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

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

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

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

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

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 $m\mid f$.

If $m=uv$ 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 $\mathrm{deg}m$. Thus $m$ is irreducible.

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