A Sufficient condition for $F$-purity

First assume $R$ is a homomorphic image of a Gorenstein local ring $S$ and write $R=S/J$. Let $x$ be a nonzerodivisor on $R$. The long exact sequence for $\operatorname{Ext}$ yields:

But $\operatorname{Ext}_{S}^{\operatorname{{ht}}J}(R/xR,S)=0$ because the first non-vanishing $\operatorname{Ext}$ occurs at $\operatorname{depth}_{(J+x)}S=\operatorname{{ht}}(J+x)=\operatorname{{ht}}J+1$. So $x$ is a nonzerodivisor on $\operatorname{Ext}_{S}^{\operatorname{{ht}}J}(R,S)\cong\operatorname{Ext}_{S}^{\dim S-\dim R}(R,S)\cong\omega_{R}$.

In the general case, for a nonzerodivisor $x$ in $R$, if we have $0\rightarrow N\rightarrow\omega_{R}\xrightarrow{x}\omega_{R}$, we may complete to get $0\rightarrow\widehat{N}\rightarrow\widehat{\omega_{R}}\xrightarrow{x}\widehat{\omega_{R}}$. It is easy to see that $\widehat{\omega_{R}}$ is a canonical module for $\widehat{R}$ and $x$ is a nonzerodivisor on $\widehat{R}$. Now since $\widehat{R}$ is a homomorphic image of a Gorenstein local ring, we have $\widehat{N}=0$ and hence $N=0$. ∎

Since $R$ is equidimensional, we know that $\omega_{R_{P}}\cong(\omega_{R})_{P}$ for every prime ideal $P$ of $R$ by Proposition 2.3. Let $W$ be the multiplicative system of $R$ consisting of all nonzerodivisors and let $\Lambda$ be the set of minimal primes of $R$. Since $R$ is equidimensional and unmixed, $W$ is simply the complement of the union of the minimal primes of $R$.

$(1)\Rightarrow(2)$: If we have $I\cong\omega_{R}$, then for every $P\in\Lambda$, $\omega_{R_{P}}\cong(\omega_{R})_{P}\cong IR_{P}\subseteq R_{P}$. But $R_{P}$ is an Artinian local ring, $l(\omega_{R_{P}})=l(R_{P})$, so we must have $\omega_{R_{P}}\cong R_{P}$. Hence $R_{P}$ is Gorenstein for every $P\in\Lambda$, that is, $R$ is generically Gorenstein.

$(2)\Rightarrow(1)$: Since $R$ is generically Gorenstein, we know that for $P\in\Lambda$, $\omega_{R_{P}}\cong R_{P}$. Now we have:

Therefore we have an isomorphism $W^{-1}\omega_{R}\cong W^{-1}R$. The restriction of the isomorphism to $\omega_{R}$ then yields an injection $j$: $\omega_{R}\hookrightarrow W^{-1}R$ because elements in $W$ are nonzero divisors on $\omega_{R}$ by Lemma 2.2. The images of a finite set of generators of $\omega_{R}$ can be written as $r_{i}/w_{i}$. Let $w=\prod w_{i}$, we have $wj$: $\omega_{R}\hookrightarrow R$ is an injection. So $\omega_{R}$ is isomorphic to an ideal $I\subseteq R$.

Finally, when these equivalent conditions hold, we know that $W^{-1}I\cong\prod_{P\in\Lambda}R_{P}$ is free. So $W^{-1}I$ contains a nonzerodivisor. But whether $I$ contains a nonzerodivisor is unaffected by localization at $W$. So $I$ contains a nonzerodivisor. ∎

By Proposition 2.4, $I$ contains a nonzerodivisor, so its height is at least one. Now we choose a height $h$ associated prime $P$ of $I$ with $h\geq 2$. We localize at $P$, $PR_{P}$ becomes an associated prime of $IR_{P}$. In particular, $R_{P}/IR_{P}$ has depth $0$ so $H^{0}_{PR_{P}}(R_{P}/IR_{P})\neq 0$.

However, by Proposition 2.3, $IR_{P}$ is a canonical ideal of $R_{P}$, which has dimension $h\geq 2$. Now the long exact sequence of local cohomology gives

We have $\operatorname{depth}R_{P}\geq 1$ ($I$ contains a nonzerodivisor) and $\operatorname{depth}IR_{P}\geq 2$ (the canonical module is always $S_{2}$ by Lemma 2.5). Hence $H^{0}_{PR_{P}}(R_{P})=H^{1}_{PR_{P}}(IR_{P})=0$. The above sequence thus implies $H^{0}_{PR_{P}}(R_{P}/IR_{P})=0$ which is a contradiction.

