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Let $\mathrm{M}$ be a finitely generated submonoid of $\mathbb{Z}^{\oplus d}$ for some $d$, let $A := k[\mathrm{M}]$ be the associated monoid algebra over a field $k$, let $\mathfrak{m} \subset A$ be a maximal ideal corresponding to a $k$-point of $A$. Is the local ring $A_{\mathfrak{m}}$ geometrically unibranch, i.e. does the strict henselization $(A_{\mathfrak{m}})^{\mathrm{sh}}$ have a unique minimal prime?

Remarks/thoughts: If the monoid $\mathrm{M}$ is "saturated" (if $x \in \mathrm{M}^{\mathrm{gp}}$ is an element such that there exists $n \in \mathbb{Z}_{\ge 1}$ for which $nx \in \mathrm{M}$, then $x \in \mathrm{M}$), then $A$ is a normal domain (see [1, Proposition 3.4.1]), so $A_{\mathfrak{m}}$ and $(A_{\mathfrak{m}})^{\mathrm{sh}}$ are also normal domains (by e.g. [2, Tag 00GY] and [2, Tag 06DI] respectively).

References:

[1] Ogus, "Lectures in Logarithmic Algebraic Geometry", version of Oct 24, 2017, link

[2] Stacks Project

Keywords: singularities of toric varieties, geometrically unibranch, completion, henselization, minimal primes, semigroup

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    $\begingroup$ By Hartshorne's connectedness theorem, it suffices to prove that the scheme is $S2$ and that it is unibranch in codimension $1$. The set of points where the scheme is not $S2$ is closed (by Auslander's theorem, 6.11.1 and 6.11.2 of EGA4). This closed set is also stable under the action of the multiplicative group $\text{Spec}\ k[M^{\text{gp}}].$ Thus, it suffices to check at the generic point of every closed orbit that the scheme is $S2$ and the scheme is unibranch at the codimension $1$ orbit closures that contain that generic point. $\endgroup$
    – Jason Starr
    Commented Oct 28, 2017 at 22:39
  • $\begingroup$ If you knew about this somewhat surprising result (note that the Stacks project uses a different definition of geometrically unibranch), it should be expected that the answer is negative in general. (Presumably the example you had in mind was $k[x^2,x^3]$; this is somehow the only type of example.) $\endgroup$
    – R. van Dobben de Bruyn
    Commented Oct 29, 2017 at 17:47
  • $\begingroup$ @R.vanDobbendeBruyn Thanks for letting me know about the surprising result. The definition I used is equivalent to that of the Stacks Project by this lemma. Could you explain your remark that the answer should be expected to be negative in general? Unless I am mistaken, the cuspidal cubic $k[x^{2},x^{3}]$ is geometrically unibranch, since there is only one prime of $k[x]$ lying over the cusp via the normalization map $k[x^{2},x^{3}] \to k[x]$. $\endgroup$
    – Minseon Shin
    Commented Oct 29, 2017 at 19:27
  • $\begingroup$ A ring $R$ is geometrically unibranch if and only if the normalisation $\operatorname{Spec} \tilde R \to \operatorname{Spec} R$ is radicial. Over an algebraically closed field or in characteristic $0$, this should land you in the situation of the surprising result. Thus there exists a filtration $R = R_0 \subseteq \ldots \subseteq R_m = \tilde R$ such that $R_i = R_{i-1}[b_i]$ for an element $b_i$ satisfying $b_i^2 , b_i^3 \in R_{i-1}$. It seems that this should not be the case for these monoid algebras, unless you have something like $k[x^2,x^3]$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Oct 29, 2017 at 21:16
  • $\begingroup$ @R.vanDobbendeBruyn OK, perhaps I see what you mean. Since the normalization of the monoid algebra corresponds to taking the monoid algebra of the saturation $\mathrm{M}^{\mathrm{sat}}$, roughly speaking the corresponding statement about monoids would be that finitely generated submonoids $\mathrm{M} \subseteq \mathbb{Z}^{\oplus d}$ can be saturated by applying the operation "adjoin an element $a \in \mathrm{M}^{\mathrm{sat}}$ for which $2a,3a \in \mathrm{M}$" a finite number of times, which seems strong. $\endgroup$
    – Minseon Shin
    Commented Oct 29, 2017 at 23:08

1 Answer 1

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Edit. As Friedrich Knop points out, these schemes are typically not even $S2$. I added an example at the end. (I believe it is the example that Friedrich Knop was suggesting.)

