Counting geometric branches via the Frobenius map and $F$-nilpotent singularities

Let $a\in(x^{n})+\mathfrak{m}^{n+1}$. Then, $a=bx^{n}+w$, where $w\in\mathfrak{m}^{n+1}$, and $a^{p^{e}}=b^{p^{e}}x^{np^{e}}+w^{p^{e}}$ for all $e\in\mathbb{N}$. But for $e\gg 0$, $w^{p^{e}}\in\mathfrak{m}^{(n+1)p^{e}}=x^{(n+1)p^{e}-N}\mathfrak{m}^{N}$, so that $a^{p^{e}}=sx^{np^{e}}$ for some $s\in R$, that is, $a\in(x^{n})^{F}$.

Now suppose $k$ is perfect, and let $y\in(x^{n})^{F}$. Since $(x^{n})$ is a homogeneous ideal, so is $(x^{n})^{F}$, thus it suffices to assume that $y$ is homogeneous. Further, we may assume $\deg y=n$, as otherwise $y\in\mathfrak{m}^{n+1}$. Then, for all $e\geq\operatorname{Fte}(x^{n})$ there is an $r\in R$ with $y^{p^{e}}=rx^{np^{e}}$, counting degrees on both sides, we must have $\deg r=0$, that is, $r\in k$. Since the field $k$ is perfect, there is an $s\in k$ with $s^{p^{e}}=r$, and so $y^{p^{e}}=(sx^{n})^{p^{e}}$. As $R$ is reduced, we get $y\in(x^{n})$, concluding the proof of (a).

Since the ideal $(x^{n})$ is principal, its tight closure $(x^{n})^{*}$ equals its integral closure $\overline{(x^{n})}$. Since $(x^{n})$ is a reduction of $\mathfrak{m}^{n}$, we must have that $\mathfrak{m}^{n}$ is contained in $\overline{(x^{n})}$. For the reverse containment, see [Smi97b, Proposition 2.1] for a general statement concerning lower bounds of the degree of an element contained in the tight closure of a homogeneous ideal.

Now we consider the containments above in the local ring $R_{\mathfrak{m}}$. The result in this case is a simple consequence of the fact that the ideals above all localize appropriately. In particular, write $(R^{\prime},\mathfrak{n})=(R_{\mathfrak{m}},\mathfrak{m}R_{\mathfrak{m}})$ and $z$ for the image of $x$ in $R^{\prime}$. First, note that the ideal equation $x\mathfrak{m}^{N}=\mathfrak{m}^{N+1}$ localizes to the ideal equation $z\mathfrak{n}^{N}=\mathfrak{n}^{N+1}$ so that $z$ continues to be a reduction of $\mathfrak{n}$ in $R^{\prime}$, and hence, a parameter of $R^{\prime}$. Then, for any $n\geq N$, we have

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  $(x^{n})^{F}R^{\prime}=(z^{n})^{F}$,

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  $((x^{n})+\mathfrak{m}^{n+1})R^{\prime}=(z^{n})+\mathfrak{n}^{n+1}$,

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  $(x^{n})^{*}R^{\prime}=\overline{(x^{n})}R^{\prime}=\overline{(z^{n})}=(z^{n})^{*}$, and²²2Tight closure does not localize in general, but for principal ideals it agrees with the integral closure which does localize.

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  $\mathfrak{m}^{n}R^{\prime}=\mathfrak{n}^{n}$.

This shows the same results hold in the local ring $R_{\mathfrak{m}}$. ∎

A map being finite and purely inseparable localizes, and the (weakly) $F$-nilpotent condition is local. Further, since $R\rightarrow S$ is purely inseparable, for each maximal ideal $\mathfrak{m}$ of $R$, there is a unique maximal ideal $\mathfrak{n}$ of $S$ containing $\mathfrak{m}$, and $\mathfrak{m}S$ is $\mathfrak{n}$-primary. So, after localizing, we may assume $(R,\mathfrak{m})\rightarrow(S,\mathfrak{n})$ is a finite, purely inseparable extension of local rings. Write $d=\dim R=\dim S$.

