Polynomial Algebras

Introduction

The polynomial algebra $k[x_1,\dots,x_n]$ is the free commutative associative algebra on $n$ generators over a field $k$. It is the universal home of the operations of evaluation and substitution, its one-variable case is the model of a principal ideal domain, and its quotients by a single polynomial produce exactly the finite-dimensional commutative algebras over $k$ that are generated by one element. This article develops the algebra and its universal property, the divisibility theory and the root theorems in one variable, the ideal theory of $k[x]$ and of $k[x_1,\dots,x_n]$, and the construction of the two-dimensional algebras of the corpus as quotients.

The ground ring is a field $k$ unless stated otherwise; several results hold for any commutative ring $R$, and where that is so it is said. The free non-commutative algebra on $n$ generators, written $k\langle x_1,\dots,x_n\rangle$, is a different object and is treated in Tensor Powers and the Free Algebra; the present article concerns the commutative case only. The general theory of ideals and quotients used below is that of Ideals and Quotients of Algebras.

The Polynomial Algebra in One Variable

Definition. The polynomial algebra $k[x]$ is the set of formal sums

$$ f = a_0 + a_1x + a_2x^2 + \dots + a_dx^d, \qquad a_j \in k, $$

with finitely many nonzero coefficients, with addition defined coefficientwise and multiplication defined by

$$ \Bigl(\sum_i a_ix^i\Bigr)\Bigl(\sum_j b_jx^j\Bigr) = \sum_m \Bigl(\sum_{i+j=m} a_ib_j\Bigr)x^m . $$

The monomials $1, x, x^2, \dots$ form a $k$-basis, so $k[x]$ is infinite-dimensional over $k$, commutative, associative and unital with unit the constant polynomial $1$.

Definition. For $f \neq 0$ the largest $d$ with $a_d \neq 0$ is the degree $\deg f$, and $a_d$ is the leading coefficient; the degree of $0$ is $-\infty$ by convention. The polynomial is monic when $a_d = 1$, and a constant when $d = 0$.

Proposition (degree). For $f, g \in k[x]$ over a field, $\deg(fg) = \deg f + \deg g$, and $k[x]$ is an integral domain.

Proof. The coefficient of $x^{d_f + d_g}$ in $fg$ is the product of the leading coefficients of $f$ and $g$, which is nonzero in a field; all higher coefficients vanish. Hence the leading coefficient of $fg$ does not vanish and the degree is the sum. A product of nonzero polynomials is then nonzero, so there are no zero divisors.

Corollary (units). The units of $k[x]$ are exactly the nonzero constants, $k[x]^\times = k^\times$.

Proof. If $fg = 1$ then $\deg f + \deg g = 0$, so $f$ and $g$ are constants; conversely every nonzero constant is a unit.

The Universal Property

Theorem (universal property). For every commutative $k$-algebra $A$ and every element $a \in A$ there is a unique $k$-algebra homomorphism

$$ \operatorname{ev}_a : k[x] \longrightarrow A, \qquad \operatorname{ev}_a(f) = f(a), $$

the evaluation at $a$; and the map $a \mapsto \operatorname{ev}_a$ is a bijection

$$ \operatorname{Hom}_{k\text{-alg}}(k[x], A) \cong A . $$

Proof. A $k$-algebra homomorphism is determined by the image of $x$, because $k$ is fixed and the monomials generate; conversely, given $a$ the assignment $x \mapsto a$ extends to a homomorphism since the monomials are linearly independent and products of monomials multiply by adding exponents. The two constructions are inverse to each other.

Corollary. $k[x]$ is the free commutative $k$-algebra on one generator, and it is the initial object among commutative $k$-algebras equipped with a chosen element: for every such pair $(A,a)$ there is a unique homomorphism $k[x] \to A$ carrying $x$ to $a$.

Example (substitution). The universal property applied to $A = k[x]$ and $a = g \in k[x]$ gives the substitution homomorphism $\operatorname{ev}_g$, which sends $f(x)$ to $f(g(x))$. Substitution is a $k$-algebra endomorphism, but it is usually not an automorphism: for $g = x^2$ the image omits $x$, so $\operatorname{ev}_g$ is not surjective.

