Kronecker’s Theorem

Theorem
Let $K$ be a field.

Let $f$ be a polynomial over $K$ of degree $n \geq 1$.

Then there exists a finite extension $L / K$ of $K$ such that $f$ has at root in $L$.

Moreover, we can choose $L$ such that the degree $\left[{L : K}\right]$ of $L / K$ satisfies $\left[{L : K}\right] \leq n$.

Proof
Let $K[X]$ be the Ring of Polynomial Forms over $K$.

By Polynomial Forms over Field form Unique Factorization Domain, $K[X]$ is a unique factorisation domain.

Therefore, we can write $f = u g_1\cdots g_r$, where $u$ a unit of $K[X]$ and $g_i$ is irreducible for $i=1,\ldots,r$.

Clearly it is sufficient to find an extension of $K$ in which the irreducible factor $g_1$ of $f$ has a root.

Let $L = K[X] / \langle g_1 \rangle$, where $\langle g_1 \rangle$ is the ideal generated by $g_1$.

By principal ideal of irreducible element $\langle g_1 \rangle$ is maximal.

Therefore by Maximal Ideal iff Quotient Ring is Field, $L$ is a field.

Moreover, writing $\overline{p(x)} = p(X) + \langle g_1 \rangle$ for $p(x) \in K[X]$:

So $\overline{X}$ is a root of $g_1$ in $L$.

It remains to show that $[L:K] \leq n$.

By the Euclidean Algorithm every polynomial $p(X) \in K[X]$ can be written as:
 * $p(X) = q(X)g_1(X) + r(X)$

with $\deg(r(x)) < \deg g_1(X) \leq \deg(f(X)) = n$.

Now we have by the definition of the quotient ring,

So if $r(x) = r_0 + r_1X + \cdots + r_{n-1}X^{n-1}$, we have
 * $\overline{p(X)} = \overline{r_0 + r_1X + \cdots + r_{n-1}X^{n-1} }$

Our choice of $p$ was arbitrary, so every polynomial can be written in this form.

In particular the set $\{\overline{1},\overline{X},\ldots,\overline{X^{n-1}}\}$ spans $L$ over $K$.

Thus by Spanning Set Contains Basis, a basis of $L$ has at most $n$ elements.

That is, $[L:K] \leq n$.