We have a short exact sequence $0\to R\to S\to S/R\to 0$, and $S/R$ is nilpotent under the usual Frobenius action by 2.10. By [MM21, Theorem 3.5], $S$ is weakly $F$-nilpotent if and only if $R$ is.

Since $R\rightarrow S$ is purely inseparable, $\sqrt{\mathfrak{m}S}=\mathfrak{n}$, so the modules $H^{d}_{\mathfrak{m}}(S)$ and $H^{d}_{\mathfrak{n}}(S)$ are the same by the change-of-rings property of local cohomology. We then use the following commutative diagram induced by the natural Frobenius action on local cohomology modules:

Since $S/R$ is nilpotent under the natural Frobenius map $\overline{F}$, its local cohomology modules $H^{j}_{\mathfrak{m}}(S/R)$ are also nilpotent under the action induced by $\overline{F}$; in particular, if $e_{0}$ is the pure inseparability. index of $R\rightarrow S$, we have $\overline{F}^{e_{0}}(H^{j}_{\mathfrak{m}}(S/R))=0$ for all $j$.

Now suppose $R$ is $F$-nilpotent, and let $\xi\in 0^{*}_{H^{d}_{\mathfrak{m}}(S)}$. Then $\beta(\xi)\in H^{d}_{\mathfrak{m}}(S/R)$, so we have $\overline{F}^{e_{0}}(\beta(\xi))=\beta(F^{e_{0}}(\xi))=0$. Thus, $F^{e_{0}}(\xi)=\alpha(\xi^{\prime})$ for some $\xi\in H^{d}_{\mathfrak{m}}(R)$. Let $c\in R^{\circ}$ have $cF^{e}(\xi)=0$ for all $e\gg 0$. Then $0=cF^{e+e_{0}}(\xi)=\alpha(cF^{e}(\xi^{\prime}))$, so that $cF^{e}(\xi^{\prime})\in\ker(\alpha)=\operatorname{im}(\delta)\subset H^{d}_{\mathfrak{m}}(R)$ for all $e\gg 0$. But $F^{e_{0}}\circ\delta=\delta\circ\overline{F}^{e_{0}}=0$, so that $c^{p^{e_{0}}}F^{e+e_{0}}(\xi^{\prime}))=0$ for all $e\gg 0$. Since $e_{0}$ is independent of $e$, this shows that $\xi^{\prime}\in 0^{*}_{H^{d}_{\mathfrak{m}}(R)}=0^{F}_{H^{d}_{\mathfrak{m}}(R)}$. Thus, $F^{e}(\xi^{\prime})=0$ for all $e\gg 0$, and thus $F^{e_{0}+e}(\xi)=\alpha(F^{e}(\xi))=0$ for all $e\gg 0$, that is, $\xi\in 0^{F}_{H^{d}_{\mathfrak{m}}(S)}$.

Finally, suppose $S$ is $F$-nilpotent and that $\xi\in 0^{*}_{H^{d}_{\mathfrak{m}}(R)}$. Then, there is a $c\in R^{\circ}$ so that $cF^{e}(\xi)=0$ for all $e\gg 0$, and consequently $\alpha(cF^{e}(\xi))=cF^{e}(\alpha(\xi))=0$ for all $e\gg 0$. Thus $\alpha(\xi)$ is in $0^{*}_{H^{d}_{\mathfrak{m}}(S)}=0^{F}_{H^{d}_{\mathfrak{m}}(S)}$, so that $F^{e}(\xi)\in\ker(\alpha)=\operatorname{im}(\delta)$. But $\operatorname{im}(\delta)$ is nilpotent under $\overline{F}$ as $S/R$ is, so $F^{e+e^{\prime}}(\xi)=0$, that is, $\xi\in 0^{F}_{H^{d}_{\mathfrak{m}}(R)}$. ∎

