Commutative Algebra/Basics on prime and maximal ideals and local rings

Prime ideals

Lemma 12.2:

Let $R$ be a ring and $I\le R$ an ideal. $I$ is prime if and only if $R/I$ is an integral domain.

Proof:

$I$ prime is equivalent to $ab\in I\Rightarrow a\in I\vee b\in I$. This is equivalent to

$ab+I=0+I\Rightarrow a+I=0+I\vee b+I=0+I$.$◻$

Proof:

Order all ideals of $R$ not intersecting $S$ by set inclusion, and let a chain

$I_1\subseteq I_2\subseteq \cdots \subseteq I_k\subseteq \cdots$

be given. The ideal

$I:=\bigcup _{k\in \mathbb{N}}{I}_k$

(this is an ideal, since $a,b\in I\Rightarrow a\in I_m,b\in I_n\Rightarrow a,b\in I_{\max \{m,n\}}$, hence $a+b\in I$, $ra\in I$) is an upper bound of the chain, since $I$ cannot intersect $S$ for else one of the $I_k$ would intersect $S$. Since the given chain was arbitrary, Zorn's lemma implies the existence of a maximal ideal among all ideals not intersecting $S$. This ideal shall be called $J$; we prove that it is prime.

Let $ab\in J$, and assume for contradiction that $a\notin J$ and $b\notin J$. Then $⟨a⟩+J$, $⟨b⟩+J$ are strict superideals of $J$ and hence intersect $S$, that is,

$s=xa+yj$,

$t=zb+w{j}^{\prime }$,

$s,t\in S$, $x,y,z,w\in R$, $j,j^{\prime }\in J$. Then $S\ni st=xzab+yzbj+xaw{j}^{\prime }+ywj{j}^{\prime }\in J$, contradiction.$◻$

Projection to the quotient ring

In this section, we want to fix a notiation. Let $R$ be a ring and $I\le R$ an ideal. Then we may form the quotient ring $R/I$ consisting of the elements of the form $r+I$, $r\in R$. Throughout the book, we shall use the following notation for the canonical projection $r\mapsto r+I$:

Maximal ideals

Lemma 12.6:

An ideal $I\le R$ is maximal iff $R/I$ is a field.

Proof:

A ring is a field if and only if its only proper ideal is the zero ideal. For, in a field, every nonzero ideal contains $1$, and if $R$ is not a field, it contains a non-unit $a$, and then $⟨a⟩$ does not contain $1$.

By the correspondence given by the correspondence theorem, $R/I$ corresponds to $R$, the zero ideal of $R/I$ corresponds to $I$, and any ideal strictly in between corresponds to an ideal $K\le R$ such that $I⊊K⊊R$. Hence, $R/I$ is a field if and only if there are no proper ideals strictly containing $I$.$◻$

Lemma 12.7:

Any maximal ideal is prime.

Proof 1:

If $R$ is a ring, $m\le R$ maximal, then $R/m$ is a field. Hence $R/m$ is an integral domain, hence $m$ is prime.$◻$

Proof 2:

Let $m\le R$ be maximal. Let $ab\in m$. Assume $a,b\notin m$. Then $1=ra+sn=tb+uk$ for suitable $n,k\in m$, $r,s,t,u\in R$. But then $1=1^2=(ra+sn)(tb+uk)=rtab+stbn+rauk+sunk\in m$.$◻$

Proof:

We order the set of all ideals $J$ such that $I\subseteq J$ and $J\ne R$ by inclusion. Let

$J_1\subseteq J_2\subseteq \cdots \subseteq J_k\subseteq \cdots$

be a chain of those ideals. Then set

$J:=\bigcup _{k\in \mathbb{N}}{J}_k$.

Clearly, all $J_k$ are contained within $J$. Since $I\subseteq J_1$, $I\subseteq J$. Further, assume $1\in J$. Then $1\in J_m$ for some $m$, contradiction. Hence, $J$ is a proper ideal such that $I\subseteq J$, and hence an upper bound for the given chain. Since the given chain was arbitrary, we may apply Zorn's lemma to obtain the existence of a maximal element with respect to inclusion. This ideal must then be maximal, for any proper superideal also contains $I$.$◻$

Proof: From the correspondence theorem.$◻$

Local rings

Proof:

1. $\Rightarrow$ 2.: Assume $a$ and $b$ are both non-units. Then $⟨a⟩$ and $⟨b⟩$ are proper ideals of $R$ and hence they are contained in some maximal ideal of $R$ by theorem 12.7. But there is only one maximal ideal $m$ of $R$, and hence $a,b\in m$, thus $a+b\in m$. Maximal ideals can not contain units.

2. $\Rightarrow$ 3.: The sum of two non-units is a non-unit, and if $a$ is a non-unit and $r\in R$, $ra$ is a non-unit (for if $sra=1$, $sr$ is an inverse of $a$). Hence, all non-units form an ideal. Any proper ideal of $R$ contains only non-units, hence this ideal is maximal.

3. $\Rightarrow$ 4.: Assume the $r_j$ are all non-units. Since the non-units form an ideal, $r_1+\cdots +r_n$ is contained in that ideal of non-units, contradiction.

4. $\Rightarrow$ 5.: Assume $r$, $1-r$ are non-units. Then $1=r+(1-r)$ is a non-unit, contradiction.

5. $\Rightarrow$ 1.: Let $m,n\le R$ two distinct maximal ideals. Then $m+n=R$, hence $1=s+t$, $s\in m$, $t\in n$, that is, $s=1-t$. $t$ is not a unit, so $1-t=s$ is, contradiction.$◻$

Localisation at prime ideals

In chapter 9, we had seen how to localise a ring at a multiplicatively closed subset $S$. An important special case is $S=R\setminus p$, where $p$ is a prime ideal.

Lemma 12.12:

Let $p\le R$ be a prime ideal of a ring. Then $S:=R\setminus p\subseteq R$ is multiplicatively closed.

Proof: Let $a,b\notin p$. Then $ab$ can't be in $p$, hence $ab\in S$.$◻$

Theorem 12.14:

Let $R$ be a ring, $p\le R$ be prime. $R_p$ is a local ring.

Proof:

Set $S:=R\setminus p$, then $R_p=S^{-1}R$. Set

$m:=\{r/s|r\in p,s\in S\}\subseteq S^{-1}R$.

All elements of $m$ are non-units, and all elements of $R_p\setminus m$ are of the form $r^{\prime }/t$, $r^{\prime }\notin p$, $t\in S$ and thus are units. Further, $m$ is an ideal since $p$ is and by definition of addition and multiplication in $S^{-1}R$ and since $S$ is multiplicatively closed. Hence $R_p$ is a local ring.$◻$

This finally explains why we speak of localisation.