Divisibility and Roots

Theorem (division algorithm). Let $f, g \in k[x]$ with $g \neq 0$. Then there exist unique $q, r \in k[x]$ with

$$ f = qg + r, \qquad \deg r < \deg g . $$

Proof. If $\deg f < \deg g$ take $q = 0$, $r = f$. Otherwise let $a$ and $b$ be the leading coefficients of $f$ and $g$ and let $m = \deg f - \deg g \geq 0$; the polynomial $f - (a/b)x^mg$ has degree less than $\deg f$, and induction on the degree produces $q$ and $r$. For uniqueness, suppose $qg + r = q'g + r'$ with $\deg r, \deg r' < \deg g$; then $(q - q')g = r' - r$. If $q \neq q'$ the left side has degree $\deg(q-q') + \deg g \geq \deg g$, while the right side has degree less than $\deg g$, a contradiction; hence $q = q'$ and then $r = r'$.

Theorem (remainder and factor). For $f \in k[x]$ and $a \in k$,

$$ f = (x-a)q + f(a) $$

for a unique $q \in k[x]$; consequently $f(a) = 0$ if and only if $(x-a)$ divides $f$.

Proof. Divide $f$ by the monic polynomial $x - a$: the remainder has degree less than $1$, hence is a constant $c$, and evaluating $f = (x-a)q + c$ at $x = a$ gives $c = f(a)$. The factor statement is immediate.

Corollary (number of roots). If $f \neq 0$ and $\deg f = d$, then $f$ has at most $d$ roots in any integral domain containing $k$.

Proof. If $a$ is a root, $f = (x-a)q$ with $\deg q = d-1$; by induction $q$ has at most $d-1$ roots, and every root of $f$ is $a$ or a root of $q$ since the ambient ring is a domain.

The bound fails over a ring with zero divisors, and this failure is precisely what allows the quotient constructions of the corpus: in $k[x]/(x^2)$ the class of $x$ is a root of $t^2$ of multiplicity two, and in $\mathbb{R}[x]/(x^2-1)$ the element $j$ is a root of $t^2 - 1$ distinct from $\pm 1$, so a degree-two polynomial has more than two roots.

Ideals and Quotients

Theorem ($k[x]$ is a principal ideal domain). Every ideal $I$ of $k[x]$ is $(f)$ for some polynomial $f$, unique up to multiplication by a unit if $I \neq 0$.

Proof. If $I \neq 0$, choose $f \in I$ of least degree; for any $g \in I$ the division algorithm gives $g = qf + r$ with $\deg r < \deg f$, and $r = g - qf \in I$, so $r = 0$ by minimality of $\deg f$. Hence $I = (f)$. Uniqueness up to units follows from $(f) = (g)$ implying $f$ and $g$ divide each other, hence are scalar multiples.

Corollary (maximal ideals and fields). For $f \neq 0$ the following are equivalent: $(f)$ is maximal; $(f)$ is prime; $f$ is irreducible. In that case $k[x]/(f)$ is a field, of degree $\dim_k k[x]/(f) = \deg f$ over $k$.

Proof. The quotient $k[x]/(f)$ is an integral domain if and only if $(f)$ is prime, and a field if and only if $(f)$ is maximal; in a principal ideal domain the two notions coincide, and $k[x]/(f)$ is a domain exactly when $f$ is irreducible, since a factorisation $f = gh$ with both factors of lower degree produces zero divisors. The dimension statement is that the images of $1, x, \dots, x^{d-1}$ form a basis, by the division algorithm.

Example (the two-dimensional algebras). The three two-dimensional commutative unital real algebras are quotients by a quadratic:

$$ \mathbb{C} = \frac{\mathbb{R}[x]}{(x^2+1)}, \qquad \mathbb{D} = \frac{\mathbb{R}[x]}{(x^2-1)}, \qquad \mathbb{D}' = \frac{\mathbb{R}[x]}{(x^2)} . $$

The first is a field because $x^2+1$ is irreducible over $\mathbb{R}$; the second is $\mathbb{R}\times\mathbb{R}$ because $x^2-1 = (x-1)(x+1)$ is a product of distinct linear factors; the third has a nilpotent because $x^2$ is a square. These are the three cases, and they exhaust the two-dimensional cases by the same discriminant argument.

Theorem (Chinese remainder, standard). If $f = f_1\cdots f_r$ with the $f_i$ pairwise coprime, then

$$ \frac{k[x]}{(f)} \cong \prod_{i=1}^{r} \frac{k[x]}{(f_i)} $$

as $k$-algebras. In particular, $k[x]/(f)$ is a product of field extensions of $k$ if and only if $f$ is squarefree; the factors are then the fields $k[x]/(f_i)$ for the distinct irreducible factors $f_i$ of $f$.

The decomposition of $\mathbb{D}$ into $\mathbb{R}\times\mathbb{R}$ is the case $f = x^2 - 1 = (x-1)(x+1)$, and the failure of $\mathbb{D}'$ to decompose is the case $f = x^2$, where the two factors coincide.

