---
title: Polynomial Rings over Fields
module: Factorization and Polynomial Rings
moduleNumber: 7
lessonNumber: 2
order: 702
summary: >
  When the coefficients form a field, polynomial long division works exactly as
  it does over the rationals, and it works with a unique quotient and remainder.
  That single fact makes F[x] a Euclidean domain, hence a PID and a UFD: every
  ideal is the multiples of one polynomial, roots correspond to linear factors,
  and F[x]/(f) is a field precisely when f is irreducible.
topics: [Factorization and Polynomial Rings]
sources:
  - book: Dummit & Foote
    ref: "Ch. 9 — Polynomial Rings; §9.1 Definitions and Basic Properties, §9.2 Polynomial Rings over Fields I"
  - book: Judson
    ref: "Ch. 17 — Polynomials; §17.1 Polynomial Rings, §17.2 The Division Algorithm"
draft: false
---

Beside $\mathbb{Z}$, the other fundamental
[Euclidean domain](/abstract-algebra/factorization-and-polynomials/euclidean-domains-pids-ufds)
is $F[x]$, polynomials in one variable over a field. Polynomial long division
terminates with a unique quotient and remainder of smaller degree, and that one
fact makes $F[x]$ Euclidean — hence a PID and a UFD, with every ideal the
multiples of a single polynomial, roots matched to linear factors, and
$F[x]/(f)$ a field exactly when $f$ is irreducible. The coefficients must lie in
a _field_: that is what lets a division algorithm exist.

## The polynomial ring

Let $R$ be a commutative ring with $1 \neq 0$.

> **Definition (Polynomial ring).** $R[x]$ is the set of formal sums
> $a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0$ with $n \geq 0$ and each
> $a_i \in R$, added componentwise and multiplied by extending
> $(a x^i)(b x^j) = ab\, x^{i+j}$ through distributivity. If $a_n \neq 0$ the
> polynomial has **degree** $n$, **leading term** $a_n x^n$, and **leading
> coefficient** $a_n$; it is **monic** if $a_n = 1$. The zero polynomial has
> leading coefficient $0$.

The constant polynomials form a copy of $R$ inside $R[x]$. The arithmetic of
degrees is where the coefficient ring first shows through.

> **Proposition (degree and units).** If $R$ is an integral domain and $p, q$
> are nonzero, then
> - $\deg(pq) = \deg p + \deg q$,
> - $R[x]$ and $R$ have the same units, and
> - $R[x]$ is again an integral domain.[^df91]

Each part fails without the domain hypothesis. Over $\mathbb{Z}/4\mathbb{Z}$ the
product $(2x)(2x) = 0$ collapses degree, and $2x + 1$ is a unit because $(2x +
1)^2 = 1$. The domain condition keeps the leading term of a product equal to the
product of the leading terms, and every statement below rests on that fact.

A **reduction homomorphism** transports ideal information from $R$ to $R[x]$.
Given an ideal $I \subseteq R$, reducing every coefficient modulo $I$ is a
surjection $R[x] \to (R/I)[x]$ with kernel $I[x]$, so

$$
R[x]/I[x] \;\cong\; (R/I)[x].
$$

If $I$ is prime in $R$ then $(R/I)[x]$ is a domain, so $I[x]$ is prime in
$R[x]$. Reducing $\mathbb{Z}[x]$ modulo a prime $p$ yields
$\mathbb{F}_p[x]$, the ring that will drive the irreducibility tests of
[the criteria lesson](/abstract-algebra/factorization-and-polynomials/irreducibility-criteria-and-groebner).[^df91]

## The division algorithm over a field

Everything special about $F[x]$ flows from one theorem, and its uniqueness clause
is what distinguishes the field case from the general one.

> **Theorem (Division algorithm in $F[x]$).** Let $F$ be a field. For
> $a(x), b(x) \in F[x]$ with $b(x) \neq 0$ there exist **unique** $q(x), r(x) \in
> F[x]$ with
> $$
> a(x) = q(x)\, b(x) + r(x), \qquad r(x) = 0 \ \text{ or } \ \deg r < \deg b.
> $$
> Consequently $F[x]$ is a Euclidean domain with norm $N(p) = \deg p$.[^df92]

The proof is induction on $n = \deg a$. If $n < m = \deg b$, take $q = 0$ and
$r = a$. Otherwise cancel the leading term: with $a = a_n x^n + \cdots$ and
$b = b_m x^m + \cdots$, the polynomial

$$
a'(x) = a(x) - \frac{a_n}{b_m} x^{n-m}\, b(x)
$$

has degree below $n$. The quotient $a_n/b_m$ exists **because $b_m$ is invertible
in the field $F$** — this is the step that breaks over a general ring. By
induction $a' = q'b + r$, and adding back gives $a = \big(q' + \tfrac{a_n}{b_m}
x^{n-m}\big)b + r$. For uniqueness, if $qb + r = q_1 b + r_1$ then
$b(q - q_1) = r_1 - r$ has degree below $\deg b$ on the right, but degree
$\geq \deg b$ on the left unless $q = q_1$; so $q = q_1$ and $r = r_1$.

