Kronecker’s Theorem

Theorem
Let $K$ be a field.

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

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

Moreover, we can choose $L$ such that the degree $\index L K$ of $L / K$ satisfies $\index L K \le n$.

Proof
Let $K \sqbrk X$ be the ring of polynomial forms over $K$.

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

Therefore, we can write $f = u g_1 \cdots g_r$, where $u$ a unit of $K \sqbrk X$ 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 $L = K \sqbrk X / \gen {g_1}$, where $\gen {g_1}$ is the ideal generated by $g_1$.

By Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal, $\gen {g_1}$ is maximal.

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

Moreover, writing $\overline {\map p x} = \map p X + \gen {g_1}$ for $\map p x \in K \sqbrk X$:

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

It remains to show that $\index L K \le n$.

By the Euclidean Algorithm every polynomial $\map p X \in K \sqbrk X$ can be written as:
 * $\map p X = \map q X \map {g_1} X + \map r X$

with $\map \deg {\map r X} < \map \deg {\map {g_1} X} \le \map \deg {\map f X} = n$.

Now we have by the definition of the quotient ring:

So if $\map r X = r_0 + r_1 X + \cdots + r_{n - 1} X^{n - 1}$, we have:
 * $\overline {\map p X} = \overline {r_0 + r_1 X + \cdots + r_{n - 1} X^{n - 1} }$

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

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

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

That is, $\index L K \le n$.