---
tag: 5GCANV7337
kind: statement
role: proposition
name: Integral extensions of fields
msc: [13B21]
uses: [B7X0KFY1S1, 0QVD7WBFGN, 04C5ZBG1WT]
about: [B7X0KFY1S1]
sources:
  - url: https://en.wikipedia.org/wiki/Integral_element
---

## Statement

Let $A \subseteq B$ be [[integral domains|0QVD7WBFGN]] with $B$ [[integral over|B7X0KFY1S1]] $A$.
Then $A$ is a [[field|04C5ZBG1WT]] if and only if $B$ is a field.

## Remarks

Suppose $A$ is a field and let $b \in B$ be nonzero.
Take a monic relation $b^n + a_{n-1} b^{n-1} + \cdots + a_0 = 0$ of least degree; since $B$ is a domain, $a_0 \ne 0$.
Then $b^{-1} = -a_0^{-1}(b^{n-1} + a_{n-1} b^{n-2} + \cdots + a_1)$ lies in $B$.
Conversely, suppose $B$ is a field and let $a \in A$ be nonzero.
Then $a^{-1} \in B$ is integral over $A$, say $a^{-m} + c_{m-1} a^{-m+1} + \cdots + c_0 = 0$, and multiplying by $a^{m-1}$ gives $a^{-1} = -(c_{m-1} + c_{m-2} a + \cdots + c_0 a^{m-1}) \in A$.

Example: $\Q \subseteq \Q(\sqrt{2})$ is integral and both are fields; $\Z \subseteq \Z[i]$ is integral and neither is a field.

The hypothesis that $B$ is a domain is needed for "$A$ field implies $B$ field": the ring $k[x]/(x^2)$ is integral over the field $k$, but it is not a field.

## Explanations

### L1 human

In an integral extension of domains, the small ring is a field exactly when the big ring is.
So integral extensions cannot turn a non-field into a field, nor the reverse.

### L2 human

This is the step that links maximal ideals across an integral extension: for a prime $\mathfrak{q}$ of $B$, apply it to $A/(\mathfrak{q} \cap A) \subseteq B/\mathfrak{q}$.
It gives that $\mathfrak{q}$ is maximal if and only if $\mathfrak{q} \cap A$ is maximal, and it is used in the proofs of [[lying over|YZ97GKYY3P]], [[incomparability|Q2XBHJ96GV]] and [[Zariski's lemma|YBJ2VQHFDQ]].