Finite Morphism to Spec k

Proposition. Suppose $\pi: X \to \text{Spec} k$ is a finite morphism. Then $X$ is a finite union of point with discrete topology. The residue at each point is a finite extension of $k$.

Proof.  
As $\pi$ is finite, it is in particular affine, so we must have $X = \text{Spec} A$ for some $A$.

Claim: Every point of $X$ is closed.

Indeed, for all point $\mathfrak{p}$ of $X$, we have $A/\mathfrak{p}$ is an integral domain that is a finite $k$-algebra (i.e. finite dimensional $k$-vector space), so it must be a field.

Thus the points of $A$ must be the irreducible components of $A$. But $X= \text{Spec} A$ has only finitely many irreducible components (being a Noetherian space). So $X$ is finite and discrete.

On the other hand, the residue field at every point $\mathfrak{m}$ of $A$ is $A/\mathfrak{m}$ which is  a finite field extension of $k$.

Note: $A$ is Artinian.
------
Lemma. Suppose $B$ is an integral domain such that $B$ is a finite $k$-algebra (i.e. finitely generated as a $k$-module). Then $B$ is a field.
Proof. Suppose $b_1, \ldots, b_n$ is a basis of $B$ as a $k$-vector space. Let $b \in B$ be non-zero. Consider the multiplication by $b$-map $B \to B$. As $B$ is an integral domain, this map must be injective. Thus linear transformation must be also surjective. In particular, $1$ must be in the image.
--------------------
Example.
$$\text{Spec}\left(\mathbb{F}_8 \times \frac{\mathbb{F}_4[x,y]}{(x^2, y^4)} \times \frac{\mathbb{F}_4[t]}{t^9}\times \mathbb{F}_2 \right) \to \text{Spec} \mathbb{F}_2.$$
LHS is disjoint union of
$\text{Spec} \mathbb{F}_8 $ which is a point,
$\text{Spec}\frac{\mathbb{F}_4[x,y]}{(x^2, y^4)}$ which is a fat point $(x,y)$,
$\text{Spec}\frac{\mathbb{F}_4[t]}{t^9}$, which is another fat point $(t)$,
and  $\text{Spec}  \mathbb{F}_2 $ which is a point.