Over what fields are finite order endomorphisms of vector spaces diagonalizable? The problem is as follows:
Let $V$ be a finite dimensional vector space over $F$. Let $T : V \to V$ be a linear endomorphism such that $T^{n} = I$ for some fixed $n$. What is a necessary and sufficient condition on $F$ for $T$ to be diagonalizable?
I have shown that $T$ will be diagonalizable if its minimal polynomial factors into distinct linear terms in $F$. I also know that the minimal polynomial must divide $x^n - 1$. From here, I am stuck, as I do not know how to gather any more information about the form of the minimal polynomial
Thank you in advance for any help.
 A: If you want conditions on $F$ that are necessary and sufficient so that any endomorphism $T$ of any finite dimensional vector space with $T^k=I$ is diagonalizable, then...


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*For characteristic zero, a necessary and sufficient condition is that $F$ contain all $k$th roots of unity for each $k\geq 1$. To see the sufficiency, note that the minimal polynomial of such a $T$ divides $x^k-1$, and over such a field this splits into distinct linear factors. Conversely, the companion matrix of $x^k-1$ has minimal polynomial $x^k-1$, which you need to factor into linear terms in order to be diagonalizable. Thus, the condition (which is weaker than being algebraically closed) is both necessary and sufficient.

*For positive characteristic, this is impossible. Let $p$ be the characteristic. The companion matrix of $x^p-1 = (x-1)^p$ has minimal polynomial $(x-1)^p$, and hence is not diagonalizable. So if $\mathrm{char}(F)\gt 0$, there is always an endomorphism of finite multiplicative order that is not diagonalizable. 
If $V$ is fixed, of dimension $n$, the situation is slightly different. It is not hard to verify that if $T$ has finite order in $V$, then the order is at most $n$. So in this case:


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*If $F$ has characteristic $0$, or characteristic $p$, $2\leq p\leq n$, a necessary and sufficient condition is that the field contain all $k$th roots of unity, $1\leq k\leq n$. An argument as above works (use the companion matrix of $x^k-1$ and then complete it with $0$s to get an $n\times n$ matrix that has minimal polynomial $x(x^k-1)$).

*If $F$ has positive characteristic $p\geq n$, then the same argument as above shows you cannot do it. 
Per the comment, we actually have a third permutation: $k$ is fixed. What is required so that every endomorphism of order (dividing?) $k$ is diagonalizable?


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*If $F$ has characteristic $0$ or characteristic $p$ that does not divide $k$, then you need $F$ to contain (i) all $k$th roots of unity if you want order exactly $k$ only; and (ii) all $m$th roots of unity for all divisors $m$ of $k$ if you want it for any endomorphism such that $T^k=I$. 

*If the characteristic of $F$ is $p$ and $p$ divides $k$ and is no larger than $\dim(V)$, then you’re still out of luck. Writing $k=pm$, the companion matrix of $x^k -1 = (x^m)^p-1 - (x^m-1)^p$ has minimal polynomial $(x^m-1)^p$, and hence is not diagonalizable. 
A: A sufficient condition is that $F$ is algebraically closed. But of course, this is not necessary, for the identity operator is trivially diagonalizable over any field (indeed, it is already a diagonal operator).
A: Since working in finite dimensional vector spaces one usually takes $n$ to stand for its dimension, I (like Arturo Magidin) will assume that the relation given is $T^k=I$. The question I will answer then is:
for a fixed integer $k>0$, under what condition on $F$ is it the case that every endomorphism $T$ of any (finite dimensional) vector space $V$ over $F$ that satisfies $T^k=I$ is diagonalisable. (I cannot assume that $V$ is given, as the formulation in the question suggests, since that would automatically fix $F$; and it will turn out that a hypothesis of finite dimension does not simplify the answer.)
If the supposed annihilating polynomial $X^k-1$ is split over $F$ with simple roots, then $T$ will necessarily be diagonalisable (with spectrum contained in the set of roots of $X^k-1$; a theorem guarantees this, and it does not require $V$ to be finite dimensional). Since we can realise $X^k-1$ as the minimal polynomial of an appropriate $T$, for instance using a companion matrix, this sufficient condition is also necessary: the minimal polynomial of a diagonalisable endomorphism$~T$ is $\prod_{\lambda\in\operatorname{Spec}T}(X-\lambda)$, which is split and (by definition of the product) has simple roots.
If $F$ is of characteristic $0$ of of a characteristic not dividing $k$, then $X^k-1$ is relatively prime to its derivative $kX^{k-1}$, so it cannot have multiple roots, and therefore the condition is equivalent to $X^k-1$ being split over $F$, in other words to the existence in $F$ of all the $k$-th roots of unity ($F$ contains the $k$-th cyclotomic field). If the characteristic of $F$ does divide $k$, then multiplication in $F^2$ by $\left(1~1\atop0~1\right)$ provides an example of an endomorphism$~T$ satisfying $X^k=I$ but which is not diagonalisable, so no field of such characteristic can satisfy the condition.
