1
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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 $p$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.