integral domains整域せいいき, zero divisors零因子れいいんし, and polynomial rings多項式環たこうしきかん

In a ring環かん, the product of two nonzero elements may become 0. This phenomenon means that information is lost through multiplication. An element that causes this phenomenon is called a zero divisor零因子れいいんし.

1Zero divisors

A nonzero element $$a$$a of a ring $$R$$R is a zero divisor零因子れいいんし if there exists a nonzero element $$b$$b such that

or

For example, in $$\mathbb{Z}/6\mathbb{Z}$$\mathbb Z/6\mathbb Z,

Both $$\left[2\right]$$[2] and $$\left[3\right]$$[3] are nonzero, so they are zero divisors零因子れいいんし.

2Integral domains

An integral domain整域せいいき is a commutative ring with $$0\ne 1$$0\ne1 in which the product of nonzero elements is never 0. Equivalently, for all $$a,b\in R$$a,b\in R,

holds.

The integer ring $$\mathbb{Z}$$\mathbb Z is an integral domain整域せいいき. The real field体たい $$\mathbb{R}$$\mathbb R is also an integral domain整域せいいき. On the other hand, $$\mathbb{Z}/6\mathbb{Z}$$\mathbb Z/6\mathbb Z is not an integral domain整域せいいき.

3Theorem: cancellation in an integral domain整域せいいき

In an integral domain整域せいいき $$R$$R, if $$a,b,c\in R$$a,b,c\in R and $$a\ne 0$$a\ne0, then

Proof. If $$ab=ac$$ab=ac, distributivity gives

Since $$a\ne 0$$a\ne0 and an integral domain has no zero divisors, $$b-c=0$$b-c=0, hence $$b=c$$b=c.

No division is used here. We are not dividing by a nonzero element; we are using the absence of zero divisors零因子れいいんし to cancel.

4Polynomial rings

Let $$R$$R be a commutative ring. A polynomial多項式たこうしき over $$R$$R is a formal finite sum of the form

Here “formal” means that the polynomial is the coefficient sequence $$(a_0,a_1,\dots )$$(a_0,a_1,\ldots) with all but finitely many coefficients equal to 0, rather than the function obtained by substituting values for $$x$$x. The set of all such polynomials is denoted by

Define addition coefficientwise and multiplication by distributivity:

With these operations, $$R\left[x\right]$$R[x] is a ring. Its additive identity is the zero polynomial, and its multiplicative identity is the constant polynomial 1. For example, $$\mathbb{Z}\left[x\right]$$\mathbb Z[x] is the ring of polynomials with integer coefficients.

We verify this claim through coefficients. If two coefficient sequences have finite support, then their sum and the product defined above also have finite support, so both operations are closed on $$R\left[x\right]$$R[x]. Associativity and commutativity of addition follow coefficientwise from $$R$$R, and the additive inverse of $$\sum _i{a}_i{x}^i$$\sum_i a_ix^i is $$\sum _i(-a_i)x^i$$\sum_i(-a_i)x^i.

Now let $$f=\sum _i{a}_i{x}^i$$f=\sum_i a_ix^i, $$g=\sum _j{b}_j{x}^j$$g=\sum_j b_jx^j, and $$h=\sum _{\ell }{c}_{\ell }{x}^{\ell }$$h=\sum_\ell c_\ell x^\ell. The coefficient of $$x^n$$x^n in both $$\left(fg\right)h$$(fg)h and $$f\left(gh\right)$$f(gh) is

so multiplication is associative. The constant polynomial 1 is the multiplicative identity. For example, distributivity follows because the coefficient of $$x^n$$x^n in $$f(g+h)$$f(g+h) is

The other distributive law is similar. Finally, multiplication is commutative because $$R$$R is commutative. Thus $$R\left[x\right]$$R[x] is indeed a commutative ring.

For a nonzero polynomial $$f\left(x\right)=a_0+\cdots +a_m{x}^m$$f(x)=a_0+\cdots+a_mx^m with $$a_m\ne 0$$a_m\ne0, the integer $$m$$m is the degree次数じすう $$\text{[PARSE ERROR: Undefined("Command("deg")")]}f$$\deg f, and $$a_m$$a_m is the leading coefficient最高次係数さいこうじけいすう. We leave the degree of the zero polynomial undefined here.

5Theorem: a polynomial ring over an integral domain is an integral domain

If $$R$$R is an integral domain整域せいいき, then $$R\left[x\right]$$R[x] is also an integral domain. Moreover, for nonzero polynomials $$f,g\in R\left[x\right]$$f,g\in R[x],

Proof. Let $$\text{[PARSE ERROR: Undefined("Command("deg")")]}f=m$$\deg f=m and $$\text{[PARSE ERROR: Undefined("Command("deg")")]}g=n$$\deg g=n, and let $$a_m$$a_m and $$b_n$$b_n be the leading coefficients of $$f$$f and $$g$$g, respectively. Since $$f$$f and $$g$$g are nonzero, $$a_m\ne 0$$a_m\ne0 and $$b_n\ne 0$$b_n\ne0. Because $$R$$R is an integral domain, $$a_m{b}_n\ne 0$$a_mb_n\ne0.

The coefficient of $$x^{m+n}$$x^{m+n} in $$fg$$fg is $$a_m{b}_n$$a_mb_n, and no term of higher degree occurs. Thus $$fg\ne 0$$fg\ne0 and $$\text{[PARSE ERROR: Undefined("Command("deg")")]}\left(fg\right)=m+n$$\deg(fg)=m+n. Therefore a product of nonzero polynomials is nonzero; since $$R\left[x\right]$$R[x] is commutative and has $$0\ne 1$$0\ne1, it is an integral domain.

6Relation with fields体たい

The formal definition of a field体たい is given in the next lecture. In this section, as a preview, we use the following minimum meaning: a field is a commutative ring with $$0\ne 1$$0\ne1 in which every nonzero element has a multiplicative inverse.

Every field体たい is an integral domain整域せいいき. If $$a\ne 0$$a\ne0 and $$ab=0$$ab=0, multiplying by $$a^{-1}$$a^{-1} gives

This step uses $$a^{-1}$$a^{-1}, so it is necessary to check that $$a\ne 0$$a\ne0.

7Exercise link

At this point, Problems 1 and 4 in the following exercise are ready. After studying fields, continue with Problems 2 and 3 and the proof exercise; after studying finite fields, complete Problem 5.

8Summary

A zero divisor零因子れいいんし is a nonzero element that annihilates another nonzero element. In an integral domain整域せいいき, such collapse does not occur, so nonzero factors can be cancelled. Polynomial addition and multiplication make $$R\left[x\right]$$R[x] a ring; when $$R$$R is an integral domain, leading coefficients show that $$R\left[x\right]$$R[x] is also an integral domain and that degrees add under multiplication. Every field is an integral domain, and the next lecture develops fields formally.