# nLab maximal ideal

Contents

### Context

#### Algebra

higher algebra

universal algebra

# Contents

## Definition

A maximal ideal (in say a commutative ring $R$) is an ideal $M$ which is maximal among proper ideals. (This is a second-order definition, as it quantifies over subsets of $R$.)

Equivalently, an ideal $M \subseteq R$ is maximal if the quotient ring $R/M$ is a field. This suggests a first-order definition: an ideal $M$ is maximal if $(\forall_x)\; \neg (x \in M) \Rightarrow \exists_y x y = 1$.

## Properties

### General

###### Proposition

(every proper ideal is contained in a maximal one)

Assuming the axiom of choice then:

Let $R$ be a commutative ring and let $I \subset R$ be a proper ideal. Then $R$ contains a maximal ideal $\mathfrak{m}$ containing $I$, i.e. $I \subset \mathfrak{m}$.

###### Proof

Write $PropIdl(R)_{\subset}$ for the set of proper ideals of $R$, partially ordered by inclusion. We claim that every chain in $PropIdl(R)_{\subset}$ has an upper bound (def.). This then implies the statement by Zorn's lemma (equivalent to the axiom of choice).

To show the claim, assume that $\mathcal{C} \subset PropIdl(R)_{\subset}$ is a chain. We have to produce an $I \in PropIdl(R)$ such that for all $c \in C$ then $c \subset I$.

We claim that such $I$ is provided by the union:

$I \coloneqq \underset{J \in \mathcal{C}}{\cup} J \,.$

It is clear that if this is indeed a proper ideal, then it is an upper bound of the chain.

To see first of all that this $I$ is an ideal, consider $x_1, x_2 \in I$. There are thus $J_1, J_2 \in \mathcal{C}$ with $x_1 \in J_1$ and $x_2 \in J_2$. Since a chain is total order by definition, either $J_1 \subset J_2$ or $J_2 \subset J_1$. We may assume the former without restriction, otherwise rename $1 \leftrightarrow 2$. Therefore now $x_1, x_2 \in J_2$ and so we may add them there and find that $x_1 + x_2 \in J_2 \subset I$. Similarly if $r \in R$ then $r x_i \in J_2 \subset I$.

Finally to see that this idea $I$ is indeed proper. But since all the $J_i$ are proper, neither of them contains $1 \in R$, and hence $I$ does not contain $1 \in R$.

###### Proposition

In classical mathematics then:

Every maximal ideal is a prime ideal.

### Relation to points in the spectrum

Assuming AC and EM, then

Maximal ideals in the spectrum of a commutative ring $Spec(R)$ correspond precisely to the closed points in the Zariski topology on $Spec(R)$ (this prop.).

Closed points are at the heart of the definition of schemes. A scheme $X$ is a sheaf with respect to the Zariski topology that admits a covering by open embeddings of affine schemes, where “covering” means that every closed point $p: Spec(F) \to X$ ($F$ a field) factors through one of the embeddings.

## References

Last revised on April 25, 2017 at 11:07:41. See the history of this page for a list of all contributions to it.