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I'm currently doing an exercise on Noether Normalization in the context of a course on commutative algebra and I'm not sure whether the solution I have come up with is correct or does even make sense. Reason for that is probably also, that I'm getting confused with the notions of:

  • a $B$-algebra $A$ that is of finite type which we defined as:
    $A$ is finitely generated as a $B$-algebra, i.e. there is a surjective morphism of $B$-algebras: $B[X_1,X_2,...,X_l] \rightarrow A$
  • a finite $B$-Algebra $A$ which we defined as:
    $A$ is finitely generated as a $B$-module, i.e. there is a surjective morphism of $B$-modules: $B^l \rightarrow A$

Ok, so the exercise is the following:
We are given a field $k$, a $k$-algebra $A$ of finite type which is also an integral domain and $k$-sub algebra $B \subset A$ of $A$ that is isomorphic with a polynomial ring over $k$ in $n$ variables, where $n=trdg(Frac(A)/k)$.
We are supposed to give an example where $A$ is not finite over $B$.

The form of the Noether Normalization Theorem from my course is the following:
(As above) $k$ is a field, $A$ a $k$-algebra of finite type which is also an integral domain, $n=trdg(Frac(A)/k)$.
Then there exists a finite, injective morphism of $k$-algebras:
$k[X_1,X_2,...,X_n] \rightarrow A$
i.e. this morphism is injective and $A$ is finitely generated as a $k[X_1,X_2,...,X_n]$-module. We only defined "finite morphism" in a side note in another proposition, so I'm not absolutely sure about the definition.

So, to be honest, if the exercise had been: "Proove that $A$ is finite over $B$.", I would've thought, yup seems like an application of the Noether Normalization lemma. But if I understand correctly the exercise exactly tells us, that finiteness does not always carry over via isomorphisms.

The example, of which I'm not sure whether it is correct, is the following:
Let $A=k[X] \simeq k[X,Y]/(Y),\; Frac(A)=Frac(k[X])=k(X)$ and thus $n=trdg(Frac(A)/k)=1$.
Set $B=k[X^2]\subset A$. I think we can say $B=k[X^2] \simeq k[T]$
Now if $A=k[X]$ was finite over $B=k[X^2]$, there would have to exist an integer $l$ and a surjective morphism of $B$-modules:
$k[X^2]^l \rightarrow k[X]$
Intuitively, that shouldn't be the case, but I'm not sure whether all the assumptions from the exercise are satisfied, as I'm quite new to the whole subject.
Is this example correct ?

I hope the question isn't all too long.
Thanks for any help :) .

Sebastian Hebold
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  • Your $A$ is generated as a $B$-module by two elements $1,x\in A$, by division algorithm. – Mohan Mar 08 '16 at 16:59
  • Oh, I see what you are saying, thanks. I'll try to find something else. Do you know an example ? Not so you could tell me, but just so that I know, I'm not looking for something that doesn't exist. Thanks :) . – Sebastian Hebold Mar 08 '16 at 17:18
  • Plenty. But, have you learned the relation to your question with integral extensions? – Mohan Mar 08 '16 at 18:22
  • Not yet, I guess. Noether Normalization was more or less the last proposition in our chapter on dimension. The chapter on integral extensions comes right after this, on which I started working after I had posted the question. Is there a condition on B such that we could actually deduce that $A$ is finite over $B$ ? I guess, I'll try to prove that $A$ is finite over $B$, and maybe see what goes wrong. – Sebastian Hebold Mar 08 '16 at 20:10

1 Answers1

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The exercise is supposed to tell you the following:

Noether Normalization tells you: If $A$ is of finite type over $k$ and $n$-dimensional, we can find some $B \cong k[X_1, \dotsc, X_n]$, such that $B \subset A$ is finite.

The exercise tells you, that one actually has to pick that $B$ wisely, as not any such $B$ does the trick. It can happen, that $k[X_1, \dotsc, X_n] \cong B \subset A$, but $B \subset A$ is not finite.

An example is $k[x] \subset k[x,x^{-1}]$ (Geometrically, we project the hyperbola on the $x$-axis, thus the origin has no pre-image).

In this example, the "correct" $B$ (or say, one possible choice of $B$) would be $B=k[x+x^{-1}] \subset k[x,x^{-1}]$ as this extension (as a module) is generated by $1,x,x^{-1}$.

MooS
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  • Ohhh, so if we were always able to deduce that $B \subset A$ must be finite, then the statement of the Noether Normalization lemma, from my lecture would say:
    Any injective morphism $k[X_1,...,X_n] \rightarrow A$ of $k$-algebras in $n=trdg(Frac(A)/k)$ many variables is already finite, as the image of such an injectiv $k$-algebra morphism is isomorphic with $k[X_1,...,X_n]$.

    But the above statement is false and we were supposed to give an example for that.

     – Sebastian Hebold Mar 09 '16 at 16:09
  • Now it took me forever to verify the statements u made in your post, but I guess the following arguments work: $k[X] \subset k[x,x^-1]$ isn't finite: If $k[x] \subset k[x,x^{-1}]=k[x][x^{-1}]$ was finite, it would be integral. Thus we would have that $x^{-1}$ is integral over $k[x]$, i.e. we would have a normed polynomial equation of degree $m$ with coefficient in $k[x]$, that has $x^{-1}$ as a zero. Multiplying by $x^{m-1}$ would give us $x^{-1} \in k[x]$, wchich isn't the case. Hence the extension isn't integral and thus not finite. – Sebastian Hebold Mar 09 '16 at 16:10
  • Sorry I'm pretty new to the forums and I couldn't post any longer comments, so I posted three. The "Answer Your Question" thingie said, I shouldn't respond to other answers that way. Thanks very much for your explanations :) .

    About your example for Noether Normalization: $k[x+x^{−1}] \subset k[x,x^{−1}]$ is finite because: $k[x+x^{−1}][x,x^{-1}]=k[x,x^{−1}]$ and $x$ and $x^{-1}$ are intregral over $k[x+x^{−1}]$ via: $-(x+x^{−1})x^3 + (x+x^{−1})^2x^2 - x^2 - 1 =0$ and a similar equation holds for $x^{-1}$.

     – Sebastian Hebold Mar 09 '16 at 16:19
  • Yes these are good thoughts. But actually the integral equation for $x$ is $$x^2-(x+x^{-1})x+1$$ – MooS Mar 09 '16 at 21:31