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Let $k$ be a field and let $A$ be a local $k$-algebra which has finite dimension over $k$. Let $\mathfrak{m}$ be the maximal ideal of $A$ and let $k' = A / \mathfrak{m}$ be the residue field.

For every $a \in A$ we define $\mathrm{Tr}(a) \in k$ the trace of the $k$-linear endomorphism $A \to A$ given by the multiplication by $a$. The map $\mathrm{Tr} \colon A \to k$ is $k$-linear. Consider the symmetric $k$-bilinear form $\langle \cdot,\cdot \rangle \colon A \times A \to k$ given by $\langle a, b \rangle = \mathrm{Tr}(ab)$.

Now consider the following statements:

  1. $A$ is reduced, or equivalently $\mathfrak{m} = 0$, i.e. $A = k'$;
  2. $\langle \cdot, \cdot \rangle$ is non-degenerate;
  3. $k' / k$ is a finite separable extension of fields.
  4. $\mathrm{Tr} \colon A \to k$ is surjective.

(Some assertions contained in this paragraph are wrong.) I know a proof of the facts "(2) implies (1)" and "(2) iff (3) iff (4)". (1) does not implies (2) because it suffices to pick $A$ as an inseparable finite field extension of $k$. Is there some fact which involves the field extension $k' / k$ and which is equivalent to (1)?

My question comes from studying the property of the discriminant and the ramification in finite extensions of Dedekind domains. Lemma 7.4.14 at page 289 of Qing Liu's Algebraic geometry and arithmetic curves concerns the problem above; but, in my opinion, the condition "every nilpotent element $\epsilon$ of $A$ verifies $\epsilon^m = 0$ for some integer $m$ prime to $\mathrm{char}(k)$" is empty. Am I wrong? It seems to me that the following proposition holds:

Proposition Let $n \geq 2$ be an integer and let $A$ be a noetherian ring with nilradical $\mathcal{N}$. There exists an integer $m$ prime to $n$ such that $\mathcal{N}^m = 0$.

EDIT. I hope to do true assertions, unlike my original post. (2) implies (1) because a nilpotent element of $A$ belongs to the radical of the form $\langle \cdot, \cdot \rangle$. If (1) holds, then (2), (3) and (4) are equivalent. (4) holds if an only if $\langle \cdot, \cdot \rangle$ is non zero. So it is obvious that (2) implies (4). Now consider the further statement:

5 . $k'/k$ is separable and every nilpotent element of $A$ has nilpotence order prime to $\mathrm{char} (k)$.

(4) and (5) are equivalent by the lemma cited above.So (4) implies (3).

Now my questions are:

  • Is there exists a proof of the lemma cited above that does not require scheme theory?
  • Are there any other implications among the statements above?
  • How the notions smooth and étale are involved with this matter?
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Andrea, you are completely right ! The correct statement is every nilpotence element $\epsilon$ of $A$ has nilpotence order prime to $\mathrm{char}(k)$. The proof of the lemma is based on this hypothesis. –  user18119 Nov 17 '11 at 12:35
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3 Answers

up vote 1 down vote accepted

First, any nilpotent element of $A$ has trace $0$ over $k$. So as you noticed, (2) implies that $A$ is a field. Then the classical theory (see Georges's answer) implies that $A$ is a finite separable field extension of $k$. Of course, the converse is also true.

Now we consider Condition (4). As $A$ is Artinian, the separable closure $k''$ of $k$ in $k'$ can be lifted in $A$ and $A$ becomes a $k''$-algebra. Moreover $$\mathrm{Tr}_{k''/k}(\mathrm{Tr}_{A/k''}(A))=\mathrm{Tr}_{A/k}(A),$$ and $\mathrm{Tr}_{k''/k}(k'')=k$. So (4) is equivalent to $\mathrm{Tr}_{A/k''}(A)=k''$.

