## 9.9 Minimal polynomials

Let $E/k$ be a field extension, and let $\alpha \in E$ be algebraic over $k$. Then $\alpha$ satisfies a (nontrivial) polynomial equation in $k[x]$. Consider the set of polynomials $P \in k[x]$ such that $P(\alpha ) = 0$; by hypothesis, this set does not just contain the zero polynomial. It is easy to see that this set is an ideal. Indeed, it is the kernel of the map

$k[x] \to E, \quad x \mapsto \alpha$

Since $k[x]$ is a PID, there is a generator $P \in k[x]$ of this ideal. If we assume $P$ monic, without loss of generality, then $P$ is uniquely determined.

Definition 9.9.1. The polynomial $P$ above is called the minimal polynomial of $\alpha$ over $k$.

The minimal polynomial has the following characterization: it is the monic polynomial, of smallest degree, that annihilates $\alpha$. Any nonconstant multiple of $P$ will have larger degree, and only multiples of $P$ can annihilate $\alpha$. This explains the name minimal.

Clearly the minimal polynomial is irreducible. This is equivalent to the assertion that the ideal in $k[x]$ consisting of polynomials annihilating $\alpha$ is prime. This follows from the fact that the map $k[x] \to E, x \mapsto \alpha$ is a map into a domain (even a field), so the kernel is a prime ideal.

Lemma 9.9.2. The degree of the minimal polynomial is $[k(\alpha ) : k]$.

Proof. This is just a restatement of the argument in Lemma 9.6.8: the observation is that if $P$ is the minimal polynomial of $\alpha$, then the map

$k[x]/(P) \to k(\alpha ), \quad x \mapsto \alpha$

is an isomorphism as in the aforementioned proof, and we have counted the degree of such an extension (see Example 9.7.6). $\square$

So the observation of the above proof is that if $\alpha \in E$ is algebraic, then $k(\alpha ) \subset E$ is isomorphic to $k[x]/(P)$.

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