I have a question regarding the above topic. Let $k$ be a (not neccesarily algebraically closed) field, $A$ a finitely generated $k$-Algebra. Show that the set of the closed points is dense in Spec $A$.
So far I have got the following. I was not quite sure, how to use the fact, that $A$ is a finitely generated $k$-Algebra.
But we know that the closed points in $A$ are the maximal ideals. Any topological space is the union of its irreducible components. So we have Spec $A = \bigcup_{i} U_i$ for some closed, irreducible subsets $U_i$. Each $U_i$ has a unique generic point $\mathfrak p_i$, so that $\overline{\{\mathfrak p_i\}} = U_i$. Now, since $A$ is a finitely generated $k$-Algebra, we have for any prime ideal $\mathfrak p$: $$\mathfrak p = \bigcap_{\mathfrak m \supset \mathfrak p} \mathfrak m,$$ where $\mathfrak m$ is a maximal ideal. So it follows, that $$ \text{Spec }A = \bigcup_i U_i = \bigcup_i \overline{\{\mathfrak p_i\}} = \bigcup_i \overline{\left(\{\bigcap_{\mathfrak m \supset \mathfrak p} \mathfrak m\}\right)_i} = \overline{\bigcup_i \left(\{\bigcap_{\mathfrak m \supset \mathfrak p} \mathfrak m\}\right)_i}$$ So it follows, that the closure of the set of maximal ideal equals Spec $A$.
I would appreciate any comments, where the proof is mistaken and what a correct proof would look like.