Kronecker’s Theorem

Theorem
Let $K$ be a field.

Let $f$ be a polynomial over $K$ of degree at least $1$.

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

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:

So $X + \langle g_1 \rangle$ is a root of $g_1$ in $L$.

Hence the rsult.