3
$\begingroup$

Let $k$ be a field, $A$ a finitely generated commutative $k$-algebra and $\mathfrak p$ a prime ideal of $A$. Let $K$ be the residue field of the local ring $A_\mathfrak{p}$. I want to show that $K$ is a field extension of $k$ and $\mathrm{tr.deg}(K/k) \leq\dim A$.

I know that $K \cong \operatorname{Quot}(A/\mathfrak p)$. Also there is a canonical homomorphism $k \rightarrow A \twoheadrightarrow A/\mathfrak p\hookrightarrow \operatorname{Quot}(A/\mathfrak p)$, but I don't know why (or if) that is injective.

Any hints?

$\endgroup$
9
  • 3
    $\begingroup$ A ring map between fields is always injective. $\endgroup$ Jul 2 '14 at 20:30
  • 2
    $\begingroup$ What do you know about dimension? Do you know the connection between transcendence degree and dimension of domains of finite type over fields? $\endgroup$ Jul 2 '14 at 20:34
  • $\begingroup$ Is $A$ commutative ? $\endgroup$ Jul 2 '14 at 21:02
  • $\begingroup$ I know that the dimension of a finitely generated, integral algebra $A$ over a field $k$ equals the transcendence degree of the $\operatorname{Quot}(A)$ over $k$. And since $\mathfrak p$ is prime, $A/\mathfrak p$ is integral, and finitely generated because $A$ is finitely generated. So $\dim A/\mathfrak p = \mathrm{tr.deg}(K/k)$. And because $\dim A = \sup_{\mathfrak p\in\operatorname{Spec}(A)} \dim(A/\mathfrak p)$, the inequality holds. Is that correct? $\endgroup$
    – Alex R.
    Jul 2 '14 at 21:05
  • $\begingroup$ @ReneSchipperus Not necessarily. $\endgroup$
    – Alex R.
    Jul 2 '14 at 21:07
4
$\begingroup$

If you know that the dimension of a domain of finite type over $k$ is equal to the transcendence degree of its fraction field, then you know that $\dim(A/\mathfrak{p})=\mathrm{tr.deg}_k(K)$. Can you show using the definition of dimension that $\dim(A/\mathfrak{p})\leq\dim(A)$?

$\endgroup$
1
$\begingroup$

As Keenan pointed out, the composition of maps you gave injects $k$ into $\text{Quot}(A/\mathfrak{p})$.

For comparing $\text{tr. deg}(K/k)$ to $\dim A$, my hint is to use Noether Normalization, a result which, if you don't know now, is well worth learning.

$\endgroup$

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.