Proof. If $\dim R=0$ , then by assumption R is Gorenstein and hence not of Teter type. Suppose now $\dim R>0$ . Then $\operatorname {height}(\operatorname {tr}(\omega _R))>0$ , since R is generically Gorenstein. On the contrary, assume R is of Teter type. Let P be a minimal prime ideal of $\operatorname {tr}(\omega _R)$ . Then $\dim R_P>0$ . Since $\omega _R$ as well as $\operatorname {tr}(\omega _R)$ localize, we may replace R by $R_P$ , and hence we may assume that R is a ring of Teter type with $\dim R>0$ and that $\operatorname {tr}(\omega _R)$ is ${\mathfrak m}$ -primary. In particular, there exists an epimorphism $\varphi \colon \omega _R\rightarrow \operatorname {tr}(\omega _R)$ . Since $\operatorname {tr}(\omega _R)$ is ${\mathfrak m}$ -primary, it follows that $R_P$ is Gorenstein for all $P\neq {\mathfrak m}$ . This implies that $\varphi _P$ is an isomorphism for all $P\neq {\mathfrak m}$ . Thus, if C is the kernel of $\varphi $ , then $C=0$ or $\operatorname {Supp}(C)=\{{\mathfrak m}\}$ . Suppose $C\neq 0$ , Then $\operatorname {depth} C=0$ , and hence also $\operatorname {depth} \omega _R=0$ . Since $\omega _R$ is a maximal Cohen–Macaulay module, it follows that $\dim R=0$ , a contradiction. Thus, we have shown that $C=0$ , which implies that $\varphi $ is an isomorphism.

As noticed in [Reference Herzog, Hibi and Stamate6], $\omega _R$ can be identified with an ideal in R, since R is generically Gorenstein and $\operatorname {tr}(\omega _R)=\omega _R\cdot \omega _R^{-1}$ , where $\omega _R^{-1}=\{x\in Q(R)\colon x\omega _R\subset R\}$ . Here, $Q(R)$ denotes the full ring of fractions of R. Thus, we have $\omega _R\cong \omega _R\cdot \omega _R^{-1}$ . This gives us

$$\begin{align*}R=\omega_R: \omega_R\cong \omega_R :(\omega_R\cdot \omega_R^{-1})=(\omega_R :\omega_R):\omega_R^{-1}=R:\omega_R^{-1}=(\omega_R^{-1})^{-1}. \end{align*}$$

Since R is not Gorenstein, [Reference Herzog and Kunz7, korollar 7.29] implies that $\omega _R\neq (\omega _R^{-1})^{-1}$ , and we obtain an exact sequence $0\rightarrow \omega _R\rightarrow (\omega _R^{-1})^{-1}\rightarrow D\rightarrow 0$ with $D\neq 0$ . Since R is Gorenstein on the punctured spectrum, the inclusion map $\omega _R\rightarrow (\omega _R^{-1})^{-1}$ becomes an isomorphism after localization with $P\in \operatorname {Spec}(R)\setminus {\mathfrak m}$ . This implies that $\operatorname {Supp}(D)=\{{\mathfrak m}\}$ . Hence, $\operatorname {depth} D=0$ . Since $(\omega _R^{-1})^{-1}\cong R$ , it follows then that $\operatorname {depth} \omega _R=1$ . Thus, $\dim R=1$ .

Since $\omega _R\subset (\omega _R^{-1})^{-1}\cong R$ , we deduce that $(\omega _R^{-1})^{-1} =(x)$ for some $x\in R$ and that $\omega _R\subset (x)$ . Hence, $\omega _R=xL$ for some ideal L. Since $\operatorname {Ann}(\omega _R)=(0)$ , the element x must be a nonzero divisor. Therefore, L is also a canonical ideal of R. Moreover, $(x)=R:(R:xL) =x(R:(R:L))$ . This shows that $R:(R:L)=R$ . Hence, replacing $\omega _R$ by L, we may assume that $R=(\omega _R^{-1})^{-1}$ . Then $\omega _R^{-1} =((\omega _R^{-1})^{-1})^{-1}=R^{-1}=R$ .

The exact sequence $0\rightarrow \omega _R\rightarrow R \rightarrow R/\omega _R\rightarrow 0$ induces the exact sequence

$$\begin{align*}0\rightarrow R\rightarrow \omega_R^{-1} \rightarrow \operatorname{Ext}_R^1(R/\omega_R, R)\rightarrow 0. \end{align*}$$

The map $R\rightarrow \omega _R^{-1}$ is the identity, since $\omega _R^{-1}=R$ . Therefore, $\operatorname {Ext}_R^1(R/\omega _R, R)=0$ . Let $y\in \omega _R$ be a nonzero divisor. Set $\overline {R}=R/(y)$ . The exact sequence

$$\begin{align*}0\rightarrow R\stackrel{y}{\rightarrow} R\rightarrow \overline{R}\rightarrow 0 \end{align*}$$

induces the exact sequence

$$\begin{align*}0\rightarrow \operatorname{Hom}_R(R/\omega_R, \overline{R})\rightarrow \operatorname{Ext}_R^1(R/\omega_R, R)\rightarrow \cdots, \end{align*}$$

since $\operatorname {Hom}_R(R/\omega _R,R)=0$ . It follows that $\operatorname {Hom}_R(R/\omega _R, \overline {R})=0$ . Notice that $R/\omega _R$ is an $\overline {R}$ -module, which implies that $\operatorname {Hom}_R(R/\omega _R, \overline {R})\cong \operatorname {Hom}_{\overline {R}}(R/\omega _R, \overline {R})$ . Thus, ${\operatorname {Hom}_{\overline {R}}(R/\omega _R, \overline {R})=0}$ . Since $\dim \overline {R}=0$ , there exists $c\in \overline {R}$ with $c\neq 0$ and $c\overline {{\mathfrak m}}=0$ . Here, $\overline {{\mathfrak m}}$ denotes the maximal ideal of $\overline {R}$ . The nonzero $\overline {R}$ -module homomorphism $\overline {R}\rightarrow \overline {R}$ , $1\mapsto c$ factorizes over $R/\omega _R$ . Thus, $\operatorname {Hom}_{\overline {R}}(R/\omega _R, \overline {R})\neq 0$ , a contradiction.

