The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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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