Several Variables

Definition. The polynomial algebra in $n$ variables $k[x_1,\dots,x_n]$ is the commutative $k$-algebra with $k$-basis the monomials $x_1^{\alpha_1}\cdots x_n^{\alpha_n}$, $\alpha_i \geq 0$, and multiplication by adding exponent vectors. Its elements are finite sums

$$ f = \sum_{\alpha} c_\alpha\, x^\alpha, \qquad c_\alpha \in k, \quad x^\alpha = x_1^{\alpha_1}\cdots x_n^{\alpha_n}, $$

and it is an integral domain.

Theorem (universal property). For every commutative $k$-algebra $A$ there is a natural bijection

$$ \operatorname{Hom}_{k\text{-alg}}\bigl(k[x_1,\dots,x_n], A\bigr) \cong A^n, \qquad \varphi \longmapsto (\varphi(x_1), \dots, \varphi(x_n)). $$

Proof. As in the one-variable case, a homomorphism is determined by the images of the generators, and any $n$-tuple of elements defines one by evaluating the monomials.

Theorem (Hilbert basis theorem, standard). If $k$ is a field, then every ideal of $k[x_1,\dots,x_n]$ is finitely generated; the algebra is therefore Noetherian.

Remark (loss of principality). For $n \geq 2$ the algebra $k[x_1,\dots,x_n]$ is not a principal ideal domain. The ideal $(x_1, x_2)$ of $k[x_1,x_2]$ is not principal: if it were $(p)$ for a single $p$, then $p$ would divide both $x_1$ and $x_2$; since $x_1$ and $x_2$ are irreducible and not associates, $p$ would be a unit, and then $(p) = k[x_1,x_2] \neq (x_1,x_2)$. The ideal $(x_1,x_2)$ is maximal, with quotient $k$, and it is the common vanishing locus of the two coordinate functions at the origin.

Theorem (Nullstellensatz, standard). If $k$ is algebraically closed, the maximal ideals of $k[x_1,\dots,x_n]$ are exactly the ideals $(x_1-a_1,\dots,x_n-a_n)$ for points $(a_1,\dots,a_n) \in k^n$, and for any ideal $I$ the radical of $I$ is the set of polynomials vanishing on the common zero set of $I$.

The polynomial algebra is thus the algebraic counterpart of affine $n$-space: its points are the $k$-algebra homomorphisms to $k$, by the universal property, and its ideals record the subvarieties.

Derivations and Automorphisms

Theorem (derivations). The $k$-derivations of $k[x_1,\dots,x_n]$ form a free $k[x_1,\dots,x_n]$-module of rank $n$,

$$ \operatorname{Der}_k\bigl(k[x_1,\dots,x_n]\bigr) = \bigoplus_{i=1}^{n} k[x_1,\dots,x_n]\,\partial_i, $$

where $\partial_i = \partial/\partial x_i$. In one variable, $\operatorname{Der}_k(k[x]) = k[x]\,\partial_x$.

Proof. Every $\sum_i f_i\partial_i$ is a derivation, since the partial derivatives are. Conversely, let $\delta$ be a derivation and put $f_i = \delta(x_i)$; then $\eta = \delta - \sum_i f_i\partial_i$ is a derivation with $\eta(x_i) = 0$ for all $i$. Since $\eta$ obeys the Leibniz rule and vanishes on the generators and on $k$, it vanishes on every monomial, hence on all of $k[x_1,\dots,x_n]$. So $\delta = \sum_i f_i\partial_i$, and the expression is unique because the $\partial_i$ are linearly independent over the algebra.

Example (the Euler derivation). The derivation $\vartheta = \sum_i x_i\partial_i$ acts on monomials by $\vartheta(x^\alpha) = |\alpha| x^\alpha$, where $|\alpha| = \sum_i\alpha_i$. Over $k = \mathbb{R}$ or $\mathbb{C}$ it is the infinitesimal generator of the one-parameter group of algebra automorphisms $x_i \mapsto \lambda x_i$, $\lambda = e^{t} \in k^\times$, whose derivative at $\lambda = 1$ is $\vartheta$; over a general field the one-parameter group need not be available, and only the derivation is available. It is not inner, because $k[x_1,\dots,x_n]$ is commutative and therefore has no nonzero inner derivations.

Proposition (automorphisms in one variable). Every $k$-algebra automorphism of $k[x]$ is of the form

$$ \varphi(x) = ax + b, \qquad a \in k^\times, \quad b \in k, $$

so that $\operatorname{Aut}_k(k[x])$ is the affine group of the line.

