Rings polynomials and fields

Groups describe one operation. Rings coordinate addition and multiplication, allowing us to study integers, matrices and polynomials in one language. Ideals specify when reducing modulo a collection of elements preserves both operations; irreducible polynomials then provide a systematic way to construct new fields.

1 Ring conventions and zero divisors

In this chapter a ring has an abelian additive group, associative multiplication, a multiplicative identity 1, and both distributive laws. Ring homomorphisms preserve 1. Some books omit the identity requirement; that convention changes some statements. A commutative ring has commutative multiplication. An integral domain is a nonzero commutative ring with no zero divisors. A field is a nonzero commutative ring in which every nonzero element has a multiplicative inverse.

The integers are a domain but not a field. $\mathbb Z/6\mathbb Z$ has zero divisors since $[2][3]=[0]$. Matrix rings usually fail commutativity. Every field is a domain: if $ab=0$ with $a\ne0$, multiply by $a^{-1}$ to get $b=0$.

A finite integral domain is a field. For nonzero $a$, the map $x\mapsto ax$ is injective by cancellation, hence surjective on the finite set, so it takes some $x$ to 1. Finiteness is essential, as $\mathbb Z$ shows.

2 Ideals and quotient rings

An ideal $I$ of a commutative ring $R$ is an additive subgroup with $ri\in I$ for all $r\in R$, $i\in I$. The absorption requirement is stronger than being a subring. In $\mathbb Z$, every ideal is $n\mathbb Z$: take the least positive element and apply division with remainder, with the zero ideal handled separately.

On additive cosets, define $(a+I)(b+I)=ab+I$. If $a'=a+i$ and $b'=b+j$, their product differs from $ab$ by $aj+bi+ij\in I$. Thus multiplication is well defined. The quotient $R/I$ is a ring. For a ring homomorphism $\phi$, its kernel is an ideal, and the same coset-map proof as for groups gives $R/\ker\phi\cong\phi(R)$ as rings.

Example. Reduction $\mathbb Z\to\mathbb Z/n\mathbb Z$ has kernel $n\mathbb Z$. A class $[a]$ is invertible exactly when $\gcd(a,n)=1$: Bézout's identity gives $ua+vn=1$ and hence an inverse $[u]$; an inverse conversely gives an integer combination equal to 1. Therefore $\mathbb Z/n\mathbb Z$ is a field exactly when $n$ is prime, for $n\ge2$.

3 Polynomials over a field

For a field $F$, the polynomial ring $F[x]$ contains finite sums $\sum a_ix^i$. Its multiplication uses coefficient convolution. For nonzero polynomials, degrees add because leading coefficients multiply to a nonzero element.

Division theorem. For $g\ne0$ and any $f$, there exist unique $q,r$ with $f=qg+r$ and either $r=0$ or $\deg r<\deg g$.

Proof. If the leading degree of $f$ is at least that of $g$, cancel its leading term using a scalar multiple of a power of $x$ times $g$. The degree strictly decreases, so repetition stops. If two decompositions exist, $g(q-q')=r'-r$; a nonzero left side has degree at least $\deg g$, while the right side has smaller degree, a contradiction. $\square$

Euclidean division produces a greatest common divisor, and back substitution gives Bézout's identity $uf+vg=\gcd(f,g)$ up to nonzero scalar normalization. A nonconstant polynomial is irreducible if it is not a product of two positive-degree polynomials over the specified field. $x^2+1$ is irreducible over $\mathbb R$ but not over $\mathbb C$; the coefficient field is part of the question.

4 Field construction by an irreducible polynomial

Theorem. For nonconstant $p\in F[x]$, $F[x]/(p)$ is a field if and only if $p$ is irreducible.

Proof. If $p$ is irreducible and $[f]\ne[0]$, then $p$ does not divide $f$, so $\gcd(f,p)=1$. Bézout gives $uf+vp=1$, hence $[u][f]=[1]$. Conversely, if $p=ab$ with both factors of smaller positive degree, the nonzero classes $[a],[b]$ multiply to zero, which cannot happen in a field. $\square$

For example, in $\mathbb F_2[x]$, $p=x^2+x+1$ has no root (evaluate at 0 and 1), hence is irreducible. The quotient has four elements $0,1,\alpha,1+\alpha$, with $\alpha^2=\alpha+1$ and $\alpha(\alpha+1)=1$. This field is not $\mathbb Z/4\mathbb Z$, which has zero divisors.

