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Let $R \subseteq S$ be two Noetherian local rings, not necessarily regular, which are integral domains, with $m_RS=m_S$, namely, the ideal in $S$ generated by $m_R$ (= the maximal ideal of $R$) is $m_S$ (= the maximal ideal of $S$).

Further assume that $R$ and $S$ are $\mathbb{C}$-algebras, $R \subseteq S$ is flat and algebraic but not integral, where algebraic non-integral means: Every element of $S$ satisfies a polynomial with coefficients in $R$, with non-invertible (in $R$) leading coefficient.

Question: Could one find an example of such rings?

Unfortunately, the examples I find are integral, for example: $R=\mathbb{C}[x(x-1)]_{(x(x-1))}$, $S=\mathbb{C}[x]_{(x)}$.

Remarks:

(i) I am interested in both cases where $R$ and $S$ have the same fields of fractions or different fields of fractions.

(ii) Recall the following results, which are not applicable here, since I assume that $R \subseteq S$ is non-integral: If $A \subseteq B$ is integral and flat, then $A \subseteq B$ is faithfully flat, and if in addition $Q(A)=Q(B)$ (same fields of fractions), then $A=B$.

Relevant questions: a, b and c. Also asked in MSE.

Any hints and comments are welcome; thank you.

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  • $\begingroup$ I assume you meant "not necessarily invertible leading coefficient"? $\endgroup$
    – Will Chen
    Commented Apr 14, 2021 at 1:38
  • $\begingroup$ @WillChen, thank you for your comment. Actually, I meant 'necessarily, not invertible leading coefficient', since I want the extension to be algebraic but not integral. For an integral extension I have found an example (localizations of polynomial rings). I wish something like $\mathbb{Z} \subset \mathbb{Z}[\frac{1}{2}]$ which is flat, algebraic, non-integral, because $f(T)=2T-1 \in \mathbb{Z}[T]$ is such that $f(\frac{1}{2})=0$. But here the rings are requiered to be $\mathbb{C}$-algebras (I wished to exclude such cases as $\mathbb{Z} \subset \mathbb{Z}[\frac{1}{2}]$). $\endgroup$
    – user237522
    Commented Apr 14, 2021 at 1:52
  • $\begingroup$ Well, the rings $\mathbb{Z}$ and $\mathbb{Z}[\frac{1}{2}]$ are not local, but we can consider localizations of them. . $\endgroup$
    – user237522
    Commented Apr 14, 2021 at 2:03
  • $\begingroup$ Every element of $R$ satisfies a polynomial with coefficients in $R$ with non-invertible leading coefficient. E.g., $r \in R$ satisfies $2T - 2r\in R[T]$. I think what you mean to say is that every element of $S$ satisfies a polynomial in $R[T]$, but not every element in $S$ satisfies a monic polynomial in $R[T]$. $\endgroup$
    – Will Chen
    Commented Apr 14, 2021 at 3:31
  • $\begingroup$ @WillChen, yes, you are right, I meant that every element of $S$ satisfies a polynomial in $R[T]$, bot not every element of $S$ satisfies a monic polynomial in $R[T]$. $\endgroup$
    – user237522
    Commented Apr 14, 2021 at 13:19

1 Answer 1

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In general if $R$ is a local ring, then its henselization $R^h$ is flat and "algebraic" over $R$, but rarely integral. The intuition is that the henselization is built out of localizations of etale extensions of $R$. Both localizations and etale extensions are flat and "algebraic" in your sense, but localizations are rarely integral.

To make a precise statement, first note that it suffices to find etale ring extensions $R\rightarrow R'\rightarrow S$ with $R$ normal, $S/R$ an etale local homomorphism of local rings, and $S/R'$ not integral.

Indeed, if we have found $R,R',S$ as above, then $S/R'$ non-integral implies $S/R$ non-integral. Moreover, etale morphisms preserve normality, so $S$ must be normal, hence a domain since $S$ is local so $\text{Spec }S$ is connected. Finally, etale ring maps are locally standard etale, so every element of $S$ is "algebraic" over $R$.

Here's a sketch of how to produce an example: Start with a connected normal scheme $Y$ over $\mathbb{C}$ and a finite flat map $f : X\rightarrow Y$ with $X$ irreducible. In characteristic 0, $f$ is generically etale, so let $y\in Y$ be a point above which $f$ is etale. Let $R := \mathcal{O}_{Y,y}$, let $X_R := X\times_Y\text{Spec }\mathcal{O}_{Y,y}$, and write $X_R = \text{Spec }R'$. Then $R'/R$ is finite etale. On the other hand, $R'$ is connected, since the generic point of $X$ maps to the generic point of $Y$ which lies in $\text{Spec }R$, so the generic point of $X$ lies in $X_R$ and specializes to every closed point of $X_R$, so $R'$ is a normal domain. However, if $X$ has multiple closed points lying over $y$, then $R'$ is not local, in which case let $\mathfrak{m}$ be a maximal ideal of $R'$, and let $S := R'_{\mathfrak{m}}$. Then $S/R$ becomes an etale local homomorphism of local rings with $S/R'$ non-integral, as desired.

For an explicit example, you can take $Y = \text{Spec }\mathbb{C}[y]$, and $$X := \text{Spec }\mathbb{C}[x,y]/(y^2-(x-a)(x-b)(x-c))$$ where $a,b,c\in\mathbb{C}$ are distinct. Then take $y\in Y$ to be the point $(y = 0)$. In particular, this shows that the henselization of $\mathbb{C}[y]_{(y)}$ also satisfies your conditions.

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  • $\begingroup$ Thank you very much! Interesting. Please, what if $S$ is a simple extension of $R$, namely, $S=R[w]$, for some $w \in S$? (perhaps your answer also answers this; I do not know..). $\endgroup$
    – user237522
    Commented Apr 14, 2021 at 13:23
  • $\begingroup$ @user237522 I don't understand your question $\endgroup$
    – Will Chen
    Commented Apr 14, 2021 at 18:10
  • $\begingroup$ Please, I meant, same question and in addition to the conditions of the question we further assume that $S=R[w]$, for some $w \in S$. $\endgroup$
    – user237522
    Commented Apr 14, 2021 at 18:56

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