Note that $\mathrm{Tr}_{A/k''}(1)=\dim_{k''} A$. So (4) holds if $\mathrm{char}(k)=0$ or is positive and prime to $\dim_{k''} A$. These conditions imply $k''=k'$ because $\dim_{k''} A$ is divisible by $[k':k'']$ (use the chain of the quotients $\mathfrak m^r/\mathfrak m^{r+1}$ which are $k'$-vector spaces). So the remaining case is when $\mathrm{char}(k)=p>0$ and $p$ divides $\dim_{k''} A$. Then let me prove that $\mathrm{Tr}_{A/k''}(A)=0$. We already have $\mathrm{Tr}_{A/k''}(k'')=0$ (because $p$ divides $\dim_{k''} A$) and $\mathrm{Tr}_{A/k''}(\mathfrak m)=0$. Let $a\in A$. As $k'/k''$ is purely inseparable, there exists $r\ge 1$ such that $a^{p^r}\in k'' + \mathfrak m$. Then $$(\mathrm{Tr}_{A/k''}(a))^{p^r}=\mathrm{Tr}_{A/k''}(a^{p^r})=0.$$ So the trace of $a$ is zero.

Conclusion: (4) is equivalent to

$k'/k$ is separable and $\dim_k A/[k':k]$ is prime to $\mathrm{char}(k)$.

In particular, the cited Lemma is correct only when $A$ is generated by one element (which is enough for the purpose of Proposition 7.4.13). Hope you agree with the above proof.

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Great answer! The only point which is obscure for me is the embedding of $k''$ in $A$; I was tempted by using the $0$-smoothness of $k''$ over $k$, but it is not true that $\mathfrak{m}^2 = 0$. Besides I don't understand why the completeness of $A$ could be useful. –  Andrea Nov 20 '11 at 14:33
    
@Andrea you did right. You have to lift a primitive element of $k''/k$ to $A$. Using smoothness you find a solution when $\mathfrak m^2=0$. Suppose $\mathfrak m^n=0$, then you find a solution in $A/\mathfrak m^{[(n+1)/2]}$ and so on. As $A$ is Artinian you end up with a solution in $A$. I talked about completeness because more generally the lifting result holds for complete local ring, but the proof uses the Artinian case. I will edit. –  user18119 Nov 20 '11 at 19:38
    
Ok! I thank you very much! –  Andrea Nov 20 '11 at 22:06
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I am confused: what do you want to show?

After all tame ramification must be allowed in Liu's lemma. See the following example: Take a field $k$ with $\mathrm{char}(k)\neq 2$ and consider the extension $k[[t]]\subset k[[\sqrt{t}]]$ of discrete valuation rings. Let $A:=k[[\sqrt{t}]]/tk[[\sqrt{t}]]$, then a basis of $A$ as a $k$-vector space is given by the residues $b_1:=1+tk[[\sqrt{t}]]$ and $b_2:=\sqrt{t}+tk[[\sqrt{t}]]$.

The coordinate matrix of the left multiplication by the element $a_1b_1+a_2b_2$, $a_1,a_2\in k$ looks like this:

$ A=\left(\begin{array}{cc}a_1 & 0\\ a_2 & a_1\end{array}\right) $

Hence $\mathrm{Tr}(a_1b_1+a_2b_2)=2a_1$ thus showing surjectivity. So reducedness of $A$ is not necessary for the equivalence of (3) and (4).

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More directly $A=k[\epsilon]/(\epsilon^2)$. I deletted my stupid remark on the reducedness of $A$. –  user18119 Nov 17 '11 at 20:37
    
Your example also shows that (4) doesn't imply (1). –  user18119 Nov 17 '11 at 20:38
    
I also am confused. I thank both of you. I'm trying to edit my post in a more precise version... –  Andrea Nov 17 '11 at 21:30
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Yes, I also think that the condition you quote is empty and that your Proposition is true:
If $\mathcal N^r=0$ (and such an $r$ exists by noetherianity), any prime $p\geq r$ not dividing $n$ will do .

More generally, a finite dimensional algebra over a field is separable ( or equivalently: étale) if and only if it isomorphic to the product of finitely many separable (automatically finite dimensional) extension fields.
If the algebra is local, there can of course be only one factor in the product!

Edit Since Andrea asks, yes there are proofs not involving scheme theory. You can look one up in Bourbaki's Algebra, Chapter 5, §8, Proposition 1. Beware that for Bourbaki "extension" is used only for algebras that are fields. You will see that the surjectivity of the trace on $A$, your point 4, implies that $A$ is separable (=étale) if $A$ is a field, but not otherwise. This explains how Hagen's counterexample is possible.

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