Corollary 4.7.16 in Artin's Algebra (2nd ed.) 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.
 A: Warm-up
I guess the ending of chapter 4 in Artin's book is basically botched. The preceding theorem 4.7.10 is hard to follow because he is not explaining the big big picture of what he tries to do, namely his proof strategy.
Theorem 4.7.14
Same goes for 4.7.14 because he takes generalized eigenvector of degree 2 and states without proof that for this particular eigenvector we have
$$Tw=\lambda w$$
Logic is $$T\underbrace{(T- \lambda)v}_{w} = (T- \lambda + \lambda)(T- 
\lambda)v=\underbrace{(T-\lambda)^2}_{\text{because v is degree 2 by assumption, this is zero}}+\lambda\underbrace{(T-\lambda)v}_{w}$$
So $$Tw=\lambda w$$
He does not put any hint as to how to prove for degrees higher than two, no mention of minimum polynomials, factoring etc. Better strategy is to pick 4.7.13(c) and prove 4.7.14 using this https://math.stackexchange.com/q/734075
Corollary 4.7.15
If we can factor the characteristic polynomial in a field, then we can pick roots of the characteristic polynomial from this same field, these roots are the eigenvalues which are not necessarily distinct. So we can go along with Jordan Decomposition of 4.7.10.
Corollary 4.7.16
Suppose we have filled the gap and proved that for any degree of the generalized eigenvector we have $T w= \lambda w$. I repeat that Artin assumed it was degree two and went along.
Then we get the key identity
$$0 = (\underbrace{ T^{r-1}+ \lambda T^{r-2} + \dots + \lambda^{r-2}T + \lambda^{r-1}}_{\text{total r terms (*)}}) \underbrace{(T- \alpha  )v}_{w} $$
Using the assumed identity $T w= \lambda w$ we can simplify (*) because T is linear
$$\begin{align*}
T^{r-1}w= \lambda^{r-1}w \\
\lambda T^{r-2}w= \lambda \lambda^{r-1}w = \lambda^{r-1}w \\
\dots \\
\lambda^{r-2} T w= \lambda^{r-2} \lambda w = \lambda^{r-1}w \\
\lambda^{r-1} w = \lambda^{r-1} w
\end{align*}$$ 
So we can rewrite the above identity as
$$ 0 = \overbrace{\underbrace{r}_{\text{Note 1}} \underbrace{\lambda^{r-1}}_{\text{Note 2}}}^{\text{Note 3}} w$$
Note 1: r is the number of terms, some positive number.
Note 2: $\lambda^{r-1}$ is non-zero in our field. Suppose it is zero, then
$$\lambda \lambda^{r-1} = \lambda^{r} = 0$$
But previously we showed in our case $\lambda^r = 1$, a contradiction.
Note 3: we have to assume that in our field the field characteristic is zero. That means that no matter how many times we add up the identity element, we never get the zero element, Artin page 83
$$1 + 1 + \dots \ne 0$$
This ensures that $r \lambda^{r-1}$, adding r times element $\lambda^{r-1}$, never gives us zero. 
Now we can state that the only factor that can be zero is $w = (T - \lambda) v$, which is thus an ordinary eigenvector.
A: 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.
