Discriminant - Nice thinking stump

[Palindrom]

This is driving me nutty.

Here's a theorem and its proof as appearing in my notes. It's short, go through it if you have the courage.


Theorem. Let $$L=K\left(\alpha\right)$$, and let $$\alpha_{1},...,\alpha_{n}$$ be the $$n$$ conjugates of $$\alpha$$ . Then if $$f\left(x\right)\in K\left[x\right]$$ is the miminal polynomial of $$\alpha$$ over $$K$$, we have $$\mbox{disc}_{L/K}\left(1,...,\alpha^{n-1}\right)=\prod_{j<i}\left(\alpha_{i}-\alpha_{j}\right)^{2}=\left(-1\right)^{{n \choose 2}}N_{L/K}\left(f'\left(\alpha\right)\right)$$

Proof. The first equality is easy. For the second, note that $$f\left(x\right)=\prod_{i=1}^{n}\left(x-\alpha_{i}\right)$$ Hence, if without loss of generality $$\alpha=\alpha_{1}$$, $$f'\left(\alpha\right)=\prod_{i=2}^{n}\left(\alpha-\alpha_{i}\right)$$

Thus $$\sigma_{j}\left(f'\left(\alpha\right)\right)=\prod_{i=2}^{n}\left(\alpha_{j}-\sigma_{j}\left(\alpha_{i}\right)\right)$$.

But this move in not kosher! $$\sigma_{j}$$ is only defined on $$\alpha$$, not on any of its conjugates! That we found an expression for the evaluation of the derivative at $$\alpha$$ is all nice and good, but I don't see why I can push $$\sigma_{j}$$ into the product...

This is the way it appears in my notes, I must have been half sleeping when my professor did this not to notice. Can anyone convince me that this can be done? Or point me in an alternate (and right!) direction?

 

[Kummer]

I do not really follow what you did exactly because you did not define your notation.

This is my understanding.
First, I assume that $$f(x)$$ has all zeros of multiplicity 1, otherwise writting,
$$f(x) = \prod (x-\alpha _ i)$$ is wrong, but I assume that this is okay in your context.

I also assume that $$\sigma$$ is an automorphism of $$L$$ leaving $$K$$ fixed. In that case $$\sigma$$ maps conjugates into conjugates and elements of $$K$$ stay fixed.

 

[Palindrom]

I was just about to add my notation:

Recall from Galois theory that the n embeddings $$\sigma_{1},...,\sigma_{n}:K\left(\alpha\right)\hookrightarrow C$$ (where C is an algebraic closure of $$K\left(\alpha\right)$$) are exactly $$\sigma_{i}\left(\alpha\right)=\alpha_{i}$$.

I'm assuming the extension is separable. Better yet, assume char(K)=0.

Now do you see? $$\sigma$$ is defined only on $$\alpha$$, the other conjugates do not lie in the field $$K\left(\alpha\right)$$...

Edit: The extension is of dimension n.

 

[Palindrom]

Guys, I'm still stumped. Is my problem not clear enough or just not interesting enough?

 

[mathwonk]

too confusing to me. i am kind of lazy, can you tell me in words what is going on? what are you trying to prove? id rather just prove it than read someone else's proof. that's my advice to you too. quit trying to understand his proof since it seems flawed and just prove it.

 

[Palindrom]

I agree with you, the thing is I don't really have any other direction (it's also how I would have started).

Here's the thing to prove: In the field L=K(a), the discriminant of 1,a,...,a^n-1 (where n is the dimension of the extension) is equal to the norm of f'(a) (the derivative of the minimal polynomial of a over K evaluated at a).

Edit: Char(K)=0.

 

[matt grime]

What do you mean that sigma is only defined on alpha? It is not. It is a Galois GROUP you're using.

 

[Palindrom]

I meant that $$\sigma$$ is only defined on K(a), and that the conjugates of a do not necessarily lie in K(a) (take, for example, a to be the real third root of 2).

I'll sharpen it, and maybe then you'll tell me how it is defined on its conjugates.

We know that given an extension L/K of degree n, there are exactly n distinct embeddings of L into its algebraic closure C.

