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Every nonconstant polynomial over a field has a root in some field extension

Statement

Every nonconstant polynomial $f \in F [x]$ has a root in some field extension of $F$ .

Facts & Assumptions

Given: A field $F$ and a nonconstant polynomial $f \in F [x]$ .

[F1]

Every nonzero nonunit polynomial over a field is a finite product of irreducible polynomials (Every nonzero nonunit polynomial over a field factors into irreducible polynomials).

[F2]

If $p$ is monic irreducible, then $F [x] / (p)$ is a field extension in which $x + (p)$ is a root of $p$ ( $F [x] / (p)$ for monic irreducible $p$ is a field extension containing the root $x + (p)$ with unique reduced representatives).

Proof

technique · direct

1.1

A nonconstant polynomial is nonzero and not a unit, so [F1] supplies an irreducible factor $q$ of $f$ .

2.1

Divide $q$ by its nonzero leading coefficient to obtain a monic irreducible factor $p$ ; $(p) = (q)$ and still $p ∣ f$ .

step 1.1algebra

3.1

By [F2], $K = F [x] / (p)$ is a field extension and $a = x + (p)$ satisfies $p (a) = 0$ .

F2step 2.1

4.1

Since $f = p h$ for some $h \in F [x]$ , evaluation gives $f (a) = p (a) h (a) = 0$ .

step 2.1step 3.1algebra∎

Depends on

Used by

Dependency tree · two levels

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Sources

T. Judson, Abstract Algebra: Theory and Applications, Extension Fields (standard reference, not scraped)