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This question already has an answer here:

So, let $K$ be an algebraic closure of $\mathbb{F}_{p}$ and consider the space $V=K\otimes_{\mathbb{F}_p}\mathbb{F}_{p^n}$ as a $K$ vector space. It seems pretty straight forward to show that this vector space is $n$ dimensional since for $\{\alpha_1,\ldots\alpha_n\}$ linearly independent elements of $\mathbb{F}_{p^n}$ (considered as itself a vector space over $\mathbb{F}_p$) we have the linearly independant tensors $1\otimes\alpha_i$. Then I think it is clear that any elementary tensor is a $K$ linear sum of these and so these must span $V$.

Anyway, what I don't understand is in what way this vector space is practically different from simply considering $\mathbb{F}_{p^n}$ as a $\mathbb{F}_p$ vector space. Studying for my algebra qualifying exam, I worked on a problem that asked to calculate the basis for $V$ for which the Froebenius automorphism is in Jordan canonical form. I don't see how this is different from doing it for $\mathbb{F}_{p^n}$. Although, we know a priori that the form exists since the underlying field is complete algebraically closed, but isn't that true for $\mathbb{F}_{p^n}$?

Thanks.

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marked as duplicate by Community May 24 '17 at 20:34

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    $\begingroup$ The required roots of unity are not necessarily available in $\Bbb{F}_{p^n}$. To see this consider the case $p=2,n=3$. The Frobenius automorphism is of order three, and it is diagonalizable in $V$, because the primitive third roots of unity are in $K$. However, we don't have those roots of unity in $\Bbb{F}_8$, because $3\nmid(8-1)$. $\endgroup$ – Jyrki Lahtonen May 24 '17 at 20:31
  • $\begingroup$ I don't think this is a duplicate, $\overline{\mathbb{F}}\otimes_{\mathbb{F}_p}\mathbb{F}_{p^n}$, the Frobenius acting on it and what is useful for is a different question $\endgroup$ – reuns May 24 '17 at 20:36
  • $\begingroup$ @user1952009: You need a normal basis. I'm fairly sure I have answered that here somewhere. Gimme a minute... here? $\endgroup$ – Jyrki Lahtonen May 24 '17 at 20:36
  • $\begingroup$ @ JyrkiLahtonen Wait I didn't understand what really means "the Frobenius $T$ in Jordan normal form". We want a basis such that $T(\alpha_i) = m_i \alpha_i$ with $m \in \mathbb{F}_p$ ? $\endgroup$ – reuns May 24 '17 at 20:48
  • $\begingroup$ The minimal polynomial of $F$ is $T^n-1$. So if $\gcd(n,p)=1$ the eigenvalues of $F$ are $n$th roots of unity. Those exist in $K$, but not necessarily in $\Bbb{F}_{p^n}$. Therefore $F$ is diagonalizable over $K$ but not necessarily over the prime field, or even over $\Bbb{F}_{p^n}$. $\endgroup$ – Jyrki Lahtonen May 24 '17 at 20:49