I'm in the process of reading Artin's Algebra, and I seem to have reached a corollary with a problematic proof.
Corollary 4.7.16 Let $T$ be a linear operator on a finite-dimensional vector space over a field $F$ of characteristic zero. Assume that $T^r=I$ for some $r\geq 1$ and that the polynomial $t^r-1$ factors into linear factors in $F$. Then $T$ is diagonalizable.
The proof is not explicitly stated, but the intention is for the corollary to generalize a preceding theorem.
Theorem 4.7.14 Let $T$ be a linear operator on a finite-dimensional complex vector space $V$. If some positive power of $T$ is the identity, say $T^r=I$, then $T$ is diagonalizable.
The proof of this theorem relies in turn on a different preceding corollary.
Corollary 4.7.13 Let $T$ be a linear operator on a finite-dimensional complex vector space. The following conditions are equivalent: (a) $T$ is a diagonalizable operator, (b) every generalized eigenvector is an eigenvector, (c) all of the blocks in the Jordan form for $T$ are $1\times 1$ blocks.
In particular, 4.7.14 relies on the implication (b)$\implies$ (a), as it proves $T$ is diagonalizable by showing all generalized eigenvectors are eigenvectors. However, the proof of (b)$\implies$ (a) follows from (b)$\implies$ (c) $\implies$ (a) which assumes the existence of Jordan form.
So, inherent in the proof of Corollary 4.7.16 is the existence of Jordan form. As I see it, the issue is that the proof of Jordan form in Artin's text requires the characteristic polynomial of $T$ to factor into linear factors in $F$, and this isn't a hypothesis in Corollary 4.7.16.
On the one hand, I'm wondering if I'm grossly overlooking something and my concern is unwarranted? And if not, I'd like to know if any slight adjustment to Artin's proofs/statements irons out the problematic proof?
I'd like to emphasize that I am not concerned with a different proof of the corollary. I know a minimal polynomial argument works. I'd just like to figure out what Artin had in mind, or get confirmation that this is a legitimate issue which can't easily be resolved.
If $F$ is not algebraically closed, then there will be linear transformations that do not have Jordan forms over $F$ because their minimal polynomials don't split.
However, when your minimal polynomial splits, you can construct a Jordan form (whether your field is algebraically closed or not). In Corollary 4.7.16, we are assuming splitting.