A Refresher in Commutative Rings

31 Dec 2018

• algebraic number theory
• commutative algebra

In this post, we review the basic results from a first course in commutative ring theory.

Rings

The archetypal example of a ring is the set of integers $\bb Z$, with the operations of addition and multiplication. Note that $(\bb Z,+)$ is an abelian group, while $(\bb Z,\times)$ or even $(\bb Z\backslash\{0\},\times)$ is not, since most integers do not have multiplicative inverses.

A commutative ring with identity (or simply ring) $(R,+,\cdot)$ is an abelian group $(R,+)$ equipped with another binary operation $\cdot$ (usually written as concatenation, ie. $a\cdot b=ab$), which is commutative, associative, and distributive over $+$ (ie. $a(b+c)=ab+ac$), with identity element $1$. An element of $R$ is called a unit if it has a multiplicative inverse.

An integral domain is a ring with no zero divisors, ie. if $ab=0$ then $a=0$ or $b=0$. A field is a ring in which every nonzero element is a unit.

Proposition: Every field is an integral domain. $\qed$

Ideals

An ideal $I$ in a ring $R$ is an additive subgroup closed under $R$-multiplication, ie.

$$r\in R,\,a\in I\implies ra\in I.$$

$I$ is a proper ideal if $I\neq R$.

For instance, the principal ideal generated by $a\in R$ is defined by

$$(a)=aR\coloneqq\{ar\,:\,r\in R\}.$$

$I$ is a maximal ideal of $R$ if there is no ideal $J$ with $I\subsetneq J\subsetneq R$. By Zorn’s lemma, we can show that every proper ideal is contained in some maximal ideal.

$I$ is a prime ideal of $R$ if

$$ab\in I\implies a\in I\text{ or }b\in I.$$

The sum and product of two ideals $I,J\subseteq R$ are defined by

$$\begin{aligned} I+J&\coloneqq\{a+b\,:\,a\in I,\,b\in J\},\\ IJ&\coloneqq\{a_1b_1+\cdots+a_nb_n\,:\,a_k\in I,\,b_k\in J\}. \end{aligned}$$

We can check that $I+J$, $IJ$, and $I\cap J$ are all ideals, which satisfy the inclusions

$$IJ\subseteq I\cap J,\quad I,J\subseteq I+J.$$

Quotient rings

It is easy to show that the (additive) quotient group $R/I$ is in fact also a ring (called the quotient ring), with operations given by

$$(a+I)+(b+I)=a+b+I,\quad(a+I)(b+I)=ab+I.$$

Proposition: Let $I\subseteq R$ be an ideal. Then:

1. $I$ is maximal if and only if $R/I$ is a field;
2. $I$ is prime if and only if $R/I$ is an integral domain. $\qed$

Corollary: Every maximal ideal is prime. $\qed$

Coprime ideals

Two ideals $I,J\subseteq R$ are coprime or relatively prime if $I+J=R$. Note that if $I,J$ are coprime ideals then

$$I\cap J=(I\cap J)(I+J)\subseteq IJ\subseteq I\cap J,$$

so $I\cap J=IJ$. Also, if $I$ is coprime to each of $J_1,J_2$, then

$$R=(I+J_1)(I+J_2)\subseteq I+J_1J_2,$$

so $I$ is coprime to $J_1J_2$. Hence by induction we have:

Proposition: If $I_1,I_2,\ldots,I_n$ are pairwise coprime ideals, then $I_1\cap\cdots\cap I_n=I_1\cdots I_n$. $\qed$

For any two ideals $I,J$, form the product of the two quotient maps to obtain a ring homomorphism $\varphi:R\to R/I\times R/J$. This has kernel $I\cap J$, so we get the injective ring homomorphism

$$\widehat\varphi:R/(I\cap J)\to R/I\times R/J.$$

If $I,J$ are coprime, there exists $a\in I$, $b\in J$ such that $a+b=1$. Now for any $x,y\in R$, note that $\varphi(bx+ay)=(x+I,y+J)$. Hence $\varphi$ is surjective, thus so is $\widehat\varphi$, ie. $\widehat\varphi$ is an isomorphism. By induction, we get:

