Kronecker’s Theorem

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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 $\left[{L : K}\right]$ of $L / K$ satisfies $\left[{L : K}\right] \le n$.



Proof

Let $K \left[{X}\right]$ be the ring of polynomial forms over $K$.

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

Therefore, we can write $f = u g_1 \cdots g_r$, where $u$ a unit of $K \left[{X}\right]$ 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 $L = K \left[{X}\right] / \left\langle{g_1}\right\rangle$, where $\left\langle{g_1}\right\rangle$ is the ideal generated by $g_1$.

By Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal, $\left\langle{g_1}\right\rangle$ is maximal.

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

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

\(\displaystyle g_1 \left({\overline X }\right)\) \(=\) \(\displaystyle \overline {g_1 \left({X}\right)}\)
\(\displaystyle \) \(=\) \(\displaystyle \overline {0_{K \left[{X}\right]} }\)
\(\displaystyle \) \(=\) \(\displaystyle 0_L\)

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


It remains to show that $\left[{L : K}\right] \le n$.

By the Euclidean Algorithm every polynomial $p \left({X}\right) \in K \left[{X}\right]$ can be written as:

$p \left({X}\right) = q \left({X}\right) g_1 \left({X}\right) + r \left({X}\right)$

with $\deg \left({r \left({X}\right)}\right) < \deg \left({g_1 \left({X}\right)}\right) \le \deg \left({f \left({X}\right)}\right) = n$.

Now we have by the definition of the quotient ring,

\(\displaystyle \overline {p \left({X}\right)}\) \(=\) \(\displaystyle \overline {q \left({X}\right) g_1 \left({X}\right) + r \left({X}\right)}\)
\(\displaystyle \) \(=\) \(\displaystyle \overline {r \left({X}\right)}\)

So if $r \left({X}\right) = r_0 + r_1 X + \cdots + r_{n-1} X^{n-1}$, we have:

$\overline {p \left({X}\right)} = \overline {r_0 + r_1 X + \cdots + r_{n-1} X^{n-1}}$

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

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

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

That is, $\left[{L : K}\right] \le n$.

$\blacksquare$


Source of Name

This entry was named for Leopold Kronecker.