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!


To see that $\ast \ast$ implies $\ast$:

First, it should be clear that $\alpha_i^p$ is divisible by $p$ (in $\Gamma_\mathbb{Z}[\alpha]$), since, for some integer $c$ $$\alpha_i^p = c \cdot \alpha_{pi}$$ and one of the factors in $c$ is $\binom{p i}{(p-1) i}$ which is divisible by $p$ (the top has more factors of $p$ than the bottom, so this follows from Lucas' theorem). This proves one direction of $\ast$.

For the other direction, we must show that if all $n_i < p$, then the indicated product is not divisible by $p$. We proceed by induction on $k$. For the base case, note that $\alpha_1^n = c \alpha_n$, and the constant $c$ is the product of binomial coefficients all of whose top entries are $\le n < p$, hence $c$ is not divisible by $p$. For the inductive step, the product under consideration can be written (by the inductive hypothesis) as $$c \cdot a_m \cdot a_{p^k}^{n_k}$$ where $m = n_0 + n_1 p + \ldots n_{k-1}p^{k-1}$, and $c$ is not divisible by $p$. But by $\ast \ast$, $a_m a_{p^k}^{n_k}$ is not divisible by $p$ because the coefficient of $p^k$ in $m$ is 0, not $p-1$. Hence the entire product is not divisible by $p$.

  • $\begingroup$ Thanks Ted! That makes sense. Do you have any idea how the $\ast$ condition is sufficient to prove what is claimed? $\endgroup$ – Juan S May 15 '12 at 6:59
  • $\begingroup$ Here's how I would proceed. I believe the claimed isomorphism is given by $\alpha_m \mapsto \otimes \alpha_{p^i}^{n_i}$, and extending linearly. To show that the map is well-defined, you must check the proposed images satisfy the defining relation of the divided algebra. The inverse map sends $\otimes \alpha_{p^i}^{n_i} \mapsto \alpha_m$, and extending linearly. To check this is well-defined, you must check the defining relations again, and also check that if $n_i \ge p$, then the image is 0. Note: You may have to fiddle with these maps by adjusting with constants to get things to work out. $\endgroup$ – Ted May 15 '12 at 7:17

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