$$
% caption: One step of polynomial long division: scale the divisor to match the
% leading term of the dividend, subtract to cancel it, and recurse on a
% strictly-lower-degree remainder.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize,
  box/.style={draw, minimum height=8mm, align=center, inner xsep=3mm}]
\definecolor{acc}{HTML}{4A6FA5}
\node[box] (a) at (0,2.4) {dividend, degree $n$};
\node[box] (scale) at (0,1.2) {scale divisor to match the leading term};
\node[box] (sub) at (0,0) {subtract: leading terms cancel};
\node[box, draw=acc, text=acc, thick] (rem) at (0,-1.2) {remainder of lower degree};
\draw[->, acc] (a) -- (scale);
\draw[->, acc] (scale) -- (sub);
\draw[->, acc] (sub) -- (rem);
\draw[->, acc, dashed] (rem.east) .. controls (5.6,-1.2) and (5.6,2.4) .. (a.east)
  node[midway, right, black, align=left] {repeat on\\the remainder};
\end{tikzpicture}
$$

> **Worked example.** Divide $a = x^3 - 2x + 1$ by $b = x^2 + x$ over
> $\mathbb{Q}$. Cancel the leading term, then repeat on the lower-degree result:
> $$
> \begin{aligned}
> x^3 - 2x + 1 &= (x)(x^2 + x) + (-x^2 - 2x + 1), \\
> -x^2 - 2x + 1 &= (-1)(x^2 + x) + (-x + 1).
> \end{aligned}
> $$
> So $q(x) = x - 1$ and $r(x) = -x + 1$, with $\deg r = 1 < 2 = \deg b$.

The degree of the remainder drops at every stage, which is why the process, and
the Euclidean algorithm built on it, terminates. Iterating the division computes
a monic gcd, exactly as in $\mathbb{Z}$.

> **Worked example.** Compute $\gcd(f, g)$ in $\mathbb{Q}[x]$ for
> $f = x^4 + 2x^3 - x - 2$ and $g = x^3 + x^2 - x - 1$ by the
> [Euclidean algorithm](/abstract-algebra/factorization-and-polynomials/euclidean-domains-pids-ufds):
> $$
> \begin{aligned}
> f &= (x + 1)\,g + (x - 1), \\
> g &= (x^2 + 2x + 1)(x - 1) + 0.
> \end{aligned}
> $$
> The last nonzero remainder is $x - 1$, so $\gcd(f, g) = x - 1$ (taken monic).
> Both polynomials carry that factor: $f = (x - 1)(x + 2)(x^2 + x + 1)$ and
> $g = (x - 1)(x + 1)^2$.

$$
% caption: Degrees in the Euclidean algorithm for F[x] fall strictly at every
% step, so the sequence of nonnegative remainder degrees cannot descend forever
% and the algorithm halts.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black, ->] (0,0) -- (0,3.3) node[anchor=south] {degree};
\draw[black, ->] (0,0) -- (7.0,0) node[anchor=west] {step};
\fill[acc!14, draw=acc] (0.9,0) rectangle (1.7,3.0);
\fill[acc!14, draw=acc] (2.3,0) rectangle (3.1,2.3);
\fill[acc!14, draw=acc] (3.7,0) rectangle (4.5,1.6);
\fill[acc!14, draw=acc] (5.1,0) rectangle (5.9,0.9);
\node[anchor=south, acc, font=\scriptsize] at (1.3,3.0) {$\deg b$};
\node[anchor=south, acc, font=\scriptsize] at (2.7,2.3) {$\deg r_0$};
\node[anchor=south, acc, font=\scriptsize] at (4.1,1.6) {$\deg r_1$};
\node[anchor=south, acc, font=\scriptsize] at (5.5,0.9) {$\deg r_2$};
\node[anchor=west, black] at (3.4,2.7) {strictly decreasing};
\end{tikzpicture}
$$

There is one subtlety absent from $\mathbb{Z}$: the outputs are stable under
field extension. If $F \subseteq E$ and $b \mid a$ in $E[x]$, then $b \mid a$
already in $F[x]$, because the unique $E$-quotient and $E$-remainder must equal
the $F$-ones. So divisibility and the monic gcd of two polynomials do not change
when the coefficient field grows.[^df92]

## Consequences: PID, UFD, and ideals

The chain ED implies PID implies UFD now applies verbatim.