The quotient and Euclidean steps with hypotheses retained

The ideal discussion in this chapter concerns commutative unital rings. In a noncommutative ring, a ring quotient requires a two-sided ideal, absorbing both $ri$ and $ir$; a one-sided ideal alone is insufficient. For a commutative $R$ and ideal $I$, addition is also independent of representatives: $(a+i)+(b+j)-(a+b)=i+j\in I$. Associativity, distributivity and the identities descend from $R$ by replacing each operation with its coset. The quotient identity is $1+I$. If $I=R$, this is the zero ring; a field quotient therefore requires $I$ proper. For a homomorphism of commutative unital rings $\phi:R\to S$, the map $a+\ker\phi\mapsto\phi(a)$ is well defined and bijective onto the image by the group argument, and it preserves multiplication and the identity by the ring homomorphism laws. This proves the ring isomorphism, including both operations.

Here is the polynomial Bézout argument behind the field construction. For $f,g\in F[x]$ not both zero, first handle a zero input. If $g=0$, take $d=f/\operatorname{lc}(f)$ with Bézout coefficients $(1/\operatorname{lc}(f),0)$; if $f=0$, take $d=g/\operatorname{lc}(g)$ with coefficients $(0,1/\operatorname{lc}(g))$. These monic polynomials divide both inputs and are divisible by every common divisor. When both inputs are nonzero, repeatedly divide the previous two nonzero remainders. Nonnegative degrees strictly decrease, so there is a last nonzero remainder $d$. The final division shows $d$ divides its predecessor; reversing every earlier equation shows $d$ divides both $f$ and $g$. Conversely every common divisor of $f,g$ divides each remainder, hence $d$. Every remainder is a combination $uf+vg$, starting from the initial two and using subtraction of a polynomial multiple at each step. Thus $d=uf+vg$. Dividing by its nonzero leading coefficient makes $d$ monic without changing its divisor property. This is the monic gcd and the required Bézout identity. The coefficient-field assumption permits each leading-term cancellation; it cannot be discarded for an arbitrary coefficient ring.

For nonconstant $p$ of degree $d$, every coset in $F[x]/(p)$ has a unique representative of degree below $d$. Existence is polynomial division. If two such representatives differ by a multiple of $p$, a nonzero difference would have degree both below $d$ and at least $d$, impossible. In particular $[1]\ne[0]$, constants embed a copy of $F$, and for finite $|F|=q$ there are exactly $q^d$ classes. A degree-two or degree-three polynomial is reducible exactly when it has a root in $F$: a proper factorization has a linear factor, and division by $x-a$ has remainder $p(a)$. This justifies the low-degree irreducibility tests in the examples without extending them to degree four or higher.

5 Graded practice

E1 Foundation. Find all units of $\mathbb Z/10\mathbb Z$ and their inverses.

E2 Polynomial calculation. Divide $x^3+2x+1$ by $x^2+1$ over $\mathbb Q$.

E3 Quotients. Is $\mathbb R[x]/(x^2-1)$ a field? Exhibit evidence. Is $\mathbb R[x]/(x^2+1)$ a field?

E4 Proof. Show the kernel of a unital ring homomorphism is an ideal. What happens if its kernel contains 1?

6 Full solutions

S1. The coprime classes are $[1],[3],[7],[9]$. Their inverses are $[1],[7],[3],[9]$ respectively, checked by multiplication modulo 10.

S2. Subtract $x(x^2+1)=x^3+x$ to leave $x+1$. Thus $q=x$, $r=x+1$, whose degree is below 2.

S3. In the first quotient, the nonzero classes $[x-1]$ and $[x+1]$ multiply to zero, so it is not a field. In the second, $x^2+1$ has no real root and a reducible quadratic would have a linear factor, so it is irreducible and the quotient is a field, isomorphic to $\mathbb C$ by $[x]\mapsto i$.

S4. Additive homomorphism properties give an additive subgroup. For $i$ in the kernel, $\phi(ri)=\phi(r)\phi(i)=0$ and $\phi(ir)=\phi(i)\phi(r)=0$, so both absorption laws hold, even for a noncommutative source ring. If 1 is in the kernel, every $r=r1$ is, so the kernel is all of $R$. With a nonzero target and a unital map this cannot happen because $\phi(1)=1\ne0$.

Summary

Rings combine additive and multiplicative structure. Ideals are exactly the right compatibility objects for quotient multiplication. Polynomial division over a field leads to Bézout identities, and irreducibility turns polynomial quotients into fields. Always retain the field and ring conventions in each statement.