Polynomial Rings over Fields

Beside $\mathbb{Z}$, the other fundamental Euclidean domain 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 $1\ne 0$.

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

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

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.¹

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.

The proof is induction on $n=\mathrm{deg}a$. If $n<m=\mathrm{deg}b$, 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

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 $a^{\prime }=q^{\prime }b+r$, and adding back gives $a=(q^{\prime }+\frac{a_n}{b_m}{x}^{n-m})b+r$. For uniqueness, if $qb+r=q_{1}b+r_1$ then $b(q-q_1)=r_1-r$ has degree below $\mathrm{deg}b$ on the right, but degree $\ge \mathrm{deg}b$ on the left unless $q=q_1$; so $q=q_1$ and $r=r_1$.

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 $\mathbb{Z}$.

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.²

Consequences: PID, UFD, and ideals

The chain ED implies PID implies UFD now applies verbatim.

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 $\mathrm{deg}r<\mathrm{deg}p$, and $r=f-qp\in I$ forces $r=0$, so $f\in (p)$.

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

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 shows, they remain UFDs.²

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 $x\mapsto a$.

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.

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 $\mathrm{deg}p$. 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.

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.

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: $F[x]/(p(x))$ 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.

For low degrees the factor theorem decides irreducibility outright.

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 $\mathbb{Q}[x]$.