Exercise 11.19

Question

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Prove the identity $\sum \lambda(f) t^{\deg(f)} = \prod (1 - \lambda(f) t^{\deg(f)})^{-1}$, where the sum is over all monic polynomials in $F[t]$ and the product is over all monic irreducible in $F[t]$. $\lambda$ is defined in Section 4.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow} \def\gcro{\mbox{[\hspace{-.15em}[}} \def\dcro{\mbox{]\hspace{-.15em}]}} ewcommand{\D}{\mathrm{d}} ewcommand{\Q}{\mathbb{Q}} ewcommand{\Z}{\mathbb{Z}} ewcommand{\N}{\mathbb{N}} ewcommand{\R}{\mathbb{R}} ewcommand{\C}{\mathbb{C}} ewcommand{\F}{\mathbb{F}} ewcommand{ e}{\,\mathrm{Re}\,} ewcommand{\ord}{\mathrm{ord}} ewcommand{ }{\mathrm{N}} ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}} \bigskip (The solution of this exercise requires external knowledge on formal power series. To learn more about formal power series, see [Niven, Formal power series], [Bourbaki, Algebra IV, ¤4], and [Wikipedia, Formal power series]. About summable families, see [Bourbaki, General Topology, III ¤5]. ) \bigskip \begin{proof} For each monic polynomial $f(x) = x^n - c_1x^{n-1} + \cdots + (-1)^n c_n \in F[x]$, $\lambda(f)$ is defined by $$\lambda(f) = \psi(c_1) \chi(c_n).$$ To complete this definition, we define $\lambda(1) = 1$. By Lemma 1, $\lambda(fg) = \lambda(f) \lambda(g)$ for all monic polynomials $f,g \in F[x]$. We must prove the following equality in the ring of formal power series $\C[[t]]$: \begin{align} \sum_{f \in M} \lambda(f) t^{\deg(f)} = \prod_{f \in I} \left(1 - \lambda(f) t^{\deg(f)} ight)^{-1}, \end{align} where $M$ is the set of monic polynomials of $F[x]$, and $I$ is the set of monic polynomials of $F[x]$ which are irreducible over $F$. Since $I$ and $M$ are infinite sets, we must give a sense at this formula. This implies to introduce a topology on the algebra $\C[[t]]$, which is given by the distance $d$ defined by $$d(\alpha, \beta) = 2^{- u(\alpha - \beta)}, \qquad \alpha, \beta \in \C[[t]],$$ where $ u : \C[[t]] o \N \cup \{\infty\}$ is the valuation on $\C[[t]]$: if $\alpha = \sum_{k=0}^\infty a_k t^k$, then $ u(0) = \infty$, and $ u(\alpha) = \min \{k \in \N \mid a_k e 0\}$. This distance is associated to the norm $|| \cdot ||$, given by $||\gamma|| = 2^{- u(\gamma)}, \gamma \in \C[[t]]$, so that $E = \C[[t]]]$ is a normed vector space. %Moreover the metric space $\C[[t]]$ is complete. As in Bourbaki, a family $(u_i)_{i \in I}$ of vectors of a normed vector space is summable, if there is some $S \in E$ such that $$\forall \varepsilon,\ \exists J_\varepsilon \in {\cal F}(I), \ \forall J \in {\cal F}(I),\ J \supset J_\varepsilon \Rightarrow \left |\left | \sum_{i \in J} u_i - S ight | ight | < \varepsilon,$$ where ${\cal F}(I)$ is the set of finite subsets of $I$. Then we write $S = \sum\limits_{i \in J} u_i$. There is a similar definition for multipliable families. In the algebra $\C[[t]]$, this is equivalent to $\lim_k u_k = 0$ under the filter of the complementaries of finite sets: ${\{A \in {\cal P}(I) \mid I \setminus A \in {\cal F}(I)\}}$ (Bourbaki, IV, 4, Lemma 1), which means, for $(u_i)_{i\in I} \in \C[[t]]^I$, $$\forall \varepsilon,\ \exists J_\varepsilon \in {\cal F}(I), \ \forall i \in I \setminus J,\ \left |\left | u_i ight | ight | < \varepsilon.$$ Moreover, if the family $(u_i)_{i \in I}$ is summable, then $(1 + u_i)_{i \in I}$ is multipliable (Bourbaki, Algebra IV, 4, Proposition 2). A summable family $(u_i)_{i \in I}$, where $I$ is a countable set, can be summed in any order (Bourbaki, General Topology, III,7 Proposition 9). If $\varphi: \N o I$ is a bijection, then \begin{align} \sum_{i \in I} u_i = \sum_{j = 0}^\infty u_{\varphi(j)}. \end{align} \bigskip After these preliminaries, we can show that the family $\left(\lambda(f) t^{\deg(f)} ight)_{f \in M}$ is summable. If $\varepsilon >0$, let $N$ be an integer such that $2^{-N} < \varepsilon$, and consider the set $J_\varepsilon$ of monic polynomials $f$ such that $\deg(f) \leq N$. Then $J_\varepsilon$ is a finite set, and for all $f \in I \setminus J$, $\deg(f) >N$, so that $ ||\lambda(f) t^{\deg(f)}|| = 2^{-\deg(f)} \leq 2^{-N} < \varepsilon$. This proves that the family $\left(\lambda(f) t^{\deg(f)} ight)_{f \in M}$ is summable, and $\sum_{f \in M} \lambda(f) t^{\deg(f)}$ makes sense. Then the sub-family $\left(\lambda(f) t^{\deg(f)} ight)_{f \in I}$ is also summable. This proves that $ (1 - \lambda(f) t^{\deg(f)})_{f\in I}$ is multiplicable, and $\prod\limits_{f \in I}(1 - \lambda(f) t^{\deg(f)})^{-1}$ makes sense. To prove (3), we use first geometric power series. For all $f \in I$, $$(1 - \lambda(f) t^{\deg(f)})^{-1} = \sum_{k=0}^\infty \lambda(f)^k t^{k \deg(f)}.$$ The set $I$ is a countable set (countable union of finite sets). We use an arbitrary numbering of $I$, $I = \{f_1,f_2,\ldots,f_n,\ldots\}$, obtained by a bijection $\varphi : \N^* o I, \varphi(n) =f_n$. Write $I_m$ the finite set $I_m = \{f_1,\ldots,f_m\}$, et $M_m$ the set of monic polynomials whose irreducible factors are in $I_m$, so that every $f \in M$ uniquely decomposes under the form $$f = f_1^{a_1}\cdots f_m^{a_m}, \qquad a_1,\ldots,a_m \in \N.$$ Write $d_i = \deg(f_i)$. Then (see Bourbaki, Algebra IV, 4, Proposition 2) \begin{align*} \sum_{f \in M_m} \lambda(f) t^{\deg(f)} &= \sum_{(a_1,\ldots,a_m) \in \N^m} \lambda(f_1)^{a_1} \cdots \lambda(f_m)^{a_m} t^{a_1d_1+ \cdots+a_md_m}\ &= \left(\sum_{a_1 = 0}^\infty \lambda(f_1)^{a_1} t^{a_1d_1} ight) \cdots \left(\sum_{a_m = 0}^\infty \lambda(f_m)^{a_m} t^{a_md_m} ight)\ &= \left(1 - \lambda(f_1)t^{a_1} ight)^{-1} \cdots \left(1 - \lambda(f_m)t^{a_m} ight)^{-1}\ &=\prod_{i=1}^m \left( 1 - \lambda(f_i) t^{\deg(f_i)} ight)^{-1}. \end{align*} Then, using (4), \begin{align*} \lim_{m o \infty} \prod_{i=1}^m \left( 1 - \lambda(f_i) t^{\deg(f_i)} ight)^{-1} &=\prod_{i=1}^\infty \left( 1 - \lambda(f_i) t^{\deg(f_i)} ight)^{-1} \ &=\prod_{f \in I} (1 - \lambda(f) t^{\deg(f)})^{-1}, \end{align*} the limit being in the metric space $\C[[t]]$ with the distance $d$. Since $M$ is the increasing union of the $M_m$, \begin{align*} \lim_{m o \infty} \sum_{f \in M_m} \lambda(f) t^{\deg(f)} &= \sum_{f \in M} \lambda(f) t^{\deg(f)}. \end{align*} We justify this statement. If $\varepsilon >0$, let $N$ be an integer such that $2^{-N} < \varepsilon$. The set of monic irreducible polynomials $f_i \in I$ such that $\deg(f) \leq N$ is finite, thus there is some integer $M$ such that, for all integers $i$, $i \geq M$ implies $\deg(f_i) > N$. For every $m\geq M$, if $f \in M \setminus M_m$, there is some irreducible monic factor $f_i$ of $f$ such that $i \geq m$, therefore $\deg(f) \geq \deg(f_i) > N$. Then $$ u \left(\sum_{f \in M\setminus M_m} \lambda(f) t^{\deg(f)} ight) \geq N,$$ so \begin{align*} \left | \left | \sum_{f \in M} \lambda(f) t^{\deg(f)} - \sum_{f \in M_m} \lambda(f) t^{\deg(f)} ight | ight | &= \left | \left | \sum_{f \in M\setminus M_m} \lambda(f) t^{\deg(f)} ight | ight | \leq 2^{-N} < \varepsilon. \end{align*} This shows the statement. Since $$ \left\{ \begin{array}{lcl} \lim\limits_{m o \infty} \sum\limits_{f \in M_m} \lambda(f) t^{\deg(f)} &= & \sum\limits_{f \in M} \lambda(f) t^{\deg(f)},\ \lim\limits_{m o \infty} \prod\limits_{i=1}^m \left( 1 - \lambda(f_i) t^{\deg(f_i)} ight)^{-1} &=& \prod\limits_{f \in I} (1 - \lambda(f) t^{\deg(f)})^{-1}, \end{array} ight. $$ where $$\sum_{f \in M_m} \lambda(f) t^{\deg(f)} = \prod_{i=1}^m \left( 1 - \lambda(f_i) t^{\deg(f_i)} ight)^{-1},$$ the unicity of the limit shows that \begin{align*} \sum_{f \in M} \lambda(f) t^{\deg(f)} = \prod_{f \in I} (1 - \lambda(f) t^{\deg(f)})^{-1}.\ \end{align*} \end{proof}

Exercise 11.19

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Exercise 11.19

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