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A Refresher in Commutative Rings

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

A commutative ring with identity (or simply ring) (R,+,)(R,+,\cdot) is an abelian group (R,+)(R,+) equipped with another binary operation \cdot (usually written as concatenation, ie. ab=aba\cdot b=ab), which is commutative, associative, and distributive over ++ (ie. a(b+c)=ab+aca(b+c)=ab+ac), with identity element 11. An element of RR is called a unit if it has a multiplicative inverse.

An integral domain is a ring with no zero divisors, ie. if ab=0ab=0 then a=0a=0 or b=0b=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 II in a ring RR is an additive subgroup closed under RR-multiplication, ie.

rR,aI    raI.r\in R,\,a\in I\implies ra\in I.

II is a proper ideal if IRI\neq R.

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

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

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

II is a prime ideal of RR if

abI    aI or bI.ab\in I\implies a\in I\text{ or }b\in I.

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

I+J{a+b:aI,bJ},IJ{a1b1++anbn:akI,bkJ}. \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+JI+J, IJIJ, and IJI\cap J are all ideals, which satisfy the inclusions

IJIJ,I,JI+J.IJ\subseteq I\cap J,\quad I,J\subseteq I+J.

Quotient rings

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

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

Proposition: Let IRI\subseteq R be an ideal. Then:

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

Corollary: Every maximal ideal is prime.  \qed

Coprime ideals

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

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

so IJ=IJI\cap J=IJ. Also, if II is coprime to each of J1,J2J_1,J_2, then

R=(I+J1)(I+J2)I+J1J2,R=(I+J_1)(I+J_2)\subseteq I+J_1J_2,

so II is coprime to J1J2J_1J_2. Hence by induction we have:

Proposition: If I1,I2,,InI_1,I_2,\ldots,I_n are pairwise coprime ideals, then I1In=I1InI_1\cap\cdots\cap I_n=I_1\cdots I_n.  \qed

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

φ^:R/(IJ)R/I×R/J.\widehat\varphi:R/(I\cap J)\to R/I\times R/J.

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

Chinese Remainder Theorem: If I1,I2,,InI_1,I_2,\ldots,I_n are pairwise coprime ideals, then

R/(I1In)R/I1××R/In.  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,m)(R,\mf m) is a local ring if RR is a local ring with maximal ideal m\mf m. In this case, the residue field of RR is k=R/mk=R/\mf m, and we also say that (R,m,k)(R,\mf m,k) is a local ring.

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

Integral domains, PIDs, UFDs

Let RR be an integral domain throughout this section.

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

ab=ac    b=cab=ac\implies b=c

for any a,b,cRa,b,c\in R with a0a\neq0.

An element aRa\in R is prime if (a)(a) is a prime ideal. An element aRa\in R is irreducible if aa is not a unit, and for any factorisation a=bca=bc, either bb or cc is a unit. Equivalently, aa is irreducible if (a)(a) is maximal among proper principal ideals.

Proposition: Every prime in RR is irreducible.  \qed

The standard counterexample for the converse is the element 22 in the integral domain Z[5]\bb Z[\sqrt{-5}], which is irreducible (by considering the norm), but not prime (since 1±5∉(2)1\pm\sqrt{-5}\not\in(2) but (1+5)(15)=6=2×3(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 aRa\in R has two factorisations into products of primes, say

a=p1p2pn=p1p2pm,a=p_1p_2\cdots p_n=p_1'p_2'\cdots p_m',

then n=mn=m, and after some reordering of the pjp_j’ we have (pj)=(pj)(p_j)=(p_j’) for all jj.

Proof: p1p_1 prime implies that some pj(p1)p_j’\in(p_1), say p1=up1p_1’=up_1. But p1p_1’ is prime, hence irreducible, so uu is a unit. Then cancellation gives

p2pn=up2pm,p_2\cdots p_n=up_2'\cdots p_m',

so we replace p2p_2’ by up2up_2’ and finish by induction.  \qed

A principal ideal domain (PID) is an integral domain in which every ideal is principal. For instance, Z\bb Z is a PID because every nonzero ideal is generated by its smallest positive element. Also, if FF is a field, then the polynomial ring F[x]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 aRa\in R is irreducible then (a)(a) is maximal among proper principal ideals, so (a)(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 aRa\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 aa 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

(a1)(a2)(a_1)\subsetneq(a_2)\subsetneq\cdots

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

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