbbk [x_i,y_i\mid 1\le i\le n-1]\). By Lemmas 3.4 and 3.5, \({{\mathrm{depth}}}(R/I_{n-2}) \ge \min \{3,{{\mathrm{depth}}}(S/{\mathcal {I}}_{K_{n-1}})\}\), \({{\mathrm{reg}}}(R/I_{n-2}) \le \max \{1, {{\mathrm{reg}}}(S/{\mathcal {I}}_{K_{n-1}})\}\) and
$$\begin{aligned} P_{R/I_{n-2}}(t)&= tP_{R/(I_{n-2}:(x_n+y_n))}(t) + P_{R/(x_n+y_n,I_{n-2})}(t)\\&= 2t(1-t)^n + 2t^2(1-t)^{2n-3} + (1-t)^2 P_{S/{\mathcal {I}}_{K_{n-1}}}(t). \end{aligned}$$
The induction hypothesis yields \({{\mathrm{depth}}}(S/{\mathcal {I}}_{K_{n-1}})\ge 3\) and \({{\mathrm{reg}}}(S/{\mathcal {I}}_{K_{n-1}}) \le 3\). Therefore \({{\mathrm{depth}}}(R/I_{n-2}) \ge 3\) and \({{\mathrm{reg}}}(R/I_{n-2})\le 3\). This is enough to conclude that \({{\mathrm{depth}}}(R/{\mathcal {I}}_{K_n}) \ge 3\) and \({{\mathrm{reg}}}(R/{\mathcal {I}}_{K_n})\le 3\). Moreover,
$$\begin{aligned} P_{R/{\mathcal {I}}_{K_n}}(t)&= 2t(1-t)^n + \Big [(n-2)t^3 + nt^2\Big ] (1-t)^{2n-3} + (1-t)^2 P_{S/{\mathcal {I}}_{K_{n-1}}}(t). \\&= 2t(1-t)^n + \Big [(n-2)t^3 + nt^2\Big ] (1-t)^{2n-3}\\&\quad + 2(1-t)^{n+1} + \left[ -1+3t + (\frac{n^2-n-6}{2})t^2 + (\frac{n^2-5n+6}{2})t^3\right] (1-t)^{2n-3}\\&=2(1-t)^n + \left[ -1+3t + (\frac{n^2+n-6}{2})t^2 + (\frac{n^2-3n+2}{2})t^3\right] (1-t)^{2n-3}, \end{aligned}$$
as required. \(\square\)
If an ideal has a square-free initial ideal, its extremal Betti numbers agree with that of the initial ideal by [3]. Although the parity binomial edge ideal of a complete graph cannot have a square-free initial ideal (see [11, Remark 3.12]), the bottom right Betti number agrees with that of the initial ideal for any term order.
Corollary 3.7
$$\begin{aligned} \beta _{2n-3,2n}(R/{\mathcal {I}}_{K_n}) =\beta _{2n-3,2n}(R/{{\mathrm{in}}}_<({\mathcal {I}}_{K_n}))= \frac{n^2-3n+2}{2}. \end{aligned}$$
In particular,
$$\begin{aligned} {{\mathrm{reg}}}(R/{\mathcal {I}}_{K_n}) = {{\mathrm{reg}}}(R/{{\mathrm{in}}}_<({\mathcal {I}}_{K_n})) = {{\mathrm{depth}}}(R/{\mathcal {I}}_{K_n}) = {{\mathrm{depth}}}(R/{{\mathrm{in}}}_<({\mathcal {I}}_{K_n})) =3. \end{aligned}$$
Proof
From Theorem 3.6 we obtain \(\beta _{p,p+r}(R/{\mathcal {I}}_{K_n}) =\frac{n^2-3n+2}{2}\ne 0\), where \(p={{\mathrm{pdim}}}(R/{\mathcal {I}}_{K_n})\) and \(r={{\mathrm{reg}}}(R/{\mathcal {I}}_{K_n})\). Thus, \(p+r=2n\). Since \(P_{R/{\mathcal {I}}_{K_n}}(t) = P_{R/{{\mathrm{in}}}_<({\mathcal {I}}_{K_n})}(t)\), we get
$$\begin{aligned} {{\mathrm{reg}}}(R/{\mathcal {I}}_{K_n}) = {{\mathrm{reg}}}(R/{{\mathrm{in}}}_<({\mathcal {I}}_{K_n})),\quad {{\mathrm{pdim}}}(R/{\mathcal {I}}_{K_n})={{\mathrm{pdim}}}(R/{{\,\mathrm{in}}}_<({\mathcal {I}}_{K_n})) \text { and } \end{aligned}$$
\(\beta _{p,p+r}(R/{\mathcal {I}}_{K_n}) =\beta _{p,p+r}(R/{{\mathrm{in}}}_<({\mathcal {I}}_{K_n}))\). On the other hand, \(r \le 3\) and \(p\le 2n-3\) by the Auslander–Buchsbaum formula. Thus, \(r = 3\) and \(p = 2n-3\). \(\square\)