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The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element

Statement

Let $K / F$ be a field extension and $a \in K$ . Evaluation is the unique $F$ -algebra homomorphism ${ev}_{a} : F [x] ⟶ K, f ⟼ f (a) .$ If $a$ is transcendental, its kernel is zero. If $a$ is algebraic, there is a unique monic irreducible polynomial $m_{a} \in F [x]$ such that $\ker ({ev}_{a}) = (m_{a}),$ and, for every $f \in F [x]$ , $f (a) = 0 ⟺ m_{a} ∣ f .$ The polynomial $m_{a}$ is the minimal polynomial of $a$ over $F$ .

Facts & Assumptions

Given: A field extension $K / F$ and an element $a \in K$ .

[F1]

Given a unital homomorphism $ϕ : R \to S$ of commutative rings and $s \in S$ , there is a unique homomorphism ${ev}_{ϕ, s} : R [x] \to S$ extending $ϕ$ and sending $x$ to $s$ (Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism).

[F2]

Every ideal of $F [x]$ is generated by one polynomial (For every field $F$ , $F [x]$ is a principal ideal domain).

[F3]

An element is algebraic precisely when some nonzero polynomial evaluates to zero at it (Algebraic and transcendental elements and algebraic extensions).

Proof

technique · direct

1.1F1

Apply [F1] to the inclusion $F ↪ K$ and $a$ ; this gives the stated evaluation homomorphism and its uniqueness.

2.1F3step 1.1

By [F3], $a$ is transcendental exactly when $\ker ({ev}_{a}) = 0$ .

2.2F2F3step 1.1

Suppose $a$ is algebraic. Then the kernel is a nonzero proper ideal, so [F2] gives $\ker ({ev}_{a}) = (m)$ for a nonzero nonconstant $m$ .

3.1step 2.2algebra

Multiplying $m$ by the inverse of its leading coefficient does not change its principal ideal, so choose the generator $m$ monic.

3.2step 2.2

For any $f \in F [x]$ , $f (a) = 0$ if and only if $f$ belongs to the kernel, which is equivalent to $f \in (m)$ and hence to $m ∣ f$ .

4.1step 2.2step 3.1algebra

If $m = u v$ with both $u$ and $v$ nonconstant, then $0 = m (a) = u (a) v (a)$ ; since $K$ is a field, one factor evaluates to zero and lies in $(m)$ , impossible because its degree is smaller than $\deg m$ . Thus $m$ is irreducible.

5.1step 3.2algebra∎

If $m^{'}$ is another monic polynomial with the same property, then $m ∣ m^{'}$ and $m^{'} ∣ m$ by step 3.2; equal degree and monicity give $m = m^{'}$ .

The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources