Let $\Gamma_R[\alpha]$ denote the divided polynomial algebra over $R$; that is, the quotient of the free $R$-algebra $R\langle \alpha_1,\alpha_2,\cdots \rangle$ by the relations $\alpha_n \cdot \alpha_m = \binom{n+m}{n}\alpha_{n+m},$ with $t_0=1$. I am particularly interested in the case where $R = \mathbb{F}_p$.
The claim is that $\Gamma_{\mathbb{F}_p}[\alpha] \simeq \bigotimes_{k \ge 0} \mathbb{F}_p[\alpha_{p^i}]/(\alpha_{p^i}^p)$
(Note: I presume the tensor product is over $\mathbb{F}_p$ here?)
The proof is given is Hatcher's algebraic topology book pp. 286-287. I can understand what he is doing (sort-of), but I can't see how it all combines to give a proof.
First he claims that $\Gamma_{\mathbb{F}_p}[\alpha] = \Gamma_{\mathbb{Z}}[\alpha]\otimes \mathbb{F}_p$. I don't see why this is the case?
Then he claims that this is equivalent to the statement
$\ast$ The element $\alpha_1^{n_0}\alpha_{p}^{n_1} \cdots \alpha_{p^k}^{n_k}$ in $\Gamma_\mathbb{Z}[\alpha]$ is divisible by $p$ iff $n_i \ge p$ for some $i$.
which I don't see follows from the above. Now, as he states, we can use the product relation above to get $\alpha_1^{n_0}\alpha_{p}^{n_1} \cdots \alpha_{p^k}^{n_k} = m \alpha_n$ for $n = n_0+n_1p+\cdots n_k p^k$ and some integer $m$, and then the question is if $p$ divides $m$.
His other fact is
$\ast \ast \alpha_n \alpha_{p_k}$ is divisible by $p$ iff $n_k=p-1$, assuming $n_i < p$ for each $i$.
I am OK with this - this follows easily from Lucas' theorem.
How does $\ast \ast$ imply $\ast$? It is meant to be via an inductive argument by multiplying on the right by $\alpha_{p^i}$, but I can't quite see exactly what I should be proving.
Any tips appreciated!