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I'm trying to prove the following (as a part of the proof of Zariski's Lemma): Let $K$ be a field. $M = K(x_1,\ldots,x_k)$, $x_1,\ldots,x_k$ algebraically independent over K. Then $M$ cannot be a finitely generated $K$-algebra.

Proof: Assume for a contradiction $M = K[y_1,\ldots,y_l]$, with $y_j = \frac{f_j(x_1,\ldots,x_k)}{g_j(x_1,\ldots,x_k)} \in M$.

Claim: $h^{-1}(x_1,\ldots,x_k) \notin M$, where $h = g_1g_2\ldots g_l + 1$.

$h^{-1}(x_1,\ldots,x_k) $ can't be a polynomial in the $y_i$ because if $h^{-1} = p(\frac{f_1}{g_1},\ldots,\frac{f_1}{g_1})$, for $p \in K[X_1,\ldots,X_l]$,
then $1 = h p(\frac{f_1}{g_1},\ldots,\frac{f_1}{g_1})$. And we can plug in a zero of $h$ in an algebraic closed extension of $K$ and we get an contradiction $1 = 0$.

My problem:

But this does not necessarily imply that $h^{-1}(x_1,\ldots,x_k) \neq p(y_1,\ldots,y_l)$ since the evaluation map $ ev_{(x_1,\ldots,x_k)}: K(X_1,\ldots,X_k) \mapsto K(x_1,\ldots,x_k)$ is not necessarily injective. So im thinking $h^{-1}(x_1,\ldots,x_n)$ could also be $\frac{f(x_1,\ldots,x_k)}{g(x_1,\ldots,x_n}$ where the above contradiction doesn't apply.

Is there a easy way to get from the contradiction in $K(X_1,\ldots,X_k)$ to a contradiction in $K(x_1,\ldots,x_n)=K[y_1,\ldots,y_l]$, or am i approaching this problem wrong?

Eric Wofsey
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mike
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  • I am rather confused about where you think the evaluation map you mention comes up at all in the argument. But in any case, that evaluation map is an isomorphism, since $x_1,\dots,x_k$ are algebraically independent over $K$... – Eric Wofsey Jan 08 '18 at 23:51
  • I need the evaluatoin map to get from the statment "$ h^{-1}(X_1,\ldots,X_k)$ is not a polynomial in $\frac{f_j(X_1,\ldots,X_k)}{g_j(X_1,\ldots,X_k)}, , 1 \leq j \leq l$" to "$h^{-1}(x_1,\ldots,x_k)$ is not a polynomial in $\frac{f_j(x_1,\ldots,x_k)}{g_j(x_1,\ldots,x_k)}, , 1\leq j \leq l$". But now since i see that the evaluation map must be injective, it seems that i have solved the problem. Thank you for the help. – mike Jan 09 '18 at 00:06
  • I am rather confused by your argument. $h^{-1}(x_1, \ldots, x_n)$ is just a number that is obviously member of $M$. I understand you are trying to prove that the reciprocal of $h$ is not expressible as a polynomial, but how you want to conclude the thesis from this is not clear to me. – xyzzyz Jan 09 '18 at 00:10
  • Anyway, what you are trying to prove is basically a form of Hilbert's Nullstellensatz, and it requires slightly more work than you do here. – xyzzyz Jan 09 '18 at 00:12
  • I assume $M$ to be a finitely generated $K$-algebra, generated by $y_1,\ldots,y_n$. By the definition of a finitely generated $K$-algebra every element of M can be expressed as a polynomial of the generators $y_1,\ldots,y_n$. So if i can find a element of $M$ that is not expressable as a polynomial in $y_1,\ldots,y_n$ $M$ cant be a finitely generated $K$-algebra – mike Jan 09 '18 at 00:19
  • Yes, but what your proof tries to argue is not that the number $h^{-1}(x_1, \ldots, x_n)$ is not expressible as a polynomial in $y_i$, but rather that the function $h^{-1}$ is not polynomial, and indeed this is true, but it's not what you need to prove. Think about it this way: where exactly are you using the fact that $x_i$ are algebraically independent? – xyzzyz Jan 09 '18 at 00:22
  • Ok: suppose $f \neq g$, with $ev_x(f) = ev_x(g)$ $\Rightarrow ev_x(f)-ev_x(g) = 0 \Rightarrow ev_x(f-g) = 0 \Rightarrow (f-g)(x_1,\ldots,x_k) = 0$, a contradicion to algebraic independence. Now i know that $ev_x$ is an isomorphism. And if i assume $h^{-1}(x_1,\ldots,x_k) = p(y_1,\ldots,y_l)$, and apply $(ev_x)^{-1}$, i get $h^{-1} = p(\frac{f_1}{g_1},\ldots,\frac{f_l}{g_l})$ as elements of $K(X_1,\ldots,X_k)$, which can't be because of the argument in the question – mike Jan 09 '18 at 00:36
  • Again, you can only prove this way that the function $h^{-1}$ is not a polynomial function, which is not what needs to be proved here, especially as it is true regardless of any assumptions on algebraic independence of anything. – xyzzyz Jan 09 '18 at 00:54
  • Although I am not quite able to see where my argument of using $(ev_x)^{-1}$ in the above comment fails, could you give me some hints on how to proof/complete the proof of the statement in the question? – mike Jan 09 '18 at 01:08
  • You'll find almost exact statement of your problem with proof in Miles Reid, "Undergraduate Commutative Algebra", in section 4.10 The weak Nullstellensatz on page 67. The proof they provide uses Noether normalization, however simpler proofs exist if you put extra assumptions on K (e.g. there's a very simple proof I could sketch to you if you assume that K is uncountable, and Noether normalization is easier to prove if you assume that K is infinite, see exercise 4.11 in aforementioned book). – xyzzyz Jan 09 '18 at 01:56

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I don't know if it still helps the OP because in the comments there are references to books with the solution which I don't have at hand. But I think you can argue as follows:

Suppose that $A=k(x_1,\ldots,x_n)$ is a finitely generated $k$-algebra. By Noether normalization, there exist $d$ algebraic independent elements (independent over $k$) $y_1,\ldots,y_d\in A$ such that $A$ is a finite module over $k[y_1,\ldots,y_d]$. But the number $d$ is known to be (1) $d=\mathrm{dim}(A)=0$ and (2) $d=\mathrm{trdeg}(A/k)=n$. Both follows since $A$ is a field. Hence, we have $n=0$ and $A=k$ which is the only possibility that $A$ is a finitely generated $k$-algebra.

Christoph Mark
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