> **Corollary.** For a field $F$, the ring $F[x]$ is a principal ideal domain and
> a unique factorization domain.[^df92]

Being Euclidean makes it a PID (every ideal is the multiples of a single
polynomial), and being a PID makes it a UFD (every polynomial factors into
irreducibles uniquely up to unit multiples — here, up to nonzero constants).
Concretely, an ideal $I \subseteq F[x]$ is generated by any nonzero element of
_least degree_ in it: if $p$ is such an element and $f \in I$, then $f = qp + r$
with $\deg r < \deg p$, and $r = f - qp \in I$ forces $r = 0$, so $f \in (p)$.

$$
% caption: An ideal of F[x] is the set of all multiples of its
% minimal-degree generator f; every polynomial in the ideal lands on the lattice
% of multiples, and division by f leaves remainder zero.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize,
  mult/.style={draw, black, circle, minimum size=8mm, inner sep=0pt}]
\definecolor{acc}{HTML}{4A6FA5}
\node[draw, acc, thick, circle, minimum size=9mm] (f) at (0,0) {$f$};
\node[mult] (mA) at (60:2.4) {$q_1 f$};
\node[mult] (mB) at (120:2.4) {$q_2 f$};
\node[mult] (mC) at (180:2.4) {$q_3 f$};
\node[mult] (mD) at (240:2.4) {$q_4 f$};
\node[mult] (mE) at (300:2.4) {$q_5 f$};
\node[mult] (mF) at (0:2.4) {$q_6 f$};
\foreach \n in {mA,mB,mC,mD,mE,mF} \draw[acc] (f) -- (\n);
\node[anchor=south, black] at (0,2.9) {the principal ideal $(f)$};
\end{tikzpicture}
$$

The distinction between a field and a non-field coefficient ring is sharp.

| Ring | Ideal $(2, x)$ | PID? | UFD? | reason |
| --- | --- | --- | --- | --- |
| $\mathbb{Q}[x]$ | $= (1) = \mathbb{Q}[x]$ | yes | yes | $2$ is a unit; coefficients form a field |
| $\mathbb{Z}[x]$ | not principal | no | yes | $\mathbb{Z}$ is not a field |
| $\mathbb{Q}[x, y]$ | $(x, y)$ not principal | no | yes | $\mathbb{Q}[x]$ is not a field |

The pattern is a theorem: $R[x]$ is a PID (equivalently Euclidean) only when $R$
is a field. If $R[x]$ were a PID, then $(x)$ would be a nonzero prime, hence
maximal, so $R[x]/(x) \cong R$ would be a field. So $\mathbb{Z}[x]$ and every
multivariable ring $F[x, y] = F[x][y]$ fall short of being PIDs — yet, as the
[Gauss's lemma lesson](/abstract-algebra/factorization-and-polynomials/gauss-lemma-and-unique-factorization)
shows, they remain UFDs.[^df92]

## Roots and linear factors

Unique factorization gives the arithmetic of roots its familiar shape. Write
$p(a)$ for the value of $p(x)$ at $a \in F$, obtained by the evaluation
homomorphism $x \mapsto a$.

> **Theorem (Factor theorem).** For $p(x) \in F[x]$ and $a \in F$, the linear
> polynomial $(x - a)$ divides $p(x)$ if and only if $p(a) = 0$.[^df94]

Dividing by the monic $(x - a)$ gives $p(x) = q(x)(x - a) + r$ with $r$ a
constant; evaluating at $a$ yields $r = p(a)$. So $(x - a) \mid p$ exactly when
$p(a) = 0$. Iterating bounds the number of roots.

> **Theorem (Root bound).** Roots $a_1, \dots, a_k$ of $p(x)$ in $F$, counted
> with multiplicity, give a factor $(x - a_1)\cdots(x - a_k)$ of $p(x)$. A
> nonzero polynomial of degree $n$ has at most $n$ roots in $F$.[^df95]

The **multiplicity** of a root $a$ is the largest $m$ with $(x - a)^m \mid p$.
Because $F[x]$ is a UFD and linear polynomials are irreducible, the product of
the linear factors $(x - a_i)$ divides $p$, so their total number cannot exceed
$\deg p$. This is the reason a degree-$n$ polynomial identity that holds at more
than $n$ points is an identity of coefficients, and the reason interpolation
through $n + 1$ points determines a degree-$n$ polynomial uniquely.

