Kronecker's Field Extension Theorem

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Recall that over the field of real numbers the polynomial $$x^2 + 1$$ has no roots. However, over the field of complex numbers, $$x^2 + 1$$ has two distinct roots, namely $$i$$ and $$-i$$ .

We will now see that given any field $(F, +, \cdot)$ and a polynomial $f \in F[x]$ that there exists a field extension of $$F$$ such that $$f$$ can be expressed as a product of linear factors.

Theorem 1 (Kronecker's Field Extension Theorem): Let

(F, +, \cdot)

be a field and let

f \in F[x]

be a nonconstant polynomial. Then there exists a field extension

(E, +, \cdot)

$F$

and an element

u \in E

such that

$f(u) = 0$

Proof: Let $f \in F[x]$ be a nonconstant polynomial and let $p \in F[x]$ be an irreducible factor of $$f$$ , that is, $$f(x) = p(x)h(x)$$ for some $h \in F[x]$ .

From The Field of Congruence Classes of Polynomials Modulo p(x) over a Field page, since $$p$$ is irreducible we have that $$E = F[x] / <p(x)>$$ is a field. Now, consider the following set:

(1)

\begin{align} \quad H = \{ [a] \in F[x] / <p(x)> : a \in F \} \end{align}

Then $$H$$ is the set of congruence classes of the elements of $$F$$ . We claim that $$H$$ is a subfield of $$E$$ . Furthermore, $H \cong F$ and an explicit isomorphism is given by the function $\phi : H \to F$ defined for all $[a] \in H$ by $\phi([a]) = a$ .

Now let $$p(x) = a_0 + a_1x + ... + a_nx^n$$ . Then plugging in $$[x]$$ to $$p$$ gives us:

(2)

\begin{align} \quad p([x]) &= a_0 + a_1[x] + ... + a_n[x]^n \\ &= a_0 + a_1[x] + ... + a_n[x^n] \\ &= [a_0 + a_1x + ... + a_nx^n] \\ &= [p(x)] \\ &= [0] \\ &= 0 \end{align}

So $$[x]$$ is a root of $$p$$ . But since $$f = ph$$ we see that $$[x]$$ is also a root of $$f$$ . So set $$u = [x]$$ . Then $u \in E$ and $$f(u) = 0$$ .

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Kronecker's Field Extension Theorem