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Question. Let $p$ be a prime. Let $q$ be a power of $p$. Let $P^0$, $P^1$, $P^2$, ... be elements of some associative $\mathbb F_q$-algebra $A$. (Here, $P^i$ does not mean $P$ to the $i$-th power; instead, $i$ is an upper index.) Asume that the power series $\sum\limits_{a,k} \left(tu\right)^a P^a P^k$ and $\sum\limits_{c,j} u^c t^{qj}P^cP^j$ are equal, where $t$ is the indeterminate of our power series and $u=\left(1-t\right)^{q-1}=1+t+t^2+...+t^{q-1}$. Prove that any nonnegative integers $a$ and $b$ such that $a < qb$ satisfy

$\displaystyle P^aP^b = \sum\limits_j \left(-1\right)^{a-qj} \binom{\left(b-j\right)\left(q-1\right)-1}{a-qj} P^{a+b-j}P^j$.

(If you don't use the same conventions about negative binomial coefficients as I do, think of this sum as going from $j=0$ to $j=\left\lfloor a/q\right\rfloor$.)

Motivation. This question is equivalent to deriving the Adem-Wu relations in the Steenrod algebra (without the Bockstein) from the Bullett-Macdonald formula. I am working in the invariant-theoretical setting, so I want a proof which does not refer to the topological interpretation of the Steenrod algebra (at least not unless it shows that this is equivalent to the invariant-theoretical one).

Larry Smith, An algebraic introduction to the Steenrod algebra, arXiv:0903.4997 gives a proof using complex integration, but (the indexing mistakes put aside) I do not really believe it. It seems to work over $\mathbb Z$ first (which allows for integration) and then project onto $\mathbb F_q$, which is okay, but I think the condition that $\sum\limits_{a,k} \left(tu\right)^a P^a P^k = \sum\limits_{b\geq j} u^{b-j}t^{qj}P^{b-j}P^j$ cannot be "lifted" to $\mathbb Z$ in a straightforward way, to begin with, which puts the whole complex-analysis approach under question. Probably it works with the right incantations being said, but I was not able to come up with these incantations (and way too confused in this topic). A pedestrian algebraic proof would be preferred.

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up vote 5 down vote accepted

Larry Smith is not really using complex integration. Instead, he is using the residue map, which can be defined algebraically by the rule $$ \text{res}\left(\sum_{k=-N}^\infty a_k z^k dz\right)=a_{-1} $$ and the fundamental transformation property that $$ \text{res}\left(\sum_{k=-N}^\infty a_k \;f(z)^k\;f'(z) dz\right) = a_{-1} $$ under appropriate conditions on $f$. For example, this works if everything is happening over a commutative ring $R$, and $f(z)=\sum_{k=0}^\infty b_kz^k$ with $b_0$ nilpotent and $b_1$ invertible. The basic idea is old and well-known, perhaps due to Cartier. One possible set of technical details is explained in Section 5.4 of my paper "Formal groups and formal schemes":

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Thanks a lot. Since this is way more complicated than I thought, I cannot check it at the moment, but I certainly will. – darij grinberg May 15 '11 at 19:15
I see now. I was as far as formulating your Corollary 5.36 and proving it in characteristic $0$ myself, but I failed to see that it is a polynomial identities in the coefficients of $f$ and $a_k$, so it would immediately project on rings of arbitrary characteristic. – darij grinberg May 15 '11 at 22:39

Bullet and Macdonald gave an algebraic proof of this in their paper On the Adem relations

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Interesting, because Smith claims that the proof they give is the same as he gives (and even that he gives more detail). Do you have the Bullett-Macdonald paper in electronic form? – darij grinberg May 15 '11 at 16:31

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