Now, the extension L=K(a) over K isn't necessarily Galois (like my above example). There are actually more embeddings of E, the splitting field of the minimal polynomial of a, into C. We can't say that the n $$\sigma$$'s are defined on the conjugates...

Now go, make me understand your point.

 

[mathwonk]

well if it makes no sense why not change it. instead of letting the sigmas be embeddings, choose one embedding, and then after your field is embedded, let the sigmas be the galois group of the noirmal closure of that embedded subfield. then they are defined everywhere, and have the same action on the subfield.

just make sense of the argument, that's all I am saying. even if you have to change it.

 

[matt grime]

I meant that $$\sigma$$ is only defined on K(a), and that the conjugates of a do not necessarily lie in K(a) (take, for example, a to be the real third root of 2).

and is that a Galois extension? The point is you have a splitting field of f to play with.

 

[Palindrom]

@mathwonk - What do you mean, change your embeddings? The norm of f'(a) is defined by the product of the n evaluations of the specific $$\sigma_1 ,...,\sigma_n$$. I can't just change them without proper justification...

@matt grime - First read my above comment (to mathwonk). I know I have the splitting field to play with, but the embeddings are not defined on it. I could extend them, sure, but there are several ways to do it, and anyway I have to justify that I can calculate the norm with use of this expansion...

I guess you two are trying to make me see something awfully clear, and I'm being dumb. Happens to me sometimes, sorry if that frustrates you. :)

 

[mathwonk]

i mean if you don't like his proof make your own, by changing the meanings of the sigmas until they ARE defined everywhere. don't be such a right wing conservative student. accept the awful possibility THAT YOU My be right ND HE IS wRONG.

 

[Palindrom]

Oh, of course, like I said there's no argument there. I'm absolutely going in my direction, and I'm not trying to restrict myself to what he did.

The only thing I AM restricted to is the definition - and in that definition, the sigmas are NOT defined everywhere. I'm sure there's a way of overcoming it, but that's exactly what I'm asking you to help me with.

Under no circumstances am I asking you to make his proof correct. I wrote it just in case I might have missed something in it and someone else will notice it, in particular because it is not very polite to present a theorem and say "prove it to me!" without at least some background.

 

[mathwonk]

you are ignoring my response to your request to point you in an alternate and correct direction.

indeed as you say, the formula above makes no sense unless the sigmas were defined on the aj. so you have to show how indeed they can be so defined.

threfore saying they are different embeddings of K(a) into some alg closure cannot be corrct unless L is normal, i.e. unless you assume further that all ai are contained in the field L.

you have not given us the ropecise hypotheses for your argument, i.e. whether L is assumed normal, what the sigmas mean, etc,...instead of just stamping your foot and crying that the formula is undefined, do something about it.

i.e. either assume L is normal, or take the normal closure of the union of the various embeddings of L, and extend the sigmas to that field.

 

[Palindrom]

I'm not stamping my foot, I'm being careful not to make a subtle mistake.

You have all my assumptions - L=K(a) for some a and Char(K)=0. Clearly then L/K is NOT necessarily normal.

Like I said before, the sigmas are the n distinct embeddings of L=K(a) into C, the algebraic closure of L=K(a). What's not correct about that? That's a theorem from Galois theory. Given a field K with Char(K)=0 and an extension (not necessarily normal!) L/K of degree n, there are exactly n distinct K - embeddings of L into an algebraic closure of L.

That said, I believe I now have a proof.

I understand you have the impression that I'm very close minded. It is true that I can't be the judge of that, but I'm fairly sure I'm not. What you take as close mindedness is actually pedantry. Wavy proofs are my biggest enemies, and on this I do stamp my foot.

It is very annoying to read a proof with a big whole in it. I assure you that when it happens, I usually tend to completely reject the proof and find one of my own rather than to try and fix it. I definitely do not stay stuck on it hoping that it resolves itself!

Anyway, your comments did help me, so thank you. I had a mind block about something, and it came loose just now.

 

[mathwonk]

the point is those sigmas do extend to be defiend on all those aj.