Lack of understanding of the proof of the existence of an irreducible polynomial of any degree $n \geq 2$ in $\mathbb{Z}_p[x]$

Solution 1:

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.