Original answer. This is not true. Let $d$ equal $2.$ Let $n\geq 2$ be an integer that is prime to the characteristic of $k.$ Inside $\mathbb{Z}^{\oplus 2},$ consider the subsemigroup $M$ generated by the following three elements, $$a=(1,0),\ \ b=(0,1), \ \ c = (n,-n).$$ Denote by $x$ and $y$ the elements of $k[M]$ corresponding to $a$ and $b.$ Then $k[M]$ is the $k$-subalgebra of $k[x,x^{-1},y,y^{-1}]$ generated by $x,$ by $y,$ and by $z=x^n/y^n.$ This has presentation, $$k[M] = k[x,y,z]/\langle y^nz -x^n \rangle.$$ Now consider the maximal ideal $\mathfrak{m} = \langle x,y,z-1\rangle.$ Because $n$ is prime to the characteristic, the ring extension $$k[M] \to k[M][u]/\langle (u+1)^n-z \rangle,$$ is étale near $\mathfrak{m}$. After adjoining $u$, it is clear the local ring near the maximal ideal $\mathfrak{n} = \langle x,y,z-1,u \rangle$ is not unibranch, i.e., $$k[x,y,u]/\langle (y(1+u))^n - x^n \rangle,$$ is not unibranch near $\langle x,y,u\rangle.$

I suspect that for every subsemigroup $M$ of $\mathbb{Z}^{\oplus d},$ the semigroup ring $k[M]$ is $S2.$ If the ring $k[M]$ is $S2,$ then examples such as the one above would be the only examples. Precisely, if $k[M]$ is $S2$ and if $\text{Spec}\ k[M]$ is unibranch at all codimension $1$ points, then by Hartshorne's Connectedness Theorem, the scheme is unibranch everywhere.

Edit. As Friedrich Knop points out, these schemes are typically not even $S2.$ For instance, let $M$ be the subsemigroup of $\mathbb{Z}_{\geq 0}^{\oplus 3}$ with the following generators, $$M = \mathbb{Z}_{\geq 0}\cdot (1,0,0) + \mathbb{Z}_{\geq 0}\cdot (0,1,0) + \mathbb{Z}_{\geq 0} \cdot (1,0,1) + \mathbb{Z}_{\geq 0}\cdot (0,1,1) + \mathbb{Z}_{\geq 0} \cdot (0,0,2).$$ Denote the respective generators of $k[M]$ as follows, $$x = \chi^{(1,0,0)}, \ y = \chi^{(0,1,0)}, \ u = \chi^{(1,0,1)}, \ v= \chi^{(0,1,1)}, \ w = \chi^{(0,0,2)}.$$ Then a presentation for the semigroup ring is, $$k[M] = k[x,y,u,v,w]/\langle u^2 - x^2w, v^2-y^2w,xv-yu,uv-xyw \rangle.$$ This is smooth if we invert either $x$, so that the fraction $z=ux^{-1}$ is in the fraction ring, or if we invert $y$, so that the fraction $z=vy^{-1}$ is in the fraction ring. Thus, the singular locus is contained in the common vanishing set of $x$ and $y$. Set-theoretically, this implies that also $u$ and $v$ equal $0$, so that the singular locus is in the affine line $\text{Spec}\ k[w].$

Since $\text{Spec}\ k[M]$ has dimension $3$ and the singular set has dimension $1$, this ring is $R1$. Yet it is not normal, since $z$ satisfies the monic polynomial $z^2 -w=0$. Therefore, by Serre's Criterion for normality, the semigroup ring $k[M]$ is not $S2.$

Please note, if you do the similar analysis as in the original example for the maximal ideal $\mathfrak{m} = \langle x,y,u,v,w-1\rangle,$ it turns out that this new example is even worse. The strict henselization is not unibranch, and it is even disconnected by removing the codimension $2$ zero scheme of $\langle x,y,u,v\rangle.$ So it appears that we should expect no form of unibranchedness / connectedness for the strict henselizations of a non-normal semigroup ring.

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    $\begingroup$ Semigroups rings are not $S2$, in general. To see this, just carry the example to a higher dimension $d\ge3$. Let's take $n=2$, for simplicity. Let $M\subseteq\mathbb N^d$ be the set of elements $(a_1,\ldots,a_d)$ with $a_1=\ldots=a_{d-1}=0\Rightarrow 2|a_d$. Then the singular set of ${\rm Spec}k[M]$ is $1$. So $k[M]$ is $R1$, it is not normal, so can't be S2. $\endgroup$
    – Friedrich Knop
    Commented Oct 29, 2017 at 12:12
  • $\begingroup$ @FriedrichKnop. I added the example showing that semigroup rings can fail to be $S2.$ $\endgroup$
    – Jason Starr
    Commented Oct 29, 2017 at 13:06

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