# Flat and algebraic (non-integral) local rings extension $R \subseteq S$ with $m_RS=m_S$

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.

• I assume you meant "not necessarily invertible leading coefficient"? Apr 14, 2021 at 1:38
• @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 $\mathbb{Z} \subset \mathbb{Z}[\frac{1}{2}]$). Apr 14, 2021 at 1:52
• Well, the rings $\mathbb{Z}$ and $\mathbb{Z}[\frac{1}{2}]$ are not local, but we can consider localizations of them. . Apr 14, 2021 at 2:03
• 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 $R[T]$. Apr 14, 2021 at 3:31
• @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 $R[T]$. Apr 14, 2021 at 13:19

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.

• 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..). Apr 14, 2021 at 13:23
• @user237522 I don't understand your question Apr 14, 2021 at 18:10
• Please, I meant, same question and in addition to the conditions of the question we further assume that $S=R[w]$, for some $w \in S$. Apr 14, 2021 at 18:56