1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ $A$ is a finitely generated $k$-algebra because $k[X]$ is a noetherian ring. $k [X] / A$ is not a ring, so you can't really talk about its Krull dimension... $\endgroup$ – Zhen Lin Jun 5 '12 at 10:09
  • $\begingroup$ You're right... Another approach: $dim(k[X])$ being 1 and $A$ being a subalgebra, isn't $dim(A) \leq dim(k[X])$ and we just have to show that $dim(A)\neq 0$? Which is true, I think, because $k \subset A$ is strict...? $\endgroup$ – Adrian Manea Jun 5 '12 at 10:15
  • $\begingroup$ Yes. But you have to justify this! Do you know the going-up theorem? $\endgroup$ – Zhen Lin Jun 5 '12 at 10:20
  • $\begingroup$ Yes, but it involved integral extensions of rings and, as a consequence, we had this result: Let $R \subseteq S$ be an integral extension of rings. Then $dim(R)=dim(S)$. I suppose this is the key, but where's the integrality? $\endgroup$ – Adrian Manea Jun 5 '12 at 10:26
  • $\begingroup$ $k[X]$ is finitely generated as an $A$-module, hence is an integral extension of $A$. $\endgroup$ – Zhen Lin Jun 5 '12 at 10:34
3
$\begingroup$

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:

  1. 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.

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

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

Proof.

  1. 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.

  2. 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 $\mathfrak{q} \cap A = \mathfrak{m} \cap A = \mathfrak{m} \cap A_\mathfrak{p} \cap A = \mathfrak{p}$.

  3. 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 $\dim A \le \dim B$. $\qquad \blacksquare$

$\endgroup$
  • $\begingroup$ Yes, thank you, Sir! $\endgroup$ – Adrian Manea Jun 5 '12 at 10:37
3
$\begingroup$

$A$ 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 $A=k$]

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.