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Let $(R,m)$ be a Noetherian local domain whose integral closure $S$ is too. Also assume that $S$ is module-finite over $R$.

Let $x\in m^k\setminus m^{k+1}$ and $u\in S^\times$ such that $ux \in R$. Is $ux$ necessarily in $m^k\setminus m^{k+1}$?

The case I care about is when $R$ is an affine semigroup ring over $\Bbb C$ with maximal (multigraded) homogeneous ideal $m$ (or rather the localization of this stuff at $m$).

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In general, this can be wrong. Consider $R=\mathbb{R}[[x,y]]/(x^2+y^4)$ (or $\mathbb{R}[x,y]_{(x,y)}/(x^2+y^4)$). Then $u=x/y^2$ is in the integral closure $S$ and it is a unit. $y^2\in\mathfrak{m}^2$, $uy^2=x\in\mathfrak{m}$, but not in its square.

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  • $\begingroup$ For posterity: Here's why this answer satisfies the requirements on $S$: The $\Bbb R$-algebra homomorphism $R[u]\to \Bbb C[[y]]$ taking $y$ to itself and $u$ to $i$ is an isomorphism. Since $\Bbb C[[y]]$ is a DVR, it's integrally closed, and therefore $R[u]=S$. We win. $\endgroup$
    – Avi Steiner
    Jun 24, 2017 at 18:44
  • $\begingroup$ The reason this counterexample exists is non-standard grading. I think that if $A$ is a finitely generated standard-graded algebra over a field (i.e. $A$ contains a field $k$ consisting of its homogeneous degree zero elements, and $A$ is generated by degree one elements over $k$), and, $\frak m$ its homogeneous maximal ideal, and $R=A_{\frak m}$, then the answer to your question should be 'yes'. In the example above, $A = \mathbb R[x,y] / (x^2 + y^4)$, there is no way to give a grading to $A$ in such a way that it is generated over $\mathbb R$ by degree one elements; deg(x) must be 2deg(y) $\endgroup$
    – Neil Epstein
    Jun 26, 2017 at 15:48

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