Proof. Let ${\mathcal A}=\operatorname {Hom}_R(\omega _R,R)$ . Let $\varphi \in {\mathcal A}$ . We denote by $V_\varphi $ the K-subspace $(\varphi (\omega _R)+{\mathfrak m}\operatorname {tr}(\omega _R))/{\mathfrak m}\operatorname {tr}(\omega _R)$ of the K-vector space $\operatorname {tr}(\omega _R)/{\mathfrak m}\operatorname {tr}(\omega _R)$ .

We first prove by induction on n the following elementary fact: if V is any K-vector space of dimension n and $V=\sum _{\lambda \in \Lambda }V_\lambda $ , then there exist $\lambda _1,\ldots ,\lambda _n\in \Lambda $ such that $V=\sum _{i=1}^nV_{\lambda _i}$ . The assertion is trivial for $n=1$ . Now, let us assume that $n>1$ . There exists $\lambda _1\in \Lambda $ such that $V_{\lambda _1}\neq 0$ . Let $W=V/V_{\varphi _1}$ . Then W is a K-vector space of dimension $< n$ , and $\sum _{\lambda \in \Lambda , \lambda \neq \lambda _1}(V_\lambda +V_{\lambda _1})/V_{\lambda _1}=W$ . By the induction hypothesis, there exist $\lambda _2,\ldots ,\lambda _n$ such that $\sum _{i=2}^n(V_{\lambda _i}+V_{\lambda _1})/V_{\lambda _1}=W$ . It follows that $V=\sum _{i=1}^nV_{\lambda _i}$ .

Applying this result to our particular case, we see that there exist $\varphi _1,\ldots ,\varphi _n$ such that $(\sum _{i=1}^n\varphi _i(\omega _R))+{\mathfrak m}\operatorname {tr}(\omega _R)=\operatorname {tr}(\omega _R)$ . Nakayama’s lemma implies that $\sum _{i=1}^n\varphi _i(\omega _R)= \operatorname {tr}(\omega _R)$ .

Proof. Suppose I is a $\tau $ -ideal. Then there exists an epimorphism $\varphi :\omega _R\rightarrow I$ . This epimorphism induces a monomorphism $\varphi ^\vee \colon I^\vee \rightarrow R$ . Let J be the image of $I^\vee \subseteq R$ . Then J is an ideal in R and J is isomorphic to $I^\vee $ . Hence, J is a companion of I.

Let J be a companion of I. Then there exists an isomorphism $\alpha : I^\vee \rightarrow J$ . The inclusion map $I\rightarrow R$ induces an epimorphism $\psi \colon \omega _R\rightarrow I^\vee $ . Then $\varphi =\alpha \circ \psi : \omega _R \rightarrow J$ is an epimorphism. Hence, J is a $\tau $ -ideal.

Moreover, if I has a companion $J'$ , then $J'\cong I^\vee $ , and so $I\cong (J')^\vee $ . Therefore, I is a companion of $J'$ . Hence, I is a $\tau $ -ideal.

Proof. Note that $\omega _R=\operatorname {Hom}_G(R,G)\cong 0:J$ . The composition of the inclusion map $0:J\subset G$ with the canonical epimorphism $G\rightarrow R$ yields an R-module homomorphism $\omega _R\rightarrow R$ whose image is $I=((0:J)+J)/J$ . Thus, I is a $\tau $ -ideal.

Assume now that $J\subset 0:J$ . Then $I=(0:J)/J$ , and we have the exact sequence

$$\begin{align*}0\rightarrow (0:J)/J\rightarrow G/J\rightarrow G/(0:J)\rightarrow 0. \end{align*}$$

Since G is Gorenstein, the functor $\operatorname {Hom}_G(-, G)$ is exact. Applying this functor to the above exact sequence, we obtain the exact sequence

$$\begin{align*}0\rightarrow\operatorname{Hom}_G(G/(0:J), G) \rightarrow \operatorname{Hom}_G(G/J, G)\rightarrow\operatorname{Hom}_G((0:J)/J, G)\rightarrow 0. \end{align*}$$

Observe that $\operatorname {Hom}_G(G/(0:J), G)\cong 0:(0:J)=J$ , since G is Gorenstein. Furthermore, $\operatorname {Hom}_G(G/J, G)\cong 0:J$ and $\operatorname {Hom}_G((0:J)/J, G)= ((0:J)/J)^\vee $ . This shows that $(0:J)/J\cong ((0:J)/J)^\vee $ , as desired.

Proof. (a) $\Rightarrow $ (b): If R is of Teter type, then $I=\operatorname {tr}(\omega _R)$ is a $\tau $ -ideal, and so I has a companion J. By Theorem 3.2, J is also a $\tau $ -ideal, which implies that $J\subseteq \operatorname {tr}(\omega _R)=I$ . Since $J\cong I^\vee $ , it follows that $\dim _KJ=\dim _KI$ . Thus, we conclude that $J=I$ . This means that I is symmetric.

(b) $\Rightarrow $ (a): $\operatorname {tr}(\omega _R)$ is symmetric, then $\operatorname {tr}(\omega _R)$ is a $\tau $ -ideal (see Theorem 3.3). This means that R is of Teter type.

Proof. For given ${\mathbf a}\in {\mathbb Z}^n$ , let ${\mathbf c}_{i_1},\ldots ,{\mathbf c}_{i_k}$ be the column vectors of $E_{\mathbf a}$ . Since these column vectors are homogeneous, it follows that ${\mathbf c}_{i_j}=(\lambda _{1j}u_1,\ldots ,\lambda _{rj}u_r)^{\mathsf {T}}$ for $j=1,\ldots ,k$ with $\lambda _{sj}\in K$ for $s=1,\ldots , r$ . Since K is infinite, we may choose $\mu _1,\ldots , \mu _k\in K$ such that $\sum _{j=1}^k\mu _j\lambda _{sj}\neq 0$ for those s for which there exists j with $\lambda _{sj}\neq 0$ . Then the ideal generated by the entries of the column vector ${\mathbf c}=\sum _{j=1}^k\mu _j{\mathbf c}_{i_j}$ is the same as the ideal generated by the entries of $E_{\mathbf a}$ , and the desired conclusion follows from Theorem 3.8.

Proof. We may assume that $a\leq b$ . Let ${\mathfrak m}_R$ be the maximal ideal of R. We claim that

$$\begin{align*}\operatorname{tr}(\omega_R)= \begin{cases} {\mathfrak m}_R^{k-1}, & \text{if } k\leq a,\\[3pt] {\mathfrak m}_R^{a-1}, & \text{if } a<k\leq b,\\[3pt] {\mathfrak m}_R^{a+b-1-k}, & \text{if } k>b, \end{cases} \end{align*}$$

and that $\operatorname {tr}(\omega _R)$ is symmetric in all these cases. This then shows that R is of Teter type.

