Abstract AlgebraPolynomial Rings

Polynomial Rings

R[x] holds the polynomials with coefficients in R; over an integral domain deg(fg) = deg f + deg g, but in ℤₙ[x] with n composite degrees can drop.

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You have multiplied polynomials with real coefficients for years. The same rules work with coefficients in any ring, and polynomials over $\mathbb{Z}_{p}$ will build the finite fields. But the coefficient ring matters: over $\mathbb{Z}_{6}$ some familiar facts fail.

Definition. For a commutative ring $R$, $R[x]$ is the set of expressions $a_{n}x^{n} + \cdots + a_{1}x + a_{0}$ with $a_{i} \in R$, added and multiplied as usual, with the coefficients computed in $R$: $$\Big(\sum a_{i}x^{i}\Big)\Big(\sum b_{j}x^{j}\Big) = \sum_{k} \Big(\sum_{i + j = k} a_{i}b_{j}\Big)x^{k}.$$ The degree is the highest power with a nonzero coefficient, the leading coefficient. Over $\mathbb{Z}_{n}$ we write coefficients from $0$ to $n - 1$.

Why it works

Theorem. If $R$ is an integral domain, then $\deg(fg) = \deg f + \deg g$ for nonzero $f, g$, and $R[x]$ is an integral domain.
Proof. If $f$ has leading term $a_{m}x^{m}$ and $g$ has $b_{n}x^{n}$, the highest term of $fg$ is $a_{m}b_{n}x^{m+n}$, and $a_{m}b_{n} \ne 0$ because $R$ has no zero divisors. So $fg \ne 0$ and its degree is $m + n$. $\square$ It follows that the units of $F[x]$, for a field $F$, are exactly the nonzero constants: if $fg = 1$, the degrees add to $0$.

Worked example: Multiplying in $\mathbb{Z}_{5}[x]$

Compute $(3x + 4)(2x^{2} + x + 3)$ in $\mathbb{Z}_{5}[x]$.
  1. Over the integers: $6x^{3} + 3x^{2} + 9x + 8x^{2} + 4x + 12 = 6x^{3} + 11x^{2} + 13x + 12$.
  2. Reduce each coefficient mod $5$: $x^{3} + x^{2} + 3x + 2$.

Answer: $x^{3} + x^{2} + 3x + 2$.

Worked example: Degrees can drop

Compute $(2x + 1)(3x + 1)$ in $\mathbb{Z}_{6}[x]$.
  1. $6x^{2} + 2x + 3x + 1 = 6x^{2} + 5x + 1$.
  2. $6 \equiv 0$, so the product is $5x + 1$, of degree $1$, not $2$.

Answer: $5x + 1$: the leading coefficients $2$ and $3$ are zero divisors.

Worked example: A strange unit

Show that $1 + 2x$ is a unit in $\mathbb{Z}_{4}[x]$.
  1. $(1 + 2x)(1 + 2x) = 1 + 4x + 4x^{2} = 1$ in $\mathbb{Z}_{4}[x]$.
  2. So $1 + 2x$ is its own inverse, a unit of degree $1$. This cannot happen over a field.

Answer: $(1 + 2x)^{2} = 1$.

Counting. Over $\mathbb{Z}_{p}$ a polynomial of degree at most $d$ has $d + 1$ coefficients, each with $p$ choices, so there are $p^{d+1}$ of them (including $0$); exactly $p^{d}$ are monic of degree $d$. These finite counts are what make $\mathbb{Z}_{p}[x]$ such a good laboratory.

Common mistake

Forgetting to reduce coefficients mod $n$, or trusting $\deg(fg) = \deg f + \deg g$ when the coefficients have zero divisors. In $\mathbb{Z}_{4}[x]$, $(2x + 1)(2x + 1) = 1$: two degree-1 polynomials with a product of degree $0$. The degree formula needs an integral domain such as $\mathbb{Z}$, $\mathbb{Q}$ or $\mathbb{Z}_{p}$.

Key idea

$R[x]$ adds and multiplies polynomials with coefficients computed in $R$. Over an integral domain degrees add and $R[x]$ is again an integral domain; over $\mathbb{Z}_{n}$ with $n$ composite, degrees can drop and nonconstant units such as $1 + 2x$ in $\mathbb{Z}_{4}[x]$ appear.

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.

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Sources and further reading

  • Thomas W. Judson, Abstract Algebra: Theory and Applications (open textbook), chapter 17, Polynomials (section 17.1, Polynomial Rings)
  • John A. Beachy and William D. Blair, Abstract Algebra, online study guide (Northern Illinois University), section 4.1, Fields; section 4.2, Polynomials
  • MIT OpenCourseWare 18.702 Algebra II (Spring 2011), lecture notes on polynomial rings
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