I am currently working my way through a proof of the following statement: Let $X$ be an integral, affine $k$-Variety of dimension $n$. Then there exists a surjective morphism $X\rightarrow \mathbb{A}_k^n$ with finite fibres.

The book containing the proof in question introduces the dimension of $X$ as the trancendence degree of the function field $K(X)$ over $k$. It then uses an anti-equivalence between the category of affine varieties over $k$ and that of finitely generated, reduced $k$-Algebras and Noether's normalization lemma for existence of such a morphism. It think I understand it except for the following: the coordinate ring $\mathcal{O}(X)$ is by Noether's normalization lemma a finitely generated $k[x_1,...,x_n]$-module. Let $M_P=(x_1-a_1,...,x_n-a_n)$ for any $P=(a_1,...,a_n)\in \mathbb{A}_k^n$. The book then claims that if follows that $\mathcal{O}(X)/M_P\mathcal{O}(X)$ is a finite-dimensional $k$-Algebra (Then $\mathcal{O}(X)/M_P\mathcal{O}(X)$ would only have finitely many maximal ideals which correspond to the points in the preimage of $P$ under the morphism). Why is that statement true though? I know that if in general if $A$ is a finitely generated $k$-Algebra and $M$ a maximal ideal in $A$ then $A/M$ is a finite-dimensional $k$-Algebra. But $M_P\mathcal{O}(X)$ isn't necessarily maximal, right? What am I missing?

  • $\begingroup$ As $k[x_1,\dots,x_n] \subset O(X)$ is a finite morphism, it is an integral extension. The fibers of a maximal ideal of $k[x_1,\dots,x_n]$ are maximal ideals of $O(X)$. As this extension is finite, there are finitely many of them. $\endgroup$ – Youngsu Feb 16 '18 at 20:57
  • $\begingroup$ I'm probably misunderstanding your comment, but if $\mathcal{O}(X)$ had finitely many maximal ideals, $X$ would have finitely many points since there's a one-to-one correspondence between them right? $\endgroup$ – Morph Feb 16 '18 at 21:10
  • $\begingroup$ Youngsu is saying that, if $A \to B$ is finite and $\mathfrak m \subseteq A$ is maximal, then there are only finitely many maximal ideals $\mathfrak M \subseteq B$ with $\mathfrak m = \mathfrak M \cap A$. In fact, this is true of all prime ideals, not just the maximal ones. $\endgroup$ – Trevor Gunn Feb 17 '18 at 0:53

If $\mathcal O(X) = k[x_1,\dots,x_n,y_1,\dots,y_m]$ where $x_1,\dots,x_n$ are a transcendence base and $y_1,\dots,y_m$ are integral over $k[x_1,\dots,x_n]$, then $$\mathcal O(X)/M_P\mathcal O(X) = k[a_1,\dots,a_n,\bar y_1,\dots,\bar y_n] = k[\bar y_1,\dots,\bar y_m]$$ where the residue classes, $\bar y_i$ are integral over $k$.

This is because if

$$ y_i^r + f_{r - 1}(x_1,\dots,x_n)y_i^{r-1} + \dots + f_0(x_1,\dots,x_n) = 0 $$


$$ \bar {y_i}^r + f_{r - 1}(a_1,\dots,a_n)\bar {y_i}^{r-1} + \dots + f_0(a_1,\dots,a_n) = 0 $$


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