Assume first that $k\leq a$ . Then $R=S/(x,y)^k$ , and

$$\begin{align*}0\rightarrow S(-k-1)^k\stackrel{A}{\longrightarrow}S(-k)^{k+1}\rightarrow (x,y)^k\rightarrow 0 \end{align*}$$

is the resolution of $(x,y)^k$ . Here, A is the $(k+1)\times k$ -matrix

$$\begin{align*}\begin{pmatrix} y & 0& \cdots&\cdots & 0\\ -x & y& \cdots & \cdots & 0\\ 0 & -x &\ddots &\cdots& 0\\ \vdots& \vdots& \ddots &\ddots &\vdots\\ 0& 0& \cdots&\ddots & y\\ 0& 0& \cdots&\cdots & -x\\ \end{pmatrix}. \end{align*}$$

By Theorem 3.6, the entries of the kernel of $R^k\stackrel {A}{\longrightarrow }R^{k+1}$ generate $\operatorname {tr}(\omega _R)$ . Let $(f_1,\ldots ,f_k)^{\mathsf {T}}$ be in the kernel of this map. We may assume that all $f_i$ are homogeneous. Since $yf_1=0$ , it follows that $\deg (f_1)\geq k-1$ , and hence $f_1\in {\mathfrak m}_R^{k-1}$ . Next, we have $-xf_1+yf_2=0$ . Since $xf_1=0$ , it follows that $yf_2=0$ . As before, we deduce that $f_2\in {\mathfrak m}_R^{k-1}$ . Proceeding in the way, we see that $f_i\in {\mathfrak m}_R^{k-1}$ for all i. This shows that $\operatorname {tr}(\omega _R)={\mathfrak m}_R^{k-1}$ . Note that ${\mathfrak m}_R^{k-1}=\operatorname {Soc}(R)$ . It is obvious that $\operatorname {Soc}(R)$ is a symmetric ideal.

In the next steps, we show that ${\mathfrak m}_R^{a-1}$ is a symmetric $\tau $ -ideal of R if $a<k\leq b$ , and that ${\mathfrak m}_R^{a+b-k-1}$ is a symmetric $\tau $ -ideal of R if $k>b$ . Indeed, if $a<k\leq b$ , then $R=S/(f, (x,y)^k)$ . Hence, we may write $R =\widetilde {G}/\tilde {{\mathfrak n}}^k$ , where $\widetilde {G}=S/(f,p)$ is a complete intersection and p is a minimal generator of $(x,y)^k$ and $\tilde {{\mathfrak n}}$ is the maximal ideal of $\tilde {G}$ . Indeed, if K is infinite, then a nonzero divisor p modulo f can be found among the generators of $(x,y)^k$ . On the other hand, if f and g are monomials, then each of them is a pure power, say $f=x^a$ . In this case, we may choose $p=y^k$ . The socle degree of this complete intersection is $a+k-2$ . It follows that $0: \tilde {{\mathfrak n}}^k=\tilde {{\mathfrak n}}^{a-1}$ . Since $\tilde {{\mathfrak n}}^k\subset 0: \tilde {{\mathfrak n}}^k$ , the desired conclusion follows from Theorem 3.4. Next, assume $k>b$ . Then, in the ring G, we have $0:{\mathfrak n}^k={\mathfrak n}^{a+b-1-k}$ , because the socle degree of G is $a+b-2$ . Since ${\mathfrak n}^k\subset {\mathfrak n}^{a+b-1-k}$ , the desired conclusion follows again from Theorem 3.4.

It remains to be shown that any $\tau $ -ideal is contained in ${\mathfrak m}_R^{a-1}$ if $a<k\leq b$ , and is contained in ${\mathfrak m}_R^{a+b-1-k}$ if $k>b$ .

Let us first treat the case in which $a<k\leq b$ . Then $R=S/(f,{\mathfrak m}^k)$ , and since ${\mathfrak m}^k:(f)={\mathfrak m}^{k-a}$ , we obtain the exact sequence

$$\begin{align*}0\rightarrow (S/({\mathfrak m}^{k-a}))(-a)\rightarrow S/{\mathfrak m}^k\rightarrow R\rightarrow 0. \end{align*}$$

Let ${\mathbb F}$ be the minimal graded resolution of ${\mathfrak m}^k$ , and let ${\mathbb G}$ be the minimal graded free resolution of $(S/({\mathfrak m}^{k-a}))(-a)$ . The multiplication map given by f can be lifted to a graded complex homomorphism ${\mathbb G}\rightarrow {\mathbb F}$ , so that we obtain a commutative diagram

Since the multiplication map $S(-a)\stackrel {f}{\longrightarrow } S$ is injective, it follows that, for any element $g\in (G_1)_k$ , we have that $\alpha (g)\neq 0$ . From this, it can be seen that $\alpha $ is injective. Similarly, $\beta $ is injective. In fact, $\alpha $ and $\beta $ are split injective, because by degree reasons, their image is a direct summand of $S(-k)^{k+1}$ and $S(-k-1)^{k}$ , respectively.

Let ${\mathbb D}$ be the mapping cone of ${\mathbb G}\rightarrow {\mathbb F}$ . Then we obtain the short exact sequence of complexes

$$\begin{align*}0\longrightarrow {\mathbb F}\otimes_SR\longrightarrow {\mathbb D}\otimes_SR\longrightarrow {\mathbb G}[-1]\otimes_SR\longrightarrow 0, \end{align*}$$

where ${\mathbb G}[-1]$ is the complex ${\mathbb G}$ with the homological shift $-1$ . In other words, ${\mathbb G}[-1]_i=G_{i-1}$ for all i. This short exact sequence gives rise to the long exact sequence

$$\begin{align*}\cdots \rightarrow H_2({\mathbb F}\otimes_SR)\longrightarrow H_2({\mathbb D}\otimes_SR)\longrightarrow H_1({\mathbb G}\otimes_SR)\rightarrow H_1({\mathbb F}\otimes_SR)\rightarrow \cdots. \end{align*}$$

The map $H_1({\mathbb G}\otimes _SR)\rightarrow H_1({\mathbb F}\otimes _SR)$ is a connecting homomorphism in the long exact sequence and is induced by $\alpha \otimes _S R$ . Since $\alpha $ is split injective, it follows that $\alpha \otimes _S R$ is split exact as well. Therefore, $H_1({\mathbb G}\otimes _SR)\rightarrow H_1({\mathbb F}\otimes _SR)$ is injective. This implies that $H_2({\mathbb F}\otimes _SR)\rightarrow H_2({\mathbb D}\otimes _S R)$ is surjective. Since ${\mathbb D}$ is a graded free S-resolution of R, it follows that $H_2({\mathbb D}\otimes _S R)\cong \operatorname {Tor}^S_2(R,R)$ as graded R-modules. Hence, we have a graded epimorphism $H_2({\mathbb F}\otimes _SR)\rightarrow \operatorname {Tor}^S_2(R,R)$ .

