In an optional course called "Finite Geometries", we most recently constructed the fields
$$K_{p,n}[x] := \{\alpha \in K_p[x]\,| \deg(\alpha) < n\},$$
where $K = \mathbb{Z}$, $p$ is prime and $n \in \mathbb{N}$. To do this, we defined the addition as usual whereas for the multiplication we let $\varphi \in K_p[x]$ be an irreducible polynomial of $n$-th degree and then defined $$\alpha \odot \beta := \alpha \cdot \beta \text{ mod } \varphi.$$
Now, as it is not obvious that for every $n \in \mathbb{N}$ there exists such an irreducible polynomial, we proved this for $n \geq 2$ (for $n=1$ the field is simply $(K_p, +, \cdot)$) using a lot of yet unknown tools.
Let $F = \{\phi_1, \phi_2, ...\}$ be a countable set and for every $i \in \mathbb{N}$ let $m_i \in \mathbb{N}$ be the so called "index" of $\phi_i$. We write $m_i = \operatorname{ind}(\phi_i)$. For all $m \in \mathbb{N}$, let $|\{i: m_i = m\}| < \infty$. Then, we call
$$\varphi(z) := \sum_{\phi_i \in F} z^{m_i}$$
the counting power-series (is this the correct english term?) of $F$. Obviously, the coefficient of $z^m$ in $\varphi(z)$ is just $|\{i: m_i = m\}|$. For two such sets $F_1 = \{\phi_1, ...\}$ and $F_2 = \{\psi_1, ...\}$ with indices $m_i$ and $n_j$ and counting power-series $\varphi_1(z)$ and $\varphi_2(z)$, we find that the counting power-series for $F_1 \times F_2$ is simply $\varphi_1(z) \cdot \varphi_2(z)$ if we let the index of $(\phi_i, \psi_j)$ be $m_i + n_j$.
Now let $F^{(n)}$ be the set of normed irreducible polynomials of $n$-the degree in $K_p[x]$ and let $f_n = |F^{(n)}|$. We write $F^{(n)} = \{\phi_1^{(n)}, ..., \phi_{f_n}^{(n)}\}$ and for $\phi_j^{(n)} \in F^{(n)}$ let $$F_j^{(n)} = \{1, \phi_j^{(n)}, \phi_j^{(n)} \cdot \phi_j^{(n)}, ...\}.$$
Also, let $\operatorname{ind}(\phi_j^{(n)}) = n$. Then, the counting power-series of $F_j^{(n)}$ is
$$\varphi_j^{(n)} = 1 + z^n + z^{2n} + ... = \frac{1}{1-z^n}.$$
Question 1: Why is $\operatorname{ind}(\phi_j^{(n)} \cdot \phi_j^{(n)}) = 2n$? We have only defined the index for $\phi_j^{(n)}$ and we have not defined that index-function to have any special properties, so how can one obtain indices for all the other elements of this set?
Next we define
$$F := \{\phi \in F_1^{(1)} \times ... \times F_{f_1}^{(1)} \times F_1^{(2)} \times ... |\; \text{only finitely many factors of } \phi \text{ are} \neq 1\}.$$
This set is bijective to the set of all normed polynomials in $K_p[x]$.
Question 2: What does it mean if only finitely many factors are $\neq 1$ and how is this set bijective to the set of all normed polynomials in $K_p[x]$?
The counting power-series of $F$ is the product of the counting power-series of its factors, i.e.
$$\varphi(z) = \prod_{n=1}^\infty \left( \frac{1}{1-z^n} \right)^{f_n}.$$
In $K_p[x]$ there are $p^n$ normed polynomials of $n$-th degree, and thus we can also write
$$\varphi(z) = 1 + pz + p^2z^2 + ... = \frac{1}{1-pz}.$$
Question 3: What does this summation have to do with $F$ and its power-series?
After this, the writer logarithmically derives both expressions for $\varphi(z)$ and after a few transformations, he arrives at
$$\sum_{n=1}^\infty \left( \sum_{d \mid n} d f_d\right) z^{n-1} = \sum_{n=1}^\infty p^n z^{n-1}$$
which can be solved using Möbius-inversion. Then, we eventually have
$$n f_n = p^n + ... + \mu(n) p = p^m ( p^{n-m} + ... \pm 1),$$
where both factors cannot be $0$ and thus for all $n \in \mathbb{N}$ we have $f_n \neq 0$.
Question 4: I don't understand this last step. What is $m$? Why is the bracket $\neq 0$?
Thanks you in advance for helping me understand this proof.
Answer
Q1: Your teacher left the index undefined early on, but later context reveals that the index of a monic polynomial is most likely intended to be equal to its degree. S/he may be planning
on using similar generating functions later on, and gave a more general definition for starters.
