This is driving me nutty.(adsbygoogle = window.adsbygoogle || []).push({});

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 [tex]L=K\left(\alpha\right)[/tex], and let [tex]\alpha_{1},...,\alpha_{n}[/tex] be the [tex]n[/tex] conjugates of [tex]\alpha[/tex] . Then if [tex]f\left(x\right)\in K\left[x\right][/tex] is the miminal polynomial of [tex]\alpha[/tex] over [tex]K[/tex], we have [tex]\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)[/tex]

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

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

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

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# Discriminant - Nice thinking stump

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