The graded minimal free S-resolution of R is given as $0\rightarrow F_2/G_2\rightarrow F_1/G_1\oplus S(-a)\rightarrow S\rightarrow 0$ , and hence it has the form

$$\begin{align*}\overline{{\mathbb D}}\colon 0\rightarrow S(-k-1)^a\rightarrow S(-k)^a\oplus S(-a)\rightarrow S\rightarrow 0. \end{align*}$$

This yields the exact sequence $0\rightarrow \operatorname {Tor}^S_2(R,R)\rightarrow R(-k-1)^a\rightarrow R(-k)^a\oplus R(-a)$ , and by using the result of Vasconcelos, we need to show that $\operatorname {Tor}^S_2(R,R)_{d+k+1}=0$ for $d<a-1$ . For this, it suffices to show that $H_2({\mathbb F}\otimes _SR)_{d+k+1}=0$ for $d<a-1$ .

Note that $H_2({\mathbb F}\otimes _S R)$ is the kernel of $R(-k-1)^k\rightarrow R(-k)^{k+1}$ . Hence, we get the exact sequence $ H_2({\mathbb F}\otimes _S R)_{d+k+1}\rightarrow R_d^k\rightarrow R_{d+1}^k. $ Since $d<a-1$ , it follows that $R_d=S_d$ and $R_{d+1}=S_{d+1}$ , so that $0\rightarrow H_2({\mathbb F}\otimes _S R)_{d+k+1}\rightarrow S_d^k\rightarrow S_{d+1}^k$ is exact. Since $S(-k-1)^k\rightarrow S(-k)^{k+1}$ is injective, we deduce that $H_2({\mathbb F}\otimes _S R)_{d+k+1}=0$ , as desired.

Now, we deal with the case $k>b$ . If $k=a+b-2$ , then R is even a Teter ring, and we are done. Thus, we may now assume that $b<k\leq a+b-1$ . Then we have the following exact sequence:

$$\begin{align*}0\rightarrow (S/(f,{\mathfrak m}^k):(g))(-b)\rightarrow S/(f,{\mathfrak m}^k)\rightarrow R\rightarrow 0. \end{align*}$$

Let $h\in (f,{\mathfrak m}^k):(g)$ be a homogeneous polynomial with $\deg (h)=c$ . Then $hg\in (f,{\mathfrak m}^k)$ . If $c+b<k$ , then $hg=h'f$ for some $h'\in S$ . Since $f,g$ is a regular sequence, it follows then that f divides h, and this implies that $h\in (f,{\mathfrak m}^k)$ . On the other hand, if $c+b\geq k$ , then $hg\in {\mathfrak m}^k$ . Therefore, $h\in {\mathfrak m}^{k-b}$ , and hence $(f,{\mathfrak m}^k):(g)=(f,{\mathfrak m}^{k-b})$ . Since $k\leq a+b-1$ , it follows that $(f,{\mathfrak m}^k):(g)=(f,{\mathfrak m}^{k-b})={\mathfrak m}^{k-b}$ , and we obtain the exact sequence

$$\begin{align*}0\rightarrow (S/({\mathfrak m}^{{k-b}}))(-b)\rightarrow S/(f,{\mathfrak m}^k)\rightarrow R\rightarrow 0. \end{align*}$$

The map $(S/({\mathfrak m}^{k-b}))(-b)\stackrel {g}{\rightarrow } S/(f,{\mathfrak m}^k)$ can be extended to a complex homomorphism ${\mathbb H}\rightarrow {\mathbb D}$ with $\alpha \colon H_1\rightarrow D_1$ and $\beta \colon H_2\rightarrow D_2$ split injective. Here,

$$ \begin{align*}{\mathbb H}: 0\rightarrow (-k-1)^{k-b} \rightarrow S(-k)^{k-b+1}\rightarrow S(-1)\rightarrow 0\end{align*} $$

is the free S-resolution of $(S/({\mathfrak m}^{k-b}))(-b)$ and ${\mathbb D}$ is the mapping cone from before. Composing ${\mathbb H}\rightarrow {\mathbb D}$ with the complex homomorphism ${\mathbb D}\rightarrow \overline {{\mathbb D}}$ onto the graded minimal free S-resolution of $S/(f,{\mathfrak m}^k)$ , we obtain a commutative diagram

with $\alpha '$ and $\beta '$ split injective. This yields a minimal free S-resolution of R which is of the form

$$\begin{align*}0\rightarrow S(-k-1)^{a+b-k}\rightarrow S(-k)^{a+b-1-k}\oplus S(-b)\oplus S(-a)\rightarrow S\rightarrow 0. \end{align*}$$

Let ${\mathbb E}$ be the mapping cone of the complex homomorphism ${\mathbb H}\rightarrow {\mathbb D}$ . Then we obtain the short exact sequence of complexes

$$\begin{align*}0\rightarrow {\mathbb D}\otimes_SR\rightarrow {\mathbb E}\otimes_SR\rightarrow {\mathbb H}[-1]\otimes_SR\rightarrow 0, \end{align*}$$

which gives rise to the long exact sequence

$$\begin{align*}\cdots \rightarrow H_2({\mathbb D}\otimes_SR)\longrightarrow H_2({\mathbb E}\otimes_SR)\longrightarrow H_1({\mathbb H}\otimes_SR)\rightarrow H_1({\mathbb D}\otimes_SR)\rightarrow \cdots. \end{align*}$$

