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Let $K$ be a field of characteristic $0$. Let $\alpha,\beta,\gamma \in K[X]$. We write $\alpha(X) = \sum_{i=0}^{n_\alpha} a_i X^i$, $\beta(X) = \sum_{i=0}^{n_\beta} b_i X^i$, $\gamma(X) = \sum_{i=0}^{n_\gamma} g_i X^i$. We impose $b_0 = g_0 = 0$. Let us denote by $(X)$ the set of all polynomials with zero constant term.

We define the following binary composition operation on $K[X]$, denoted by $\circ$.

$$ \circ : K[X] \times (X) \to K[X], (\alpha,\beta) \mapsto \sum_{i=0}^{n_\alpha} a_i \beta(X)^i $$

I would like to show that this operation is associative, i.e. that $\alpha \circ (\beta \circ \gamma) = (\alpha \circ \beta) \circ \gamma$.

I tried doing this by comparing coefficients but this rapidly becomes too unwieldy. I also tried solving it by recursion on the degrees of the polynomial involved, but couldn't make it work.

kiyopi
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    Maybe I'm missing something, but does the more general fact that function composition is associative not suffice? – PrincessEev Aug 23 '23 at 15:55
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    @PrincessEev Not by the definition of composition here, no. These polynomials are defined as "formal polynomials", not as functions. Whether or not equality of two formal polynomials is equivalent to equality of the functions $K\to K$ that they represent is a nontrivial question whose answer would also need proving. – Franklin Pezzuti Dyer Aug 23 '23 at 17:16
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    @Franklin: $K$ has characteristic zero here so that isn't hard to establish. Even if that weren't true the only issue is if $K$ is finite and in that case you can pass to the algebraic closure. – Qiaochu Yuan Aug 23 '23 at 17:17

2 Answers2

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There's no reason to restrict to $(X)$ in the second argument (you would only need that for formal power series). Conceptually this is true because polynomials are functions (on $K$ in this case, although if $K$ were finite you'd have to pass to $\bar{K}$ or similar), polynomial composition is function composition, and function composition is associative, but it's also possible to give a direct proof, as follows:

  • Show that composition is $K$-linear in the first variable: $(c_1 \alpha_1 + c_2 \alpha_2) \circ \beta = c_1 (\alpha_1 \circ \beta) + c_2 (\alpha_2 \circ \beta)$.
  • Show that composition is also multiplicative in the first variable: $(\alpha_1 \alpha_2) \circ \beta = (\alpha_1 \circ \beta) (\alpha_2 \circ \beta)$. It's convenient here to use linearity in the first variable which shows that both sides are bilinear in $\alpha_1$ and $\alpha_2$; this reduces us to the case $\alpha_1 = X^i, \alpha_2 = X^j$ which just amounts to $\beta^{i+j} = \beta^i \beta^j$.
  • By applying each of the above $3$ times, conclude that $\alpha \circ (\beta \circ \gamma)$ and $(\alpha \circ \beta) \circ \gamma$ are both $K$-linear and multiplicative in $\alpha$.
  • This reduces the general case to the case $\alpha = X$, where we just get $\beta \circ \gamma = \beta \circ \gamma$.

Actually polynomial composition is uniquely determined by being $K$-linear and multiplicative in the first argument, and satisfying $X \circ \beta = \beta$. In other words the assignment $f_{\beta} : \alpha \mapsto \alpha \circ \beta$ is the unique homomorphism $K[X] \to K[X]$ satisfying $f_{\beta}(X) = \beta$.

Qiaochu Yuan
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    FWIW, the algebraic composition of elements in $K\left[X\right]$ is literal (underlying of ring morphism) function composition conjugated through the standard bijection from the set of ring morphisms $K\left[X\right]\to K\left[X\right]$ to the set of elements of $K\left[X\right]$ via evaluation at $X$. – Rafi Aug 23 '23 at 19:56
  • Is this same reasoning sufficient to show that composition is also associative in $K[[X]]$? Say if we can show that a FPS $\beta$ uniquely defines an endomorphism $f_\beta$, and that there's a bijection between elements of $K[[X]]$ and the (so-defined) morphisms, then we get associativity of composition of elements of $K[[X]]$ "for free" from the associativity of the composition of endomorphisms of $K[[X]]$? – kiyopi Sep 22 '23 at 16:48
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    @grove: yes, the same style of argument can work although there are a few more details ($\beta$ must have no constant term and we want to specifically consider endomorphisms which are continuous with respect to the $X$-adic topology). – Qiaochu Yuan Sep 22 '23 at 19:09
  • Thanks. If I'm not wrong, every such $f_\beta$ with $\beta \in (X)$ is continuous with respect to the $X$-adic topology. Could you hint at how continuity is used for the proof of associativity? – kiyopi Sep 22 '23 at 19:23
  • Yes, that's right. It's used to prove uniqueness. Above we use the fact that an endomorphism of $K[X]$ is determined by where it sends $X$; for $K[[X]]$ the analogous fact is that a continuous endomorphism is determined by where it sends $X$. – Qiaochu Yuan Sep 22 '23 at 19:37
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Since your field $K$ is infinite, polynomials with coefficients in $K$ may be naturally identified to functions from $K$ to $K$ and you can follow @PrincessEev's advice. However, here is a direct proof in the more general case where $K$ is an arbitrary ring:

By linearity, you may restrict to the case where $\alpha$ is a monomial $X^n.$ Then, you just have to prove that $$(\beta\circ\gamma)^n=\beta^n\circ\gamma.$$ This results from the following: $$\begin{align}(PQ)\circ\gamma&=(\sum_k(\sum_{i+j=k}p_iq_j)X^k)\circ\gamma\\ &=\sum_k(\sum_{i+j=k}p_iq_j)\gamma^k\\ &=(\sum p_i\gamma^i)(\sum q_j\gamma^j)\\ &=(P\circ\gamma)(Q\circ\gamma). \end{align}$$

Anne Bauval
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