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How do I go about proving that the composition of formal power series is associative?

I've tried proving the result directly, but the resulting expressions are quite unwieldy. Currently, I'm trying to make use of the topology on $\mathbb{C}[[x]]$, but I can't quite get it to work.

More precisely, I want to prove that if $g(0)=0$ and $h(0)=0$, then $f\circ(g\circ h)=(f\circ g)\circ h$, where $$f\circ g(x) :=\sum_{n=0}^\infty f_n (g(x))^n.$$

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@Rod: – joriki Sep 11 '12 at 18:56
Do you have any evidenct to suggest that this algebraic property would need a topological justification? – rschwieb Sep 11 '12 at 19:17
@rschwieb I don't know about "need," but the idea may be to use associativity for finite polynomials and then take limits, which would use the topology. – anon Sep 11 '12 at 19:27
@anon I suppose so :) – rschwieb Sep 11 '12 at 19:33
$g(0)\ne0$ and $h(0)\ne0$ prevent the composition from being defined at all. What is $g\circ h$ supposed to be if $g(x)=\sum_{k=0}^\infty x^k$ and $h(0)=1$? – Hagen von Eitzen Sep 11 '12 at 19:51
up vote 6 down vote accepted

A formal power series $g(x)$ such that $g(0)=0$ gives you a continuous ring homomorphism $g^*:K[[x]]\to K[[x]]$ uniquely defined by $x\mapsto g(x)$. In fact, $g^*(f)=f\circ g$ (it is clear if $f$ is a polynomial and extends by continuity to power series). As $h^*(g^*(x))=h^*(g(x))=g(h(x))=(g\circ h)^*(x)$, we have $h^*\circ g^*=(g\circ h)^*$. Evaluating both sides on $f$ we get $(f\circ g)\circ h=f\circ (g\circ h)$.

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Right. This is the way to do it, just as the most efficient way to show associativity of matrix multiplication is to use the associativity of the maps that the matrices induce on the vector spaces involved. – Lubin Sep 11 '12 at 19:49
The result is true for arbitrary coefficient rings, so continuity should not be needed. – Hans Engler Sep 11 '12 at 20:05
@HansEngler: I do use continuity (neighbourhood's of $0$ are $x^n K[[x]]$); there is no topology on $K$ itself (i.e. it is discrete) – user8268 Sep 11 '12 at 20:08
Also see or for the topology $K[[x]]$ comes equipped with (any topology that $K$ may have is irrelevant to this topology; it is present for arbitrary coefficient rings). – anon Sep 11 '12 at 20:50

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