Kronecker's Field Extension Theorem

# Kronecker's Field Extension Theorem

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 $(E, +, \cdot)$ of $F$ 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:

\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:

\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 $f(u) = 0$.