$$
% caption: Each root $a_i$ of $p$ contributes a distinct linear factor $(x - a_i)$;
% their product divides $p$, so a degree-$n$ polynomial has at most $n$ roots.
\begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize,
  rt/.style={draw, circle, fill=acc!12, minimum size=8mm, inner sep=0pt},
  ft/.style={draw, minimum width=20mm, minimum height=8mm}]
\definecolor{acc}{HTML}{4A6FA5}
\node[rt] (r1) at (0,1.2) {$a_1$};
\node[rt] (r2) at (2.6,1.2) {$a_2$};
\node[rt] (r3) at (5.2,1.2) {$a_3$};
\node[ft] (f1) at (0,-0.2) {factor for $a_1$};
\node[ft] (f2) at (2.6,-0.2) {factor for $a_2$};
\node[ft] (f3) at (5.2,-0.2) {factor for $a_3$};
\draw[->, acc] (r1) -- (f1) node[midway, right, black, font=\scriptsize] {root};
\draw[->, acc] (r2) -- (f2);
\draw[->, acc] (r3) -- (f3);
\node[ft, draw=acc, text=acc, thick, minimum width=44mm] (prod) at (2.6,-1.7)
  {product of factors divides $p$};
\draw[->, acc] (f1) -- (prod);
\draw[->, acc] (f2) -- (prod);
\draw[->, acc] (f3) -- (prod);
\end{tikzpicture}
$$

## Quotients and irreducibility

The maximal ideals of $F[x]$ carry the whole theory of field extensions, and the
factor theorem is the low-degree end of it.

> **Theorem.** The maximal ideals of $F[x]$ are the ideals $(p(x))$
> generated by irreducible polynomials, and no others. Hence $F[x]/(p(x))$ is a field if and
> only if $p(x)$ is irreducible.[^df95]

This is the PID fact "nonzero prime equals maximal" translated to $F[x]$: an
irreducible $p$ generates a prime ideal, which is automatically maximal, so the
quotient is a field. It is the construction that builds
[field extensions by adjoining a root](/abstract-algebra/field-theory/field-extensions-and-algebraic-elements):
$F[x]/(p(x))$ is a field containing $F$ in which $p$ acquires the root
$\bar{x}$. For a degree-$n$ irreducible $p$, the division algorithm shows
$\{1, x, \dots, x^{n-1}\}$ is a basis of $F[x]/(p(x))$ over $F$, so the quotient
is an $n$-dimensional $F$-vector space.

> **Worked example.** Build a four-element field as
> $\mathbb{F}_2[x]/(x^2 + x + 1)$. Over $\mathbb{F}_2$ the polynomial
> $x^2 + x + 1$ has no root ($0 \mapsto 1$, $1 \mapsto 1$), so by the corollary
> below it is irreducible and the quotient is a field. Write $\alpha = \bar x$;
> the relation $\alpha^2 + \alpha + 1 = 0$ becomes $\alpha^2 = \alpha + 1$ in
> characteristic $2$. The four elements $\{0, 1, \alpha, \alpha + 1\}$ multiply
> by reducing with that relation, for example
> $$
> \alpha(\alpha + 1) = \alpha^2 + \alpha = (\alpha + 1) + \alpha = 1,
> $$
> so $\alpha$ and $\alpha + 1$ are inverse. This quotient is the field
> $\mathbb{F}_4$, built from $\mathbb{F}_2$ by adjoining a root of an irreducible
> quadratic.

For low degrees the factor theorem decides irreducibility outright.

> **Corollary.** A polynomial of degree $2$ or $3$ over $F$ is reducible if and
> only if it has a root in $F$.[^df94]

A degree-$2$ or degree-$3$ polynomial factors nontrivially only if one factor is
linear, which by the factor theorem means a root exists. The corollary fails at
degree $4$: over $\mathbb{R}$ the polynomial $x^4 + 2x^2 + 1 = (x^2 + 1)^2$ is
reducible with no real root. Higher degrees need the sharper tools of the next
lesson — the rational root test, reduction modulo a prime, and Eisenstein's
criterion — together with Gauss's lemma to move between $\mathbb{Z}[x]$ and
$\mathbb{Q}[x]$.

[^df91]: **Dummit & Foote**, _Abstract Algebra_, §9.1 — Definitions and Basic Properties: degree additivity and units of $R[x]$ over a domain, and the reduction homomorphism $R[x]/I[x] \cong (R/I)[x]$ carrying prime ideals to prime ideals.
[^df92]: **Dummit & Foote**, _Abstract Algebra_, §9.2 — Polynomial Rings over Fields I: the division algorithm with unique quotient and remainder (Theorem 3), $F[x]$ Euclidean hence a PID and UFD (Corollary 4), independence of division under field extension, and that $R[x]$ a PID forces $R$ to be a field.
[^df94]: **Dummit & Foote**, _Abstract Algebra_, §9.4 — Irreducibility Criteria: the factor theorem (a linear factor corresponds to a root) and reducibility of degree-2 and degree-3 polynomials by the presence of a root.
[^df95]: **Dummit & Foote**, _Abstract Algebra_, §9.5 — Polynomial Rings over Fields II: the root bound (a degree-$n$ polynomial has at most $n$ roots), root multiplicity, and maximal ideals of $F[x]$ as those generated by irreducibles, so $F[x]/(p)$ is a field iff $p$ is irreducible.