Hence we have shown that every associated prime of $I$ has height one. Since $R$ is equidimensional, this proves $I$ has height one and $R/I$ is equidimensional and unmixed. ∎

By Proposition 2.7, to show $H^{d}_{\mathfrak{m}}(I)\to H^{d}_{\mathfrak{m}}(R)$ is not injective, it suffices to show

is not surjective.

It suffices to show (3.3.1) is not surjective after we localize at a height one minimal prime $P$ of $I$. Since $I$ contains a nonzerodivisor and $P$ is a height one minimal prime of $I$, it is straightforward to see that $R_{P}$ is a one-dimensional Cohen-Macaulay ring with $IR_{P}$ a $PR_{P}$-primary ideal. And by Proposition 2.3, $(\omega_{R})_{P}$ is a canonical module of $R_{P}$. Hence to show $\operatorname{Hom}_{R}(R,\omega_{R})_{P}\to\operatorname{Hom}_{R}(I,\omega_{R})_{P}$ is not surjective, we can apply Proposition 2.7 (taking Matlis dual of $E_{R_{P}}$) and we see it is enough to prove that

is not injective. But this is obvious because we know from the long exact sequence that the kernel is $H^{0}_{PR_{P}}(R_{P}/IR_{P})$, which is nonzero because $I$ is $PR_{P}$-primary. ∎

First we note that $I$ is a height one ideal by Proposition 2.6. In particular we know that $\dim R/I<\dim R=d$. We have a short exact sequence:

Moreover, if we endow $I$ with an $R\{F\}$-module structure induced from $R$, then the above is also an exact sequence of $R\{F\}$-modules. Hence the tail of the long exact sequence of local cohomology gives an exact sequence of $R\{F\}$-modules, that is, a commutative diagram (we have $0$ on the right because $\dim R/I<d$):

where the vertical maps denote the Frobenius actions on each module.

Since $R$ is equidimensional and $S_{2}$, we know that $H_{\mathfrak{m}}^{d}(I)\cong H_{\mathfrak{m}}^{d}(\omega_{R})\cong\operatorname{Hom}_{R}(\omega_{R},\omega_{R})^{\vee}\cong R^{\vee}\cong E_{R}$ by Proposition 2.7 and Lemma 2.5. We want to show that, under the hypothesis, the Frobenius action on $H_{\mathfrak{m}}^{d}(I)\cong E_{R}$ is injective. Then we will be done by Theorem 3.2.

Suppose the Frobenius action on $H_{\mathfrak{m}}^{d}(I)$ is not injective, then the nonzero socle element $x\in H_{\mathfrak{m}}^{d}(I)\cong E_{R}$ is in the kernel, i.e., $F(x)=0$. From Proposition 2.6 and Lemma 3.3 we know that $\varphi_{3}$ is not injective. So we also have $\varphi_{3}(x)=0$. Hence $x=\varphi_{2}(y)$ for some $y\in H_{\mathfrak{m}}^{d-1}(R/I)$. Because $0=F(x)=F(\varphi_{2}(y))=\varphi_{2}(F(y))$, we get that $F(y)\in\operatorname{{im}}\varphi_{1}$. Using the commutativity of the diagram, it is straightforward to check that $\operatorname{{im}}\varphi_{1}$ is an $F$-compatible submodule of $H_{\mathfrak{m}}^{d}(R/I)$. Since $R/I$ is $F$-pure, $H_{\mathfrak{m}}^{d-1}(R/I)$ is anti-nilpotent by Theorem 3.1. Hence $F$ acts injectively on $H_{\mathfrak{m}}^{d-1}(R/I)/\operatorname{{im}}\varphi_{1}$. But clearly $F(\overline{y})=\overline{F(y)}=0$ in $H_{\mathfrak{m}}^{d-1}(R/I)/\operatorname{{im}}\varphi_{1}$, so $\overline{y}=0$. Therefore $y\in\operatorname{{im}}\varphi_{1}$. Hence $x=\varphi_{2}(y)=0$ which is a contradiction because we assume $x$ is a nonzero socle element. ∎