Proof. An automorphism $\varphi$ is determined by $g = \varphi(x)$, and $\varphi$ is surjective exactly when $g$ generates $k[x]$ as a $k$-algebra. If $\deg g = d \geq 1$, every element of $k[g]$ is a $k$-linear combination of $1, g, g^2, \dots$, hence has degree a multiple of $d$; if $d \geq 2$ then $x \notin k[g]$, so $g$ does not generate. Hence $d = 1$ and $g = ax+b$ with $a \neq 0$; conversely every such $g$ generates $k[x]$ and defines an automorphism.

For $n \geq 2$ the automorphism group is much larger; already the triangular maps $x_i \mapsto x_i + h_i(x_1,\dots,x_{i-1})$ are automorphisms, and the structure of $\operatorname{Aut}_k(k[x_1,x_2])$ is known (it is generated by the affine and triangular maps), whereas for $n \geq 3$ the question of whether every automorphism is a composition of these is a substantial open problem of affine algebraic geometry.

Comparison with the Free Algebra

The commutative polynomial algebra is not the free associative algebra. In $k\langle x_1,\dots,x_n\rangle$ the monomials are words in the generators and the product is concatenation, so $x_1x_2 \neq x_2x_1$; the polynomial algebra is the quotient of the free algebra by the two-sided ideal generated by the commutators $x_ix_j - x_jx_i$:

$$ k[x_1,\dots,x_n] \cong \frac{k\langle x_1,\dots,x_n\rangle}{(x_ix_j - x_jx_i : 1 \leq i < j \leq n)} . $$

This is the general quotient construction of Quotients of the Tensor Algebra, and it is the reason the polynomial algebra is called the symmetric algebra of an $n$-dimensional vector space; that symmetric-algebra viewpoint belongs to category 06 and is not pursued here.

Summary

The polynomial algebra $k[x]$ has $k$-basis the monomials $x^m$, is a commutative integral domain, and has units $k^\times$ and degree function satisfying $\deg(fg) = \deg f + \deg g$. Its universal property is that $k$-algebra homomorphisms $k[x] \to A$ correspond bijectively to elements of $A$ by evaluation, and the same statement in $n$ variables gives $\operatorname{Hom}_{k\text{-alg}}(k[x_1,\dots,x_n], A) \cong A^n$. The division algorithm and the remainder theorem give $f = (x-a)q + f(a)$, so that $a$ is a root of $f$ exactly when $(x-a)$ divides $f$, and a nonzero polynomial of degree $d$ has at most $d$ roots in a domain. Every ideal of $k[x]$ is principal, and $k[x]/(f)$ is a field exactly when $f$ is irreducible, of dimension $\deg f$; the Chinese remainder theorem decomposes $k[x]/(f)$ over the coprime factorisation of $f$. The two-dimensional real algebras are the quotients by $x^2+1$, $x^2-1$ and $x^2$. In $n$ variables the algebra is Noetherian but not principal for $n \geq 2$, its maximal ideals over an algebraically closed field are the ideals of points, and it is the quotient of the free algebra by the commutators, hence the symmetric algebra of $k^n$. Its $k$-derivations form the free module $\bigoplus_i k[x_1,\dots,x_n]\partial_i$ on the partial derivatives, and the automorphisms of $k[x]$ are the affine maps $x\mapsto ax+b$, so that $\operatorname{Aut}_k(k[x])$ is the affine group.

Summary of Notation

Symbol	Meaning
$k[x]$	Polynomial algebra in one variable
$k[x_1,\dots,x_n]$	Polynomial algebra in $n$ variables
$k\langle x_1,\dots,x_n\rangle$	Free associative algebra, non-commutative
$x^\alpha$	Monomial $x_1^{\alpha_1}\cdots x_n^{\alpha_n}$
$\deg f$	Degree, with $\deg 0 = -\infty$
$\operatorname{ev}_a(f) = f(a)$	Evaluation at $a$
$(f)$	Principal ideal generated by $f$
$k[x]/(f)$	Quotient algebra, dimension $\deg f$
$(x_1,\dots,x_n)$	Maximal ideal of the origin
$\partial_i = \partial/\partial x_i$	Partial derivative
$\operatorname{Der}_k(k[x_1,\dots,x_n])$	Module of $k$-derivations
$\vartheta = \sum_i x_i\partial_i$	Euler derivation
$\operatorname{Aut}_k(k[x])$	Affine group of the line