The map $H_1({\mathbb H}\otimes _SR)\rightarrow H_1({\mathbb D}\otimes _SR)$ is a connecting homomorphism in the long exact sequence which is induced by $\alpha \otimes _S R$ . Since $\alpha $ is split injective, it follows that $\alpha \otimes _S R$ is split exact as well. Therefore, $H_1({\mathbb H}\otimes _SR)\rightarrow H_1({\mathbb D}\otimes _S R)$ is injective. This implies that $H_2({\mathbb D}\otimes _SR)\rightarrow H_2({\mathbb E}\otimes _S R)$ is surjective. Since ${\mathbb E}\otimes _SR\cong R(-k-1)^{a+b-k}$ , we need to show that $H_2({\mathbb E}\otimes _SR)_{d+k+1}=0$ for $d<a+b-1-k$ . We have the graded epimorphisms $H_2({\mathbb D}\otimes _SR)\rightarrow H_2({\mathbb E}\otimes _S R)$ and $H_2({\mathbb F}\otimes _SR)\rightarrow H_2({\mathbb D}\otimes _S R)$ . Thus, it suffices to show that $H_2({\mathbb F}\otimes _SR)_{d+k+1}=0$ for $d<a+b-1-k$ . Note that $R(-k-1)_{d+k+1}=R_d=S_d$ and that $R(-k)_{d+k+1}=R_d=S_{d+1}$ , since $d<a+b-k-1<a-1$ . The last equation follows since $k>b$ . Thus, we have an exact sequence $0\rightarrow H_2({\mathbb F}\otimes _SR)_{d+k+1}\rightarrow S^k_d\rightarrow S^{k+1}_{d+1}$ . It follows that $ H_2({\mathbb F}\otimes _SR)_{d+k+1}=0$ , because $S(-k-1)^k\rightarrow S(-k)^{k+1}$ is exact.

Proof. Let ${\mathcal I}$ and ${\mathcal J}$ be companions. Then there is an isomorphism $\varphi : {\mathcal J}^* \rightarrow {\mathcal I}$ . By Theorem 4.1, there is a dual map $\varphi ^*: {\mathcal I}^* \rightarrow {\mathcal J}$ with $\deg (\varphi )=\deg (\varphi ^*)$ . Then we define a map $\psi : {\mathcal I}^*\cup {\mathcal J}^* \rightarrow {\mathcal I}\cup {\mathcal J}$ by setting

$$\begin{align*}\psi(u^*)= \begin{cases} \varphi(u^*), & \text{if}\ u^* \in {\mathcal J}^*, \\ \varphi^*(u^*), & \text{if}\ u^* \in {\mathcal I}^*. \end{cases} \end{align*}$$

Since $\deg (\varphi )=\deg (\varphi ^*)$ , these maps agree on $I^*\cap J^*$ and therefore $\psi $ is well defined. Moreover, $\psi $ is an isomorphism and $\deg (\psi )=\deg (\varphi )=\deg (\varphi ^*)$ .

Proof. $\Rightarrow :$ Let $\varphi : {\mathcal J}^* \rightarrow {\mathcal I}$ be an isomorphism, and let m be the monomial with $\deg (m)=\deg (\varphi )$ . Then $\varphi $ is given by $\varphi (v^*)=m/v$ for all $v\in {\mathcal J}$ . In particular, this map defines a bijection $\operatorname {Gen}({\mathcal J}^*)\rightarrow \operatorname {Gen} ({\mathcal I})$ . Note that $\operatorname {Gen}({\mathcal J}^*)=(\operatorname {Soc}({\mathcal J}))^*=\{d^*_{1},\ldots , d^*_{k'}\}$ . This implies $k=k'$ , and, after a suitable relabeling, for all $1\le i\le k$ , we have $\varphi (d^*_i)=m/d_i=a_i$ , that is to say, $a_id_i=m$ . A similar argument can be applied to the socles to conclude $l=l'$ and $b_jc_j=m$ for all $1\le j\le l$ .

$\Leftarrow :$ Let $k=k'$ , let $l=l'$ , and let $a_id_i=b_jc_j=m$ for $1\le i\le k$ and $1\le j\le l$ . We show that the map $\varphi $ with $\deg (\varphi )=\deg (m)$ defines an isomorphism ${\mathcal J}^*\rightarrow {\mathcal I}$ . We have $v^*\in {\mathcal J}^*$ if and only if there exist a socle of $c^*_j$ of ${\mathcal J}^*$ and a generator $d^*_i$ of ${\mathcal J}^*$ such that $c^*_j\preceq v^*\preceq d^*_i$ , or, in other words, $c_j|v$ and $v|d_i$ . Then $v|m$ and

$$ \begin{align*}(m/d_i)|(m/v) \text{ and } (m/v)|(m/c_j)\Leftrightarrow a_i|(m/v) \text{ and } (m/v)|b_j\Leftrightarrow b_j \preceq m/v\preceq a_i,\end{align*} $$

which is equivalent to saying that $m/v\in {\mathcal I}$ . This proves that the map is well defined on all of ${\mathcal J}^*$ . Clearly, $\varphi $ is injective. For surjectivity, one applies the argument above in the opposite direction: if $u\in {\mathcal I}$ , then $u|m$ and $v^*:=(m/u)^*\in {\mathcal J}^*$ with $\varphi (v^*)=m/v=u$ .

Proof. One claims that the ideal ${\mathcal I}_k \subseteq {\mathcal P}$ is symmetric. Since $\operatorname {Soc}({\mathcal I}_k)$ consists of those monomials $x_1^{a_1 - b_1}x_2^{a_2 - b_2}\ldots x_n^{a_n - b_n}$ with $\sum _{i=1}^n b_i = k$ , it follows from Theorem 4.8 that ${\mathcal I}_k \subseteq {\mathcal P}$ is symmetric. Hence, ${\mathcal I}_k \subseteq \operatorname {tr}({\mathcal P}^*)$ .

