# Brute Force Algebraic Geometry: interpretation of algebraic extensions of field of rational functions

I am trying to affirm the following (without use of dimension theory):

Let $V_{1},V_{2}$ be the zero-loci of prime ideals $I_{1},I_{2}<\mathbb{K}[x_{1},\ldots,x_{d}]=:R$ such that $V_{2}\subsetneqq V_{1}$, i.e.: $$\{z\in\mathbb{K}^{d};f(z)=0\forall f\in I_{2}\}\subsetneqq\{z\in\mathbb{K}^{d};f(z)=0\forall f\in I_{2}\}$$ then the transcendence degree of $\mathbb{K}(V_{2})$, the field of fractions of $\mathbb{K}[V_{2}]:=R/I_{2}$, over $\mathbb{K}$ is strictly smaller than the transcendence degree of $\mathbb{K}(V_{1})$, the field of fractions of $\mathbb{K}[V_{1}]:=R/I_{1}$, over $\mathbb{K}$.

What have I done? As $I_{2}\subseteq I_{1}$, we note that $\mathbb{K}[V_{2}]$ is a quotient of $\mathbb{K}[V_{1}]$ and hence the transcendence degree of $\mathbb{K}(V_{2})$ is at most the transcendence degree of $\mathbb{K}(V_{1})$ (I skipped a few details but it follows basically from both being finitely generated over $\mathbb{K}$).

Assume now that the transcendence degrees are equal, then both rings $\mathbb{K}[V_{1}]$ and $\mathbb[V_{2}]$ are integral extensions of the ring $\mathbb{K}[t_{1},\ldots,t_{r}]$, where $r$ is the transcendence degree, by the Noether normalization lemma. So we have $\mathbb{K}[V_{1}]$ and $\mathbb{K}[V_{2}]$ are finitely generated $\mathbb{K}[t_{1},\ldots,t_{r}]$-modules, which induces an additive isomorphism with $\mathbb{K}[t_{1},\ldots,t_{r}]^{n}$ for respective $n\in\mathbb{N}$. In particular we get: $$I_{2}/I_{1}\cong\bigoplus_{i=1}^{n} S(f_{i}+I_{1})$$ as $S$-modules, where the $f_{i}$s are elements in $I_{2}$ and $S$ is the preimage of $\mathbb{K}[t_{1},\ldots,t_{r}]$ in $\mathbb{K}[V_{1}]$ under the isomorphism $$\mathbb{K}[V_{1}]\cong\mathbb{K}[t_{1},\ldots,t_{r}][\alpha_{1},\ldots,\alpha_{s}]$$ with $\alpha_{1},\ldots,\alpha_{s}$ integral over $\mathbb{K}[t_{1},\ldots,t_{r}]$.

Question: Can I deduce a contradiction from this? May I even be able to show that $I_{j}$ being prime implies that $\mathbb{K}[V_{j}]=\mathbb{K}[t_{1},\ldots,t_{r}]$?

What was the idea? The integral extension adds finitely many copies of $\mathbb{K}[t_{1},\ldots,t_{r}]$ (thinking of finite algebraic extensions of fields) which seem to me to act like constants. This should not contain any interesting information about the variety. Truth be told: it is a bit more complicated than that but this was the idea I had when I decided to proceed like that. Another idea was that the layers corresponded to different components of the variety, which is excluded by the ideals being prime.

Anyway: it would be interesting to give an interpretation to $\mathbb{K}(V)$ being an algebraic extension of $\mathbb{K}(t_{1},\ldots,t_{r})$.

Edit: Just as a piece of information: I am going through that because I need dimension in a context, which is in fact not so much concerned with algebraic geometry. One can extract Noether normalization quite explicitely with not a much too evolved commutative algebra background (see Hungerford's Algebra book). So it would be nice if one could finish it in the rather explicit setting I am in. The goal is to have a minimal set of knowledge about algebraic groups available in order to discuss orbits of certain algebraic groups on homogeneous spaces. The latter by itself is a lot to discuss and by requiring formal algebraic geometry, any discussion becomes tremendeously huge.

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## 1 Answer

Here is how you can see the contradiction: First of all $V_2 \subset V_1$ implies $I_2 \supset I_1$. The fact that $I_1 \neq I_2$ implies that $\operatorname{height}(I_2) > \operatorname{height}(I_1)$. Now look at Theorem 1.8A in Hartshorne. Both $R/I_1, R/I_2$ are integral domains and finitely generated algebras over $k$ and so their Krull-dimension is equal to their transcendence degree. Moreover, $d = \dim R = \dim R/ I_1 + \operatorname{height}(I_1)=\dim R/I_2 + \operatorname{height}(I)_2$. If $\dim R /I_1 = \dim R/ I_2$ then $\operatorname{height}(I_1) = \operatorname{height}(I_2)$, contradiction.

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Thank you. This would however involve knowing what Krull-dimension and what height are. At this stage one would have to discuss Noetherian rings and many more. What I have done above requires very little commutative algebra and the goal was to stay as simple as possible at the risk of doing explicit things. I will check these again in order to see whether I can extract a direct proof from this. – M. Luethi Jan 31 '14 at 19:56
But you also need Noetherian rings and Krull-dimension to discuss about Noether normalization. – Manos Jan 31 '14 at 19:58
I have the impression I don't. I was following Hungerford's Algebra and there he gives a very simple proof. I was wondering whether I could get this to work out in about the same framework. – M. Luethi Jan 31 '14 at 20:03
My apologies then. The problem is you need to connect the transcendence degree of an integral domain to the integral domain itself. One way i know is to do that using the concept of Krull dimension. – Manos Jan 31 '14 at 20:23
No problem. Thank you for sure. – M. Luethi Jan 31 '14 at 20:24