Q2: The context here is that in the ring $K_p[x]$ we have unique factorization. In other words, every monic polynomial $\phi(x)\in K_p[x]$ can be written as a product of powers of irreducible monic polynomials $\phi_i(x)$ in an essentially unique way in the form
$$
\phi(x)=\prod_i \phi_i(x)^{n_i}.
$$
There are infinitely many irreducible polynomials $\phi_i(x)$ (Have you shown this in class? The usual proof about the infinitude of the set of prime numbers works for polynomials also!). But because $\phi(x)$ has a finite degree, only finitely many irreducible polynomials $\phi_i(x)$ can appear as factors here. Therefore only finitely many exponents $m_i$ are non-zero. Compare integer factorization of $n$:
$$
n=\pm\prod_ip_i^{m_i},
$$
where $p_i$ are all primes. There also only finitely many exponents $m_i$ are non-zero. Namely those, where the corresponding prime $p_i$ is a factor of $n$.
Q3: Here we simply count the number of monic polynomials of degree $n$ in two ways. One way is the direct way: $n$ unknown coefficients, $p$ choices for each. The other is to use the factorization of Question 2. The power series keeps track of the degree of the product $\prod_i\phi_i(x)^{m_i}$. We easily see that the degree of that is $\sum_i m_i\deg \phi_i(x)$. Let us fix an irreducible polynomial $\phi_j(x)$. The series $\varphi_j(z)$ related to the set $F_j=\{\phi_j^n\mid n=0,1,\ldots\}$ is then
$$
\varphi_j(z)=1+z^{m_j}+z^{2m_j}+z^{3m_j}+\cdots,
$$
where $m_j$ is the degree of $\phi_j(x)$.
Here the term $z^{km_j}$ corresponds to having a factor $\phi_j^k$. Furthermore, $km_j$ is the degree of $\phi_j^k$, so the polynomial $\phi(x)$ with the above factorization will contribute towards the term $z^{\deg f}$ as it should. For example, if $p=3$, and $\phi(x)=(x+1)^3(x^2+1)$, then the irreducible factors $\phi_1(x)=x+1$ and $\phi_2(x)=x^2+1$ appear. From the series
$$
\varphi_1(z)=1+z+z^2+\cdots
$$
we pick the term $z^3$, because $\phi_1$ appeared to the third power in the factorization of $\phi$. From the series
$$
\varphi_2(z)=1+z^2+z^4+\cdots,
$$
we pick the term $z^2$, because $\phi_2$ was a simple factor. From all the other series
$\varphi_j(z)$, where $j\neq1,2$ we pick the term $1$, because $\phi_j$ did not appear as a factor of $\phi(x)$. Multiplying all those power series together we get the term $z^3\cdot z^2=z^5$ to represent $\phi$, which is just what we wanted, because the degree of $\phi$ is equal to five. Do this for all the polynomials $\phi$, and you get the result.
Compare this with the Euler product of the $\zeta$-function:
$$
\zeta(s)=\sum_{n=1}\frac1{n^s}=\prod_{p\ \text{prime}}\frac1{1-p^{-s}}.
$$
For example, when $n=72=2^3\cdot3^2$, we get the term $1/72^s$ from the right hand side
as follows.
$$
\frac1{1-2^{-s}}=1+2^{-s}+4^{-s}+8^{-s}+\cdots
$$
has the term $1/8^s$. Similarly the series
$$
\frac1{1-3^{-s}}=1+3^{-s}+9^{-s}+27^{-s}+\cdots
$$
has the term $1/9^s$. In the product
$$
\prod_{p\ \text{prime}}\frac1{1-p^{-s}}=\prod_{p\ \text{prime}}(1+p^{-s}+p^{-2s}+p^{-3s}+\cdots)
$$
we get the term $72^{-s}$ if and only if we select the factor $8^{-s}$, when $p=2$, the factor $9^{-s}$, when $p=3$, and the factor $1$, when $p>3$.
Here we are doing something similar:
$$
\sum_{\phi\in K_p[x],\ \phi\ \text{monic}}z^{\deg \phi}=\prod_{\phi_j\ \text{monic irreducible}}\frac1{1-z^{\deg\phi_j}}.
$$
Unique factorization gives both of these power series identities.
Q4: Here $m$ is the smallest factor of $n$ such that $\mu(n/m)\neq1$. It may happen that $m=1$, but we don't know for sure ($n$ may not be square-free). The factor in parentheses is non-zero, because it is an alternating sum of distinct powers of $p$. The largest power cannot be cancelled by lower ones.
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