Assume that $(R,\mathfrak{m})$ is local. We first show that $R\rightarrow\overline{R}$ is a purely inseparable map of rings. Let $x/y$ be an element of the total quotient ring of $R$ which is integral over $R$. Then, $y$ is either a unit of $R$ (in which case the remainder of the argument is trivial) or a regular element of $R$, and $x\in\overline{(y)}$, the integral closure of the ideal $(y)$ in $R$. However, by [HH90, Corollary 5.8], $\overline{(y)}=(y)^{*}$ as $(y)$ is principal. But since $S$ is an excellent $F$-nilpotent local ring and $y$ is a parameter element, we have $(y)^{*}=(y)^{F}$ by [PQ19, Corollary 5.15].

Thus, there is a natural number $e$ so that $x^{p^{e}}\in(y)^{\left[p^{e}\right]}=(y^{p^{e}})$, that is, $x^{p^{e}}=ry^{p^{e}}$ for some $r\in R$. Note that this means $x/y\in S$, the weak normalization of $R$. Since $x/y$ was chosen to be an arbitrary element of $\overline{R}$, we get that $S=\overline{R}$, that is, the weak normalization of $R$ is normal. In particular, $R\rightarrow\overline{R}$ is a purely inseparable map of rings.

Since a purely inseparable map of rings induces a homeomorphism on spectra, we have that $\overline{R}$ is also a local ring. Let $\mathfrak{n}$ be the unique maximal ideal of $\overline{R}$ and $K:=\overline{R}/\mathfrak{n}$. Clearly, the purely inseparable ring map $R\rightarrow\overline{R}$ induces a purely inseparable map of residue fields $k\rightarrow K$, so we have $[K:k]_{\text{sep}}=1$. Thus, $R$ is geometrically unibranced, as claimed.

Finally, note that since $\overline{R}$ is local, the henselization $R^{h}$ of $R$ has a unique minimal prime ideal. As $R$ is reduced and $R^{h}$ is a filtered colimit of étale, hence smooth $R$-algebras, so $R^{h}$ is also reduced. It follows that $R^{h}$ is a domain, and therefore that $R$ is a domain. ∎

If $R$ is normal, then it is regular, so the result is trivial. Otherwise, the conductor ideal $\mathfrak{c}$ of $R\rightarrow\overline{R}$ is an $\mathfrak{m}$-primary ideal of $R$. Let $e$ be the smallest natural number so that $\mathfrak{m}^{\left[p^{e}\right]}$ is contained in $\mathfrak{c}$. Further, recall the tight closure of a principal ideal agrees with its integral closure.

let $\xi$ be an element of the $R$-module $0^{*}_{H^{1}_{\mathfrak{m}}(R)}$. By [Smi97a, Proposition 2.5], for a regular parameter $x\in R$, we have $\xi=[a+(x^{n})]$ for some $a$ lying in $(x^{n})^{*}=\overline{(x^{n})}$. Thus, the element $a/x^{n}$ of the total ring of quotients of $R$ is inside $\overline{R}$, and for any element $b\in\mathfrak{m}$, we have $b^{p^{e}}(a/x^{n})^{p^{e}}\in R$ since $\mathfrak{m}^{\left[p^{e}\right]}$ is contained in $\mathfrak{c}$. Hence, $(ba)^{p^{e}}\in(x^{np^{e}})=(x^{n})^{\left[p^{e}\right]}$, thus $ba\in(x^{n})^{F}$. This implies $b[a+(x^{n})]\in 0^{F}_{H^{1}_{\mathfrak{m}}(R)}$ so that the $R$-module $\mathfrak{m}0^{*}_{H^{1}_{\mathfrak{m}}(R)}$ is contained in $0^{F}_{H^{1}_{\mathfrak{m}}(R)}$, as required. ∎

