Residue field of a local ring as field extension

Question

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?

Answer 1

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

Answer 2

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.