Now, it is shown that every $\tau $ -ideal of ${\mathcal P}$ is contained in ${\mathcal I}_k$ . Let ${\mathcal I}$ be a $\tau $ -ideal of ${\mathcal P}$ with ${\mathcal I} \nsubseteq {\mathcal I}_k$ . Let u be a monomial of $\operatorname {Gen}({\mathcal I})$ of the smallest degree. Then $d = \deg u < a_1$ . Let ${\mathcal J}$ be a companion of ${\mathcal I}$ . Since each monomial belonging to $\operatorname {Soc}({\mathcal J})$ is of degree $\sum _{i=1}^{n} a_i - k$ , it follows that each monomial of $\operatorname {Gen}({\mathcal I})$ is of degree d. Since $d < a_1$ , each monomial of $\operatorname {Gen}({\mathcal I})$ possesses exactly n lower neighbors. It then follows that each monomial belonging to $\operatorname {Soc}({\mathcal J})$ possesses exactly n upper neighbors in ${\mathcal J}$ . In other words, if $u = x_1^{c_1}x_2^{c_2}\ldots x_n^{c_n} \in \operatorname {Soc}({\mathcal J})$ , then $c_i> 0$ for all i and $u/x_i \in {\mathcal J}$ . Let $<_{\mathrm {lex}}$ denote the lexicographic order induced by $x_1 < x_2 < \cdots < x_n$ . Let $w = x_1^{c_1}x_2^{c_2}\ldots x_n^{c_n}$ be the monomial belonging to $\operatorname {Soc}({\mathcal J})$ , which is the largest with respect to $<_{\mathrm {lex}}$ . Let $1 \leq i_0 \leq n$ be the largest integer for which $c_{i_0} < a_{i_0}$ . If $i_0> 1$ and $w_0 = x_{i_0}(w/x_{i_0-1})$ , then $w <_{\mathrm {lex}} w_0$ and $w_0 \in {\mathcal J}$ , a contradiction. Hence, $i_0 = 1$ and $w = x_1^{a_1-k}x_2^{a_2}\ldots x_n^{a_n}$ . Then $x_1^{a_1 - k + \sum _{i=2}^{n} e_i}(w/x_2^{e_2} \cdots x_n^{e_n}) \in {\mathcal J}$ , where $\sum _{i=2}^{n} e_i \leq k$ . Hence, $\operatorname {Soc}({\mathcal J}) = \operatorname {Soc}({\mathcal I}_k)$ . However, since $|\operatorname {Gen}({\mathcal J})| < |\operatorname {Soc}({\mathcal I}_k)|$ , it follows that ${\mathcal J}$ cannot be a companion of ${\mathcal I}$ , as desired.

Proof. If $k \leq n - k$ , then ${\mathcal I}_{k_0} \subseteq {\mathcal P}$ is symmetric. In fact, since $\operatorname {Gen}({\mathcal I}_{k_0})$ consists of square-free monomials in $x_1, \ldots , x_n$ of degree $k_0$ and $\operatorname {Soc}({\mathcal I}_{k_0})$ does of those of degree $n - k_0$ , Theorem 4.8 says that ${\mathcal I}_{k_0}$ is symmetric. If $k> n - k$ , then $\operatorname {Gen}({\mathcal I}_{k_0}) = \operatorname {Soc}({\mathcal I}_{k_0})$ . Thus, for each $1 \leq k \leq n - 1$ , one has ${\mathcal I}_{k_0} \subseteq \operatorname {tr}({\mathcal P}^*)$ .

Let ${\mathcal I}$ be a $\tau $ -ideal of ${\mathcal P}$ with ${\mathcal I} \nsubseteq {\mathcal I}_{k_0}$ . Let u be a monomial of $\operatorname {Gen}({\mathcal I})$ of the smallest degree. Then $d = \deg u < k_0$ . Let ${\mathcal J}$ be a companion of ${\mathcal I}$ . Since each monomial belonging to $\operatorname {Soc}({\mathcal J})$ is of degree $n - k$ , it follows that each monomial of $\operatorname {Gen}({\mathcal I})$ is of degree d. Hence, each monomial of $\operatorname {Gen}({\mathcal I})$ possesses exactly $n - d$ lower neighbors. It then follows that each monomial belonging to $\operatorname {Soc}({\mathcal J})$ possesses exactly $n - d$ upper neighbors in ${\mathcal J}$ . On the other hand, each monomial belonging to $\operatorname {Soc}({\mathcal I}_{k_0})$ possesses exactly $n - k$ upper neighbors. Hence, $n - k \geq n - d$ . Thus, $k \leq d < k_0$ , which contradicts $k_0 = \min \{k, n - k\}$ . Hence, every $\tau $ -ideal is contained in ${\mathcal I}_{k_0}$ . Thus, $\operatorname {tr}({{\mathcal P}^*}) = {\mathcal I}_{k_0}$ , as required. In particular, ${\mathcal P}$ is of Teter type if $k \leq n - k$ .

Now, suppose that $k> n - k$ . Then $\operatorname {tr}({\mathcal P}^*) = {\mathcal I}_{n-k}$ , where $\operatorname {Gen}({\mathcal I}_{n-k})$ consists of square-free monomials in $x_1, \dots , x_n$ of degree $n - k$ . One has $\operatorname {Gen}({\mathcal I}_{n-k}) = \operatorname {Soc}({\mathcal I}_{n-k})$ . If ${\mathcal P}$ is of Teter type, then one can find a bijection $\delta : \operatorname {Gen}({\mathcal I}_{n-k}) \rightarrow \operatorname {Gen}({\mathcal I}_{n-k})$ for which there is $c = (c_1, \ldots , c_n)\in {\mathbb Z}^n$ with $\deg (u) + \deg (\delta (u)) = c$ for all $u \in \operatorname {Gen}({\mathcal I}_{n-k})$ . In particular, $c_i \geq 1$ for all i. However, since $\sum _{i=1}^{n} c_i = 2(n - k) < n$ , it follows that $c_j = 0$ for some $1 \leq j \leq n$ . This contradiction shows that ${\mathcal P}$ cannot be of Teter type.

Proof. Let $J=(x_1^{a_1},\ldots ,x_n^{a_n})$ , $L=J:w=(x_1^{a_1-b_1},\ldots ,x_n^{a_n-b_n})$ , and ${\mathbf b}=(b_1,\ldots ,b_n)$ . Then we have the multigraded short exact sequence

$$\begin{align*}0\rightarrow (S/L)(-{\mathbf b})\stackrel{\varphi}{\longrightarrow}S/J\rightarrow S/I\rightarrow 0, \end{align*}$$

where the map $\varphi $ is multiplication by w. Let ${\mathbb F}=K(x_1^{a_1},\ldots ,x_n^{a_n};S)$ be the Koszul complex attached to the regular sequence $x_1^{a_1},\ldots ,x_n^{a_n}$ , which is indeed a minimal graded free S-resolution of $S/J$ , and let ${\mathbb G}=K(x_1^{a_1-b_1},\ldots ,x_n^{a_n-b_n};S)$ , which is a minimal graded free S-resolution of $S/L$ . Then the map $\varphi $ can be lifted to a multigraded complex homomorphism $\{\varphi _i:{\mathbb G}(-{\mathbf b})\rightarrow {\mathbb F}\}$ , where $\varphi _i(e_{j_1\cdots j_i})=(w/x_{j_1}^{b_{j_1}}\cdots x_{j_i}^{b_{j_i}})e^{\prime }_{j_1 \cdots j_i}$ for any basis element $e_{j_1\cdots j_i}$ of $G_i(-{\mathbf b})$ . So we have the commutative diagram

