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Can anyone help with this:

If $L/K$ is a finite field extension, and we have a $K$-bilinear form given by $$(x,y)\mapsto Tr_{L/K}(xy)$$ then the form is either non-degenerate or $Tr_{L/K}(x)=0$ for every $x\in L$.

So far, I feel like I've run into something a little nonsensical. Suppose the form is degenerate, i.e. $\exists \alpha\in L$, $\alpha\neq 0$ such that $(\alpha,\beta)=0$ for all $\beta$. Then specifically, $Tr_{L/K}(\alpha\alpha^{-1})=Tr_{L/K}(1)=0$. But it is a theorem from Morandi's "Field and Galois Theory" that if $\alpha\in K$ (the base field) then $Tr_{L/K}(\alpha)=n\alpha$ where $n$ is the dimension of the field extension, so in this case we get that $[L:K]=0$ which doesn't make any sense. Am I doing something wrong here?


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No, you get that $[L : K] = 0$ in the field $K$. What does that tell you about the characteristic of $K$? –  Qiaochu Yuan Apr 9 '11 at 18:23
So $K$ has characteristic n then, or some divisor of n, I guess. So maybe we can use this to show something about separability of the field extension. –  Jon Beardsley Apr 9 '11 at 18:25
Qiaochu, do you agree with the below solution? That is, if trace is identically zero, obviously the form is degenerate, and if the form is degenerate for $\alpha$, then $Tr(\alpha\alpha^{-1}\beta)=Tr(\beta)=0$ for every $\beta$? –  Jon Beardsley Apr 9 '11 at 19:53
what Bryan White writes is correct and is, I think, a nice way of looking at it. –  Pete L. Clark Apr 9 '11 at 20:45
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2 Answers

up vote 3 down vote accepted

I don't know about the separability of the extension but I can see how to prove the proposition. Take your $\alpha$ and hit it with $\alpha^{-1}\beta$ and the conclusion follows quickly.

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I think you're right. I've been trying all kinds of Galois theory stuff, but I think you're right, it's really basic algebra. I'm really dumb.... –  Jon Beardsley Apr 9 '11 at 19:51
Use $\alpha$ to produce $\alpha$, $\alpha^{-1}\beta$ to produce $\alpha^{-1}\beta$. –  Arturo Magidin Apr 9 '11 at 19:51
Oh nice. Thanks for tip. I'm a Tex retard. –  bcwhite Apr 10 '11 at 19:44
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For what it's worth, a complete proof of the following fact can be found in $\S 7$ of these notes.

Theorem: For a finite degree field extension $K/F$, the following are equivalent:
(i) The trace form $(x,y) \in K^2 \mapsto \operatorname{Tr}(xy) \in F$ is a nondegenerate $F$-bilinear form.
(ii) The trace form is not identically zero.
(iii) The extension $K/F$ is separable.

The specific answer to the OP's question appears in there, but here it is: always be careful to read the fine print when dealing with inseparable extensions. In this case, it turns out that when $K/F$ is inseparable, the trace from $K$ down to $F$ of $x \in K$ comes out as a power of $p$ (the characteristic of $F$) times the more familiar sum of Galois conjugates. But in characteristic $p$ multiplying something by a power of $p$ is the same as multiplying it by zero...so the trace is identically zero in inseparable extensions.

[Note also that the proof of another part of the theorem makes use of the Primitive Element Theorem, which is currently to be found in $\S 8$ of the notes, i.e., the following section. This is an obvious mistake which will have to be remedied at some point. There are plenty of other issues with these notes, which are still quite rough and incomplete.]

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Thanks. Is Bryan's answer above incorrect? –  Jon Beardsley Apr 9 '11 at 20:43
Nope, it looks perfectly correct to me. –  Pete L. Clark Apr 9 '11 at 20:47
Okay, thanks. Yeah, that's a helpful fact that I hadn't thought of. Basically, you're saying the inseparable degree here is a power of $p$. –  Jon Beardsley Apr 9 '11 at 20:49
@Jbeardz: that's exactly what I'm saying. –  Pete L. Clark Apr 9 '11 at 20:51
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