Since $R$ is Cohen-Macaulay, it is equidimensional and the direct limit system defining $H=H^{1}_{\mathfrak{m}}(R)$ is injective. For a parameter element $x\in R$, we have $0^{*}_{H}=\varinjlim\,(x^{n})^{*}/(x^{n})$ and $0^{F}_{H}=\varinjlim\,(x^{n})^{F}/(x^{n})$. Then, $0^{*}_{H}/0^{F}_{H}=\varinjlim\,(x^{n})^{*}/(x^{n})^{F}$, and we show that for all $n\gg 0$, $\dim_{k}(x^{n})^{*}/(x^{n})^{F}=[\ell:k](b(R)-1)$. The fact that the direct limit system is injective then proves the claimed equality.

Since $R$ is one dimensional, the conductor ideal $\mathfrak{c}$ of $R\rightarrow\overline{R}$ is $\mathfrak{m}$-primary, so there is an $N$ such that $\mathfrak{m}^{N}$ lies in $\mathfrak{c}$. Thus, by renaming $x^{N}$ to $x$, we may safely assume that $x$ lies in $\mathfrak{c}$.

By [SH06, Proposition 1.6.1] and [HH90, Corollary 5.8], we have $(x)^{*}=\overline{(x)}=x\overline{R}\cap R$, but $x\overline{R}\subset R$ since $x\in\mathfrak{c}$, that is, $\overline{(x)}=x\overline{R}$ as $R$-submodules of $\overline{R}$. Similarly, $(x)^{F}=xS\cap R=xS$. Further, $x$ is a regular element on $R$ which implies $x\overline{R}/xS\simeq\overline{R}/S$, so we must compute the length of the $R$-module $\overline{R}/S$.

Let $J\subset\overline{R}$ be the Jacobson radical of $\overline{R}$, which is the intersection of the finitely many maximal ideals $\mathfrak{M}_{1},\ldots,\mathfrak{M}_{b}$ of $\overline{R}$. Write $K_{i}=\overline{R}/\mathfrak{M}_{i}$; by 2.3, we have

For convenience, write $[\ell:k]=w$. Now, we have a sequence of finite field extensions $k\rightarrow\ell\rightarrow K_{i}$ for each $i$, and $k\rightarrow\ell$ is the perfect closure of $k$ in $K_{i}$. This implies $\ell\rightarrow K_{i}$ is a separable extension, and thus

for each $i$.

Note that since the conductor ideal $\mathfrak{c}$ of $R\rightarrow\overline{R}$ has height $1$, every element of $J$ has a power which is inside $\mathfrak{c}$. Thus, for some natural number $e$ we must have $F^{e}(J)$ is inside $R$, and so $J$ lies in $S$. Furthermore, viewed as an ideal of $S$, we must have that $J=J\cap S=\mathfrak{n}$ is the unique maximal ideal of the local ring $S$ since $J$ is the Jacobson radical of $\overline{R}$ and $S\rightarrow\overline{R}$ is an integral extension.

The map $S\rightarrow\overline{R}$ induces an injective map $\ell=S/\mathfrak{n}\rightarrow\overline{R}/J$. Further, by [LV81, Corollary 1.5] we must have that the conductor $S\rightarrow\overline{R}$ is radical, and is $\mathfrak{n}$-primary since $\dim S=1$ and $S$ is reduced, so the conductor must be $\mathfrak{n}$. This implies $J\overline{R}=\mathfrak{n}S$ as $R$-submodules of $\overline{R}$, so we may apply the isomorphism theorems to see that the $R$-module $(\overline{R}/J)/(S/\mathfrak{n})$ is isomorphic to $\overline{R}/S$. Consequently, we get a short exact sequence of $k$-vector spaces

This gives the dimension equality

from which we get $\dim_{k}\overline{R}/S=w\sum_{i}[K_{i}:k]_{\text{sep}}-w=w(b(R)-1)$, as required. ∎