Minimal polynomial
Definition
Suppose $P\in k[T]$ is irreducible of degree $\geq 1$. Then
- $(P)$ is a maximal ideal in $k[T]$.
- $k_P\coloneqq k[T]/(P)$ is a field
- $[k_P:k]=\deg P$. The polynomial $P$ is called the minimal polynomial of $k_P$.
To see (1), recall $k[T]$ is a PID, so any ideal containing $(P)$ is principal (generated by a single element). In particular that single element would generate $P$, and since $P$ is irreducible this implies that the single element is $P$ itself.
To see (2), consider the Euclidean algorithm. Any polynomial $f\in k[T]$ can be written as $f(T)=g(T)P(T)+h(T)$, where $0\leq \deg h(T)<\deg P(T)$. The possible values of $h(T)$ is isomorphic to $k[T]/(P)$. It is spanned by $1,T,\dots,T^{\deg P-1}$.
Properties
Let $E/F$ be a finite Field extension. Let $\alpha\in E$.
- The minimal polynomial of $\alpha$ is unique.
Proofs 1 Consider the ring homomorphism
\begin{gather}
F[X]\to E \
f(X)\mapsto f(\alpha).
\end{gather}
Denote the kernel by $J_\alpha$; it is the set of all polynomials in $F[X]$ that have $\alpha$ as a root. By the properties of ring homomorphisms, $J_\alpha$ is an ideal. But also $F[X]$ is a principal ring, hence there exists a polynomial $P$ that generates $J_\alpha$. Clearly this must have minimal degree among all nonzero elements of $J_\alpha$. Now suppose $Q\in J_\alpha$ has the same degree as $P$. Then $P-Q\in J_\alpha$ has degree less than $P$, but since $P$ has minimal degree this means $P-Q=0$ and so $P=Q$.