(Theorem) Every nonconstant polynomial over a field has a root in some extension field

Overview

The claim that every nonconstant polynomial over a field $F$ has a root in some extension field is a consequence of an important construction relating field extensions and irreducible polynomials: if $f\in F[x]$ is irreducible, we can identify $F$ as a subfield of $E=F[x]/(f)$ is a field, and hence $E$ as an extension of $F$, by the embedding

$$a\mapsto a+(f),a\in F,$$

and define a root of $f$ in $E$ by

$$\alpha =x+(f).$$

For example, this construction relates simple extensions with minimal polynomials: if $E=F(\alpha )$, where $\alpha$ is algebraic over $F$ and $f=\text{irr}(\alpha ,F)$ is its minimal polynomial, then it can be shown that the projection $\alpha \mapsto x+(f)$ as above defines an inverse for the isomorphism

$${\overline{\text{ev}}}_{\alpha }:F[x]/(f)\to F(\alpha ),g+(f)\mapsto g(\alpha ),$$

induced by evaluation $g\mapsto g(\alpha )$ for all $g\in F[x]$, where

$$(f)=\{g\in F[x]:g(\alpha )=0\}=\ker ({\text{ev}}_{\alpha }).$$

By induction, the theorem implies that for every nonconstant polynomial $f\in F[x]$, there is some extension field $E$ of $F$ such that $f$ splits into linear factors in $E$. This result is related to, but not sufficient to prove the existence of a splitting field for $f\in F[x]$, which is the smallest extension $E$ of $F$ where $f$ splits completely.

---

Statement and proof of the theorem

Modern Algebra II 3.4: Constructing larger fields to find polynomial roots

If $f\in F[x]$ is a nonconstant polynomial in $F[x]$, then there exists a field $E$ containing a subfield isomorphic to $F$ such that $f$ has a root in $E$, i.e., there is $\alpha \in E$ such that $f(\alpha )=0$.

Proof from Modern Algebra II. The key insight is that if $p$ is an irreducible factor of $f$ in $F[x]$, we can take $E=F[x]/(p)$ and $\alpha =x+(p)$, identifying $F$ with the subfield $\{a+(p):a\in F\}$ of $E$.

Let $p$ be an irreducible factor of $f$. Then $E=F[x]/(p)$ is a field (since $p$ irreducible implies $(p)\subseteq F[x]$ is maximal). Further, the projection $F\to F[x]/(p)$ defined by $a\mapsto a+(p)$ has kernel $\{a+(p)=0:a\in F\}=\{a\in (p):a\in F\}=\{0\},$ so it is injective and hence gives an isomorphism identifying $F$ with a subfield of $F[x]/(p)$. Finally, the coset $\alpha =x+(p)$ is clearly a root of $p$ in $E$ (as $p(\alpha )=p+(p)=0$), so if $g\in F[x]$ is another factor such that $f=pg$, then $f(\alpha )=p(\alpha )g(\alpha )=0$ and we conclude that $\alpha$ is a root of $f$ in $E$. $\square$

---

Corollary: Every nonconstant polynomial splits into linear factors in some extension

Modern Algebra II 3.5: Every nonconstant polynomial splits into linear factors in some extension

If $f\in F[x]$ is a nonconstant polynomial, then there exists a field $E$ containing a subfield isomorphic to $F$ such that $f$ factors into linear factors in $E[x]$, i.e., there exist ${\alpha }_1,\dots ,{\alpha }_n\in E$ and $c\in F$ such that $f=c(x-{\alpha }_1)\dots (x-{\alpha }_n)$ in $E[x]$. In this case, we say that $f$ splits completely in $E[x]$.

Proof from Modern Algebra II. We proceed by induction on $\mathrm{deg}f$. If $\mathrm{deg}f=1$, this is obvious. Now suppose we know that $\mathrm{deg}f=n$, and we have proven this claim for all fields $F$ and all polynomials in $F[x]$ of degree $n-1$. By the theorem above, we can define a field $E_1=F[x]/(f)$ so that $F\le E$ by the injective mapping

$$F\to E_1=F[x]/(f),a\mapsto a+(f)\text{ for all }a\in F$$

and $\alpha =x+(f)$ is a root of $f$ in $E$. Thus $f$ has at least one linear factor in $E_1[x]$, meaning there exists some $g\in E_1[x]$ such that

$$f=(x-\alpha )g,\mathrm{deg}g=n-1.$$

Applying the inductive hypothesis to the field $E_1$ and the polynomial $g$, there exists a field $E$ containing a subfield isomorphic to $E_1$ (i.e., an injection $E_1\to E$ that embeds $E_1$ as a subfield) such that $g$ splits into linear factors in $E$. Then $f$ is a product of linear factors in $E$, and the composition of injections

$$F\to E_1\to E$$

gives an isomorphism between $F$ and a subfield of $E$, completing the proof. $\square$