# Residue field of a local ring as field extension

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?

• A ring map between fields is always injective. Jul 2 '14 at 20:30
• What do you know about dimension? Do you know the connection between transcendence degree and dimension of domains of finite type over fields? Jul 2 '14 at 20:34
• Is $A$ commutative ? Jul 2 '14 at 21:02
• 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? Jul 2 '14 at 21:05
• @ReneSchipperus Not necessarily. Jul 2 '14 at 21:07

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 $\dim(A/\mathfrak{p})\leq\dim(A)$?
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.