Polynomial algebra 
Let $k$ be a field and $k \subset A \subseteq k[X]$ be a $k$-subalgebra of $k[X]$. Prove that $\dim(A)=1$ (Krull dim) and that $A$ is a finitely-generated $k$-algebra.

My initial thought: Consider the short exact sequence $0 \rightarrow A \hookrightarrow k[X] \rightarrow k[X]/A \rightarrow 0$ and see how the Krull dimension behaves under this sequence. It's just a spark, I don't know how to continue exactly.
 A: $A$ is finitely generated: See here. The dimension is the transcendence degree of $\mathrm{Quot}(A)$. This field is $k$ or $k(x)$, so the dimension is $0$ or $1$. [it's $0$ only when $A=k$]
A: You've probably figured this out by now, but for the record:
Lemma. The polynomial algebra $k[X]$ is finitely generated as an $A$-module.
Proof. Since $k \subsetneq A$, $A$ contains a non-constant polynomial $f(X)$. Obviously, $T = X$ satisfies the polynomial equation
$$f (T) - f (X) = 0$$
and therefore $k[X]$ is finitely generated as an $A$-module. $\qquad \blacksquare$
Theorem. If $B$ is an integral domain and $A \subseteq B$ is a subring such that $B$ is a finitely generated $A$-module, then:


*

*For every prime $\mathfrak{q}$ of $B$, $\mathfrak{p} = \mathfrak{q} \cap A$ is a prime of $A$, and $\mathfrak{q}$ is a maximal ideal if and only if $\mathfrak{p}$ is a maximal ideal.

*For every prime $\mathfrak{p}$ of $A$, there is a prime $\mathfrak{q}$ of $B$ such that $\mathfrak{p} = \mathfrak{q} \cap A$.

*Given primes $\mathfrak{q} \subseteq \mathfrak{q}'$ of $B$, if $\mathfrak{q} \cap A = \mathfrak{q}' \cap A$, then $\mathfrak{q} = \mathfrak{q}'$.
Proof. 


*

*If $\mathfrak{p}$ is a maximal ideal, then $B / \mathfrak{q}$ is an integral domain that is finite-dimensional as a $A / \mathfrak{p}$-vector space, so $B / \mathfrak{q}$ is a field. Conversely, if $B / \mathfrak{q}$ is a field that is finitely generated as an $A / \mathfrak{p}$-module, then $A / \mathfrak{p}$ must be a field too.

*Let $\mathfrak{p}$ be a prime of $A$, and consider the localisations $A_\mathfrak{p} \subseteq B_\mathfrak{p}$. By Krull's theorem, $B_\mathfrak{p}$ has a maximal ideal $\mathfrak{m}$, and by (1) we have $\mathfrak{m} \cap A_\mathfrak{p} = A_\mathfrak{p} \mathfrak{p}$; so $\mathfrak{q} = \mathfrak{m} \cap B$ is a prime of $B$ such that $\mathfrak{q} \cap A = \mathfrak{m} \cap A = \mathfrak{m} \cap A_\mathfrak{p} \cap A = \mathfrak{p}$.

*Let $\mathfrak{p} = \mathfrak{q} \cap A$, and consider the localisations $A_\mathfrak{p} \subseteq B_{\mathfrak{q}'}$. The integral domain $B_\mathfrak{q'}$ is still finitely generated as an $A_\mathfrak{p}$-module, so we may apply (1) to conclude $B_{\mathfrak{q}'} \mathfrak{q} = B_{\mathfrak{q}'} \mathfrak{q}'$, from which it follows $\mathfrak{q} = \mathfrak{q}'$. $\qquad \blacksquare$
Corollary. If $B$ is an integral domain and $A \subseteq B$ is a subring such that $B$ is a finitely generated $A$-module, then $\dim A = \dim B$. 
Proof. From (3), we know any strictly ascending chain of primes of $B$ gives a strictly ascending chain of primes of $A$, so $\dim A \ge \dim B$. But by (2) we know any strictly ascending chain of primes of $A$ comes from a strictly ascending chain of primes of $B$, so $\dim A \le \dim B$.  $\qquad \blacksquare$