Chinese Remainder Theorem: If $I_1,I_2,\ldots,I_n$ are pairwise coprime ideals, then

$$R/(I_1\cap\cdots\cap I_n)\cong R/I_1\times\cdots\times R/I_n.\ \qed$$

Local rings

A local ring is a ring with a unique maximal ideal. We say that $(R,\mf m)$ is a local ring if $R$ is a local ring with maximal ideal $\mf m$. In this case, the residue field of $R$ is $k=R/\mf m$, and we also say that $(R,\mf m,k)$ is a local ring.

Note that if $(R,\mf m)$ is a local ring, then every $x\in R\backslash\mf m$ is not contained in any maximal ideal of $R$, ie. $x$ is a unit. Conversely, if the set of non-units of $R$ form an ideal $\mf m$, then $\mf m$ contains all proper ideals of $R$, and hence is the unique maximal ideal of $R$, so $R$ is a local ring.

Integral domains, PIDs, UFDs

Let $R$ be an integral domain throughout this section.

The most important elementary property of integral domains is the cancellation law:

$$ab=ac\implies b=c$$

for any $a,b,c\in R$ with $a\neq0$.

An element $a\in R$ is prime if $(a)$ is a prime ideal. An element $a\in R$ is irreducible if $a$ is not a unit, and for any factorisation $a=bc$, either $b$ or $c$ is a unit. Equivalently, $a$ is irreducible if $(a)$ is maximal among proper principal ideals.

Proposition: Every prime in $R$ is irreducible. $\qed$

The standard counterexample for the converse is the element $2$ in the integral domain $\bb Z[\sqrt{-5}]$, which is irreducible (by considering the norm), but not prime (since $1\pm\sqrt{-5}\not\in(2)$ but $(1+\sqrt{-5})(1-\sqrt{-5})=6=2\times3$).

The prime factorisation of an element, if it exists, is unique up to unit multiples and the order of factors:

Proposition: If $a\in R$ has two factorisations into products of primes, say

$$a=p_1p_2\cdots p_n=p_1'p_2'\cdots p_m',$$

then $n=m$, and after some reordering of the $p_j’$ we have $(p_j)=(p_j’)$ for all $j$.

Proof: $p_1$ prime implies that some $p_j’\in(p_1)$, say $p_1’=up_1$. But $p_1’$ is prime, hence irreducible, so $u$ is a unit. Then cancellation gives

$$p_2\cdots p_n=up_2'\cdots p_m',$$

so we replace $p_2’$ by $up_2’$ and finish by induction. $\qed$

A principal ideal domain (PID) is an integral domain in which every ideal is principal. For instance, $\bb Z$ is a PID because every nonzero ideal is generated by its smallest positive element. Also, if $F$ is a field, then the polynomial ring $F[x]$ is a PID, because every nonzero ideal is generated by a nonzero element of minimal degree.

Proposition: Every nonzero prime ideal in a PID is maximal. $\qed$

Proposition: Every irreducible element of a PID is prime.

Proof: If $a\in R$ is irreducible then $(a)$ is maximal among proper principal ideals, so $(a)$ is a maximal ideal, hence prime. $\qed$

A unique factorisation domain (UFD) is an integral domain in which every nonzero, non-unit element has a prime factorisation (which must then be essentially unique, as we have shown).

Proposition: Every PID is a UFD.

Proof: Given a nonzero non-unit element $a\in R$, start with the trivial one-term factorisation and repeatedly split a term into a product of two non-unit elements. If this process terminates, we have a factorisation of $a$ into irreducibles, which are primes in a PID.

Otherwise, Kőnig’s lemma implies that every infinite binary tree has an infinite path, so there is a sequence of ideals

$$(a_1)\subsetneq(a_2)\subsetneq\cdots$$

The union of these ideals is a new ideal, say $(a)$; but $a\in(a_n)$ for some $n$, so $(a_k)=(a_n)$ for all $k\geq n$, contradiction. $\qed$