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I struggle on this since two days, and still found no answer. My course states the following:

If the characteristic of $K \neq p$, then they are exactly $p-1$ different $p$-th roots of unity in the decomposition field $L$ from the polynomial $X^p-1$ over $K$, because $(X^p-1)'=pX^{p-1}$.

That I am not sure I totally understand. The formal derivative is not constant and has $X=0$ as root with multiplicity $p-1$. If a polynomial has one if its roots also in the derivative, then this root is of multiplicity at least 2. This fact cannot be used here as the roots are different and the roots should all have multiplicity 1 to prove the statement I suppose. But I do not see how this derivative be used to prove this first statement.

Leaving field theory, I do understand that there are $p$ complex roots to the polynomial $X^p-1$ including $1$. If we suppress $1$ as it is not primitive, we do obtain $p-1$ roots primitive roots of unity. Still I cannot make the link with the latter.

The second statement I even less understand.

If the characteristic of $K=p$, then we have $X^p-1 = (X-1)^p$ en there does not exist any $p$-th primitive root of unity in any possible field extension of $K$.

I do understand that because of the freshman's dream $X^p-1=(X-1)^p$, the only possible root is $1$, which is not primitive. Also, using the fact above we have $(X^p-1)'\equiv 0\mod p$, but I don't succeed in interpreting this.

Leaving field theory again, if we consider the different complex roots of the polynomials $\alpha_i$ (excluding non-primitive root $1$, we would have $1\leq i \leq p-1$). Why can't we consider $L=K(\alpha_1,...,\alpha_{p-1})$ as a valid field extension?

I already found this topic (No field of characteristic $p > 0$ contains a primitive $p^{th}$ root of unity.), but it didn't answer my questions, maybe it can still help somebody.

Thank you in advance,

Best regards,

Henri

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  • $\begingroup$ I don't see what the question is. In any field of characteristic $p$, the only solution of $x^p=1$ is $x=1$. $\endgroup$ – Lord Shark the Unknown Jun 22 '18 at 19:31
  • $\begingroup$ For the first statement, Any multiple root of the original polynomial would have to be a root of the derivative, but $0$ is the only root of the derivative so, as $0$ is not a root of the original, the original has no multiple roots. $\endgroup$ – lulu Jun 22 '18 at 19:32
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    $\begingroup$ In a general field of characteristic $7$, what does $.6235$ mean? $\endgroup$ – lulu Jun 22 '18 at 19:40
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    $\begingroup$ Your $.6235$ isn't even a rational number...it's short hand for $\cos \left( \frac {2\pi}7\right)\approx 0.623489801858\cdots $. Functions like $\cos $ aren't even defined in general fields. $\endgroup$ – lulu Jun 22 '18 at 19:55
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    $\begingroup$ You need to revisit what characteristic $p$ means. What happens over complex numbers has very little to do with what happens in fields of positive characteristic. $\endgroup$ – Torsten Schoeneberg Jun 22 '18 at 20:16

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