# Intersection of maximal ideal with subring

1. Jun 11, 2014

### coquelicot

Let R be an integral ring (eventually can be supposed integrally closed), and R' an integral extension of R. Assume that M is a maximal ideal of R'. Must the intersection of M with R be a maximal ideal of R ?

Thx.

2. Jun 11, 2014

### Erland

Is an integral ring the same as an integral domain? And what's is an integral extension? (I get the impression that you mean somrthing like the extension of the natural numbers to the integers.)

3. Jun 11, 2014

### micromass

This follows from the following two lemma's:

Lemma 1: If $R\subseteq R^\prime$ are rings such that $R^\prime$ is integral over $R$, and if $J$ is an ideal of $R^\prime$ and if $I=R\cap J$, then $R^\prime/J$ is integral over $R/I$.

Proof: Take $x\in R^\prime$, then we have some equation
$$x^n + a_1x^{n-1} + ... + a_n = 0$$
with $a_i \in R$. Reducing this modulo $J$ then yields that $x+J$ is integral over $R/I$.

Lemma 2: Let $R\subseteq R^\prime$ be integral domains such that $R^\prime$ is integral over $R$. Then $R$ is a field if and only if $R^\prime$ is a field.

Proof: If $R$ is a field, then let $y\in R^\prime$ be nonzero. Then there is some equation
$$y^n + a_1y^{n-1} + ... + a_n = 0$$
with $a_i \in R$. We can take this equation of smallest possible degree. But then
$$y^{-1} = -a_n^{-1}(y^{n-1} + a_1 y^{n-2} + ... + a_{n-1})$$
note that $a_n\neq 0$ because of the integral domain requirement. Thus $R^\prime$ is a field.

Conversely, if $R^\prime$ is a field and if $x\in R$ is nonzero, then $x^{-1}$ is integral over $R$ and thus satisfies an equation
$$x^{-n} + a_1x^{-n+1} + ... + a_n=0$$
It follows that $x^{-1} = - (a_1 + a_2x + ... + a_nx^{n-1})\in R$.

4. Jun 11, 2014

### coquelicot

FOR ERLAND : 1) Yes, I meant integral domain. 2) I meant that every element x of R' is integral over R.
MICROMASS : You are the best of the best. thx a lot.

5. Jun 13, 2014

### mathwonk

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