Let ${\mathbb D}$ be the mapping cone of ${\mathbb G}(-{\mathbf b})\rightarrow {\mathbb F}$ . Then ${\mathbb D}$ is a graded free S-resolution of R with the differential maps $d_i:F_i\oplus G_{i-1}(-{\mathbf b})\rightarrow F_{i-1}\oplus G_{i-2}(-{\mathbf b})$ defined as

$$ \begin{align*}d_i(f,g)=(\varphi_{i-1}(g)+\partial^{\prime}_i(f),-\partial_{i-1}(g))\end{align*} $$

for $1\leq i\leq n+1$ . For any set $B=\{i_1,\ldots ,i_s\}\subseteq [n]$ , let $e_B=e_{i_1\cdots i_s}$ and $e^{\prime }_B=e^{\prime }_{i_1\cdots i_s}$ . Then $\{e_{[n]\setminus i}:\ 1\leq i\leq n\}$ is a basis for $G_{n-1}(-{\mathbf b})$ . Let

$$ \begin{align*}\overline{d}_n:G_{n-1}(-{\mathbf b})\rightarrow F_{n-1}\oplus G_{n-2}(-{\mathbf b})\end{align*} $$

be the S-module homomorphism with $\overline {d}_n(e_{[n]\setminus i})=d_n(e_{[n]\setminus i})=(x_i^{b_i}e^{\prime }_{[n]\setminus i},-\partial _{n-1}(e_{[n]\setminus i}))$ for any $1\leq i\leq n$ . Then, clearly, $\overline {d}_n$ is an injective map. Moreover, from the equalities

$$ \begin{align*}d_n(e^{\prime}_{[n]})=\partial^{\prime}_n(e^{\prime}_{[n]})=\partial^{\prime}_n \varphi_n(e_{[n]})=\varphi_{n-1} \partial_n(e_{[n]})=d_n\partial_n(e_{[n]}),\end{align*} $$

we obtain $d_n(F_n)\subseteq d_n(G_{n-1}(-{\mathbf b}))=\overline {d}_n(G_{n-1}(-{\mathbf b}))$ . This implies that $\operatorname {Im} \overline {d}_n=\operatorname {Im} d_n$ . Therefore,

is a free S-resolution of $S/I$ .

Consider the complex $\overline {{\mathbb D}}\otimes _SR$ with the differential maps $\sigma _i$ . We have $\overline {D}_{n}=\bigoplus _{i=1}^n Se_{[n]\setminus i}$ . Hence, $\overline {D}_{n}\otimes _SR=\bigoplus _{i=1}^n Re_{[n]\setminus i}$ and $\sigma _n(e_{[n]\setminus i})=(x_{i}^{b_i}e^{\prime }_{[n]\setminus i},-\partial _{n-1}(e_{[n]\setminus i}))$ for $1 \leq i \leq n$ . Here, for simplicity, the residue class of the monomial $x_i^{b_i}$ in R is denoted again by $x_i^{b_i}$ . Let C be the kernel of $\sigma _n:\overline {D}_n\otimes _SR\rightarrow \overline {D}_{n-1}\otimes _SR$ . Then, by Theorem 3.6, $\operatorname {tr}(\omega _R)$ is the ideal in R generated by the entries of a generating set of C. For any $1\leq i\leq n$ , we have $\sigma _n((w/x_i^{b_i})e_{[n]\setminus i})=0$ . Indeed, for any $j\neq i$ , the coefficient of $e_{[n]\setminus \{i,j\}}$ in $\partial _{n-1}(e_{[n]\setminus i})$ is $\pm x_{j}^{a_j-b_j}$ and $(w/x_i^{b_i})x_{j}^{a_j-b_j}=0_R$ , since $(w/x_i^{b_i})x_{j}^{a_j-b_j}$ as a monomial in S is divisible by $x_j^{a_i}$ . So $(w/x_i^{b_i})e_{[n]\setminus i}\in C$ for any $1\leq i\leq n$ . Now, consider the element $z=\partial _n(e_{[n]})=\sum _{i=1}^n (-1)^i x_i^{a_i-b_i} e_{[n]\setminus i}\in \overline {D}_{n}\otimes _SR$ . We have $\sigma _n(z)=(\sum _{i=1}^n (-1)^i x_i^{a_i}e^{\prime }_{[n]\setminus i},-\partial _{n-1}\partial _n(z))=0$ , which implies that $z\in C$ . Without loss of generality, assume that $b_i>0$ for $1\leq i\leq r$ and $b_{r+1}=\cdots =b_n=0$ . Then, by Theorem 3.6,

$$ \begin{align*}(x_i^{a_i-b_i}:\ 1\leq i\leq r)+(w/x_i^{b_i}:\ 1\leq i\leq r)\subseteq \operatorname{tr}(\omega_R).\end{align*} $$

Now, consider an arbitrary multigraded element $h=\sum _{i=1}^n s_i u_ie_{[n]\setminus i}\in C$ , where $u_i$ ’s are monomials and $s_i\in K$ . Then $\sum _{i=1}^n s_i x_i^{b_i}u_ie_{[n]\setminus i}=0$ and so $x_i^{b_i}u_i\in I$ for any $1\leq i\leq n$ . This means either $x_i^{a_i-b_i}|u_i$ or $(w/x_i^{b_i})|u_i$ . Therefore,

$$ \begin{align*}\operatorname{tr}(\omega_R)\subseteq (x_i^{a_i-b_i}:\ 1\leq i\leq r)+(w/x_i^{b_i}:\ 1\leq i\leq r).\end{align*} $$

Thus, the equality holds.

