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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 $$f\circ g(x) :=\sum_{n=0}^\infty f_n (g(x))^n.$$

user8268
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Victor
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1 Answers1

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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 $(f\circ g)\circ h=f\circ (g\circ h)$.

user8268
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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
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    @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 http://planetmath.org/encyclopedia/IAdicTopology.html or http://en.wikipedia.org/wiki/Completion_(ring_theory)#Krull_topology 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
  • @user8268 In order to identify polynomials with polynomial functions, you need the field $K$ to be infinite. Is there an easy way to conclude that composition of polynomials is associative over finite fields? – Amritanshu Prasad Sep 30 '21 at 04:14
  • @AmritanshuPrasad This proof works also for polynomials, not just for power series (just remove the continuity assumption, which now makes no sense, and also the assumption that $g(0)=0$, which is not needed anymore) – user8268 Sep 30 '21 at 09:47
  • @user8268 I now realize that I had misunderstood your proof. I thought that you were using the associativity of functional composition of polynomial functions to conclude the associativity of composition of formal polynomials. But the main idea behind your proof is already necessary to give a slick proof of associativity of composition of formal polynomials. Lovely. – Amritanshu Prasad Oct 01 '21 at 11:31