Vtyjkyuk

Let $A$ be a $k$-algebra of finite type and an integal domain. Futhermore set $X=operatornameSpec(A)$ the affine scheme.

Let $U subset X$ an arbitrary open subscheme of $X$. The main goal is to show that $dim U = dim X$ holds and i know that there are many ways to do it but this question refers concretely to following argument presented in Bosch's "Algebraic Geometry and Commutative Algebra" (p. 376):

There are two steps which aren't clear to me there:

1. In first step he says that the set of closed points of $U$ is dense in $X$. His argument was an application of Hilbert's Nullstellensatz. Can anybody explain to me how this step works. I don't see how the HN is incorporated here.




2. After having shown 1. according to the proof the statement $dim U = dim X$ should follow instantly. Why?

Remark: The prolem here isn't just to see why $dim U = dim X$ holds but to understand concretely to two steps used in the argument above.

I can at least answer question one. The version of Hilbert's Nullstellensatz being used here is often called Zariski's lemma, and says: if $B$ is a field which is finitely generated as a $k$-algebra, then $B$ is a finite field extension of $k$ (i.e., $B$ is finite as a $k$-module.) With this in mind, let's proceed.

Let $A$ be a finitely generated $k$-algebra. First, we will show that the closed points of $X = mathrmSpec(A)$ are dense in $A$. To do so, it suffices to show that for any $f in A$ such that the distinguished open $D(f)$ is nonempty, there is a closed point of $A$ in $D(f)$. Let $f in A$ be such that $D(f)$ is nonempty; we recall that $D(f)$ is the image of the open embedding of schemes $varphi colon mathrmSpec(A_f) to mathrmSpec(A)$ induced by the canonical localization morphism $alpha colon A to A_f$.

Note that $A_f$ is not the zero ring, since $D(f)$ is nonempty. Hence, $A_f$ has a maximal ideal $M$, i.e. there is a point $M in mathrmSpec(A_f)$ which is closed. If we can show that $varphi(M)$ is a closed point of $A$, then we're done. This will be our strategy. (As an aside, it's worth noting that this step is where our argument breaks down if $A$ is not finitely generated as a $k$-algebra; consider $A = k[X]_langle X rangle, f = X$, for example.)

Indeed, $A_f$ is also a finitely generated $k$-algebra, and therefore so is $A_f/M$. Since $A_f/M$ is a field, $A_f/M$ is a finite field extension of $k$ by the Hilbert Nullstellensatz. Let $P = varphi(M) = alpha^-1(M) in mathrmSpec(A)$. Then $P$ is precisely the kernel of the composition $A to A_f to A_f/M$, and hence we obtain an injection of $k$-algebras $A/P hookrightarrow A_f/M$. But $A_f/M$ is a finite-dimensional $k$-vector space, and therefore $A/P$ is an integral domain which is a finite-dimensional $k$-vector space. By a well known lemma, $A/P$ is thus a field, and $P$ is a closed point of $mathrmSpec(A)$ in $D(f)$, as desired.

Now, let $U subset X$ be a (nonempty) open subscheme. Note that any closed point of $X$ contained in $U$ is a closed point of $U$ in the subspace topology, so it suffices to show that for any nonempty open subset $V subset X$, $V cap U$ contains a closed point of $X$.

First, note that $X$ is an irreducible topological space, since $A$ is a domain. This is because $(0)$ is a prime ideal of $A$, and the closure of $(0)$ in $X$ is $V((0)) = mathrmSpec(A)$. It is a well known fact from topology that any two nonempty open subsets of an irreducible topological space have nonempty intersection. Now, suppose $V$ is a nonempty open subset of $X$. Since $U$ is also nonempty, the intersection $U cap V$ is a nonempty open subset of $X$. Since the closed points of $X$ are dense in $X$, it follows that $U cap V$ must contain a closed point of $X$, as desired.

As a final observation, note that the hypothesis that $A$ is a domain is crucial here. It is always true that the closed points of $X$ are dense in $X$ when $A$ is a finitely generated $k$-algebra, whether or not $A$ is a domain. However, it is not true that the closed points of any nonempty open $U subset X$ are dense in $X$ without the assumption that $A$ is a domain. A counterexample is given by choosing any $A$ such that $mathrmSpec(A)$ contains two nonempty open subsets which do not intersect, and picking one of these to be $U$. For a concrete counterexample, recall that the spectrum of any Artinian ring is finite and discrete, and so every point is simultaneously open and closed. Taking $A$ to be an Artinian ring which is finitely generated as a $k$-algebra but not a domain and $U$ to be a point $mathrmSpec(A)$ then gives a counterexample; for example, to be explicit, one could take $A = k[X]/langle X^2 rangle times k[Y]/langle Y^3 rangle$. Of course, any Artinian domain is a field, and so has spectrum consisting of a point, so this is not a problem under our hypotheses.