Proof. Without loss of generality, we can assume that $a_i\ge 2$ for all $i=1,\ldots ,n$ . We can also assume that $b_1b_2\cdots b_r\not =0$ and $b_{r+1}=\cdots =b_n=0$ for some $2\le r\le n$ . Let $w=x_1^{b_1}\cdots x_n^{b_n}$ , and let ${\mathcal P}$ be the divisor poset of R. From Theorem 5.1, we know that

$$ \begin{align*}\operatorname{tr}({\mathcal P}^*)=(x_1^{a_1-b_1},\ldots, x_n^{a_n-b_n},\frac{w}{x_1^{b_1}},\ldots,\frac{w}{x_n^{b_n}})=(x_1^{a_1-b_1},\ldots, x_n^{a_r-b_r},\frac{w}{x_1^{b_1}},\ldots,\frac{w}{x_r^{b_r}}).\end{align*} $$

Let $v=x_1^{a_1-1}x_2^{a_2-1}\cdots x_n^{a_n-1}$ . Then one can check that $\operatorname {Soc}({\mathcal P})=\{\frac {v}{x_1^{a_1-b_1}}, \ldots , \frac {v}{x_r^{a_r-b_r}}\}$ . We consider several cases:

(1) There exist $j\not =j'$ such that $2b_j\ge a_j$ and $2b_{j'}\ge a_{j'}$ . Without loss of generality, we may assume that $2b_1\ge a_1$ and $2b_2 \ge a_2$ . Then $\frac {w}{x_1^{b_1}}$ is divisible by $x_2^{b_2}$ and therefore by $x_2^{a_2-b_2}$ . Similarly, $\frac {w}{x_2^{b_2}}$ is divisible by $x_1^{b_1}$ , and therefore by $x_1^{a_1-b_1}$ . Each $\frac {w}{x_i^{b_i}}$ , $3\le i\le r$ , is divisible by $x_1^{b_1}x_2^{b_2}$ and thus by $x_1^{a_1-b_1}x_2^{a_2-b_2}$ . In other words, $\operatorname {tr}({\mathcal P}^*)=(x_1^{a_1-b_1},\ldots , x_r^{a_r-b_r})$ . We have $\operatorname {Gen}(\operatorname {tr}({\mathcal P}^*))=\{x_1^{a_1-b_1},\ldots , x_r^{a_r-b_r}\}$ and $\operatorname {Soc} (\operatorname {tr}({\mathcal P}^*))=\{\frac {v}{x_1^{a_1-b_1}}, \ldots , \frac {v}{x_r^{a_r-b_r}}\}$ . By Theorem 4.8, this is a symmetric ideal. In fact, the desired isomorphism has multidegree v. Therefore, R is of Teter type.

(2) There exists at most one j such that $2b_j\ge a_j$ . Without loss of generality, we assume that $2b_2<a_2, \ldots , 2b_r<a_r$ and no conditions on $b_1$ .

(a) $r>2$ . Then, among monomials $x_1^{a_1-b_1},\ldots , x_r^{a_r-b_r},\frac {w}{x_1^{b_1}}:=w_1$ , no pair of monomials divides each other. Indeed, $\operatorname {supp}(w_1)=\{x_2,x_3,\ldots , x_r\}$ , which contains at least two variables, which implies that $w_1$ does not divide any other monomial. On the other hand, the $x_i$ exponent of $w_1$ is $b_i<a_i-b_i$ for all $2\le i\le r$ , and thus no other monomial divides $w_1$ . Therefore, $|\operatorname {Gen}(\operatorname {tr}({\mathcal P}^*))|>|\operatorname {Soc}(\operatorname {tr}({\mathcal P}^*))|$ , which by Theorem 4.8 implies that $\operatorname {tr}({\mathcal P}^*)$ is not symmetric and thus R is not of Teter type.

(b) $r=2$ . In this case, $I=(x_1^{a_1},\ldots , x_n^{a_n}, x_1^{b_1}x_2^{b_2})$ with $2b_2<a_2$ . Let $e:=\min \{b_1,a_1-b_1\}$ . We have

$$ \begin{align*}\operatorname{tr}({\mathcal P}^*)=(x_1^{a_1-b_1}, x_2^{a_2-b_2}, x_1^{b_1}, x_2^{b_2})=(x_1^e, x_2^{b_2})=:(t_1,t_2)\end{align*} $$

and

$$ \begin{align*}\operatorname{Soc}(\operatorname{tr}({\mathcal P}^*))=\{x_1^{b_1-1}x_2^{a_2-1}, x_1^{a_1-1}x_2^{b_2-1}\}=:\{s_1,s_2\}.\end{align*} $$

The only possible isomorphisms can be given by $\varphi _1: s_1 \mapsto t_1, s_2 \mapsto t_2$ and $\varphi _2: s_1 \mapsto t_2, s_2 \mapsto t_1$ . In order for $\varphi _1$ to be an isomorphism, we need to have $(e+b_1-1, a_2-1)=(a_1-1,2b_2-1)$ , which is impossible since $2b_2<a_2$ . Similarly, in order for $\varphi _2$ to be an isomorphism, we need to have $(b_1-1, a_2+b_2-1)=(e+a_1-1,b_2-1)$ , which is also impossible since $a_2\not =0$ .

Proof. Let, say, $w_0 = x_1x_2 \cdots x_{i_0}$ , and set $v_0 = x_1x_2 \cdots x_n$ . Then $\operatorname {Soc}({\mathcal P}^*)$ consists of the monomials $v_0/x_1, v_0/x_2, \ldots , v_0/x_{i_0}$ . Theorem 4.8 guarantees that the poset ideal ${\mathcal I}_0$ with $\operatorname {Gen}({\mathcal I}_0) = \{x_1, x_2, \ldots , x_{i_0}\}$ is symmetric. Hence, ${\mathcal I}_0 \subset \operatorname {tr}({\mathcal P}^*)$ . Let ${\mathcal I}$ be a $\tau $ -ideal of ${\mathcal P}$ and suppose that ${\mathcal I} \nsubseteq {\mathcal I}_0$ . Let, say, $u = x_{i_0+1}x_{i_0+2} \cdots x_{i_0+d} \in \operatorname {Gen}({\mathcal I})$ . Let ${\mathcal J}$ be a companion of ${\mathcal I}$ , and $v \in {\mathcal J}$ is accompanied by u. Then there is $1 \leq j_0 \leq i_0$ with $v = v_0/x_{j_0}$ . Since u divides each of the monomials $v_0/x_1, v_0/x_2, \ldots , v_0/x_{i_0}$ , it follows that each of the monomials $v_0/x_1u, v_0/x_2u, \ldots , v_0/x_{i_0}u$ divides $v = v_0/x_{j_0}$ . Thus, $x_iu/x_{j_0} \in {\mathcal P}$ for each $1 \leq i \leq i_0$ . However, $x_iu/x_{j_0} \in {\mathcal P}$ only for $i = j_0$ . Hence, ${\mathcal I}$ cannot be a $\tau $ -ideal. Thus, ${\mathcal I}_0 = \operatorname {tr}({\mathcal P}^*)$ , as desired.