The title has it all. I'm looking for a proof/disproof of the fact that an algebraically closed field, say $\mathbb K$, has characteristic zero iff the following property (R) holds: For all $n,k \in \mathbb N^+$, every invertible $n$-by-$n$ matrix with entries in $\mathbb K$ has at least one $k$-th root. The question is certainly well-known, and boils down to the case of Jordan blocks. I myself have a sense, but not a proof, that it must have an answer in the positive. I'm not interested in the discussion of special cases (e.g., the complex case is quite standard, and can be treated even analytically), unless of course the inspection of a finite, small number of them leads to a general conclusion. In case of an affirmative answer, I'd appreciate much a reference to the result in its full generality. As for motivation, the question is 'naturally' related to another that I've recently posted. Thanks in advance for any help.
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asked Mar 17, 2013 at 19:14
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3$\begingroup$ Why don't you try the obvious thing of thinking of a $2\times 2$ Jordan block with 1s on the diagonal and raising it to the power $p$ ($p$ the characteristic of the field), and then thinking about what this tells you. $\endgroup$– user30035Commented Mar 17, 2013 at 19:19
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$\begingroup$ The p-th power of any n-by-n invertible Jordan block with entries in a field of characteristic p is obviously the identity of the relevant matrix ring. But this counts just as a special (and simple) case, doesn't it? And I'm not interested, as I said, in special cases. $\endgroup$– Salvo TringaliCommented Mar 17, 2013 at 19:30
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$\begingroup$ To clarify the previous comment, I'm saying that the case of an algebraically closed field $\mathbb K$ of finite characteristic, though relevant, is trivial, once noticed that every $n$-by-$n$ matrix with entries in $\mathbb K$ has a Jordan normal form. The interesting case is that of a field of zero characteristic (I guess that I should have phrased the OP in a different form). $\endgroup$– Salvo TringaliCommented Mar 17, 2013 at 19:44
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1$\begingroup$ @Salvo: because it's the identity, however are you going to find a $2\times2$ matrix whose $p$th power is not semisimple? QED! $\endgroup$– user30035Commented Mar 17, 2013 at 20:38
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3$\begingroup$ The other way: you know a proof for the complexes, and so it's true for all algebraically closed fields of characteristic zero by general nonsense. $\endgroup$– user30035Commented Mar 17, 2013 at 20:40
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1 Answer
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wccanard's comment gives one direction: If the field has characteristic $p$, then there is no matrix $A$ with $A^p=\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}$.
For the other direction, let the field have characteristic $0$. We want to show that $A^k=B$ has a solution $A$ for any $B$. Without loss of generality, $B$ is a Jordan block with eigenvalue $1$, so $B-1$ is nilpotent. Then $A=\sum_{i\ge0}\binom{1/k}{i}(B-1)^i$ is a finite sum with $A^k=1+(B-1)=B$.
answered Mar 17, 2013 at 19:45
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$\begingroup$ Wonderful! Do you know of any reference to (the 2nd part of) your proof (other than the link to this answer)? $\endgroup$ Commented Mar 17, 2013 at 20:01
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$\begingroup$ I don't know a reference, but I'm pretty sure that it should be at many places. Maybe mostly only over the complex numbers, but that doesn't make a difference. All we need is that the binomial coefficients $\binom{1/k}{i}$ exist in the base field. $\endgroup$ Commented Mar 17, 2013 at 20:08
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$\begingroup$ I had a look at Higham's Functions of Matrices - Theory and Applications, before asking. And even if the 7th chapter of the book is entirely focused on $k$-th roots of complex matrices, nothing like your (absolutely brilliant) proof seems to be there. And the 3rd edition of the classical Matrix Computations by Golub and Van Loan spends no word for arbitrary $k$-th roots, while square roots are just mentioned in relation to the Cholesky and polar decompositions (and again, there's nothing in the lines of your slick argument). $\endgroup$ Commented Mar 17, 2013 at 21:18