I am reading "Lectures on Algebra" by Teiji Takagi.
The following proposition by Eisenstein is in this book.
Let $f(x)=c_0+c_1x+c_2x^2+\dots+c_lx^l$ be a polynomial such that $c_i\in\mathbb{Z}$ and $p\mid c_0,p^2\nmid c_0, p\mid c_1,\dots,p\mid c_{l-1},p\nmid c_l$ for some prime number $p$.
Then, $f(x)$ is irreducible in $\mathbb{Q}[x]$.
Proof:
If $f(x)$ is reducible in $\mathbb{Q}[x]$, we can write $f(x)=\phi(x)\psi(x)$, where $\phi(x),\psi(x)\in\mathbb{Z}[x]$ by a famous theorem.
Let $\phi(x)=a_0+a_1x+a_2x^2+\dots+a_mx^m$.
Let $\psi(x)=b_0+b_1x+b_2x^2+\dots+b_nx^n$.
Then, $p\nmid a_i$ for some $i\in\{0,\dots,m\}$ and $p\nmid b_j$ for some $j\in\{0,\dots,n\}$.
Let $h:=\min\{i\in\{0,\dots,m\}\mid p\nmid a_i\}$.
Let $k:=\min\{j\in\{0,\dots,n\}\mid p\nmid b_j\}$.
Then, $c_{h+k}=a_hb_k+a_{h-1}b_{k+1}+a_{h-2}b_{k+2}+\dots+a_{h+1}b_{k-1}+a_{h+2}b_{k-2}+\dots$.
Since $a_hb_k$ is not divisible by $p$ and the other terms is divisible by $p$, $p\nmid c_{h+k}$.
So, $c_{h+k}=c_l$.
Therefore, $h=m$ and $k=n$.
So, $p\mid a_0$ and $p\mid b_0$.
So, $p^2\mid c_0=a_0b_0$.
Why do $p\nmid a_i$ for some $i\in\{0,\dots,m\}$ and $p\nmid b_j$ for some $j\in\{0,\dots,n\}$ hold?
My answer to this question is here:
If $p\mid a_i$ for all $i\in\{0,\dots,m\}$, then $p\mid a_mb_n=c_l$, but $p\nmid c_l$ by assumption.
If $p\mid b_j$ for all $j\in\{0,\dots,n\}$, then $p\mid a_mb_n=c_l$, but $p\nmid c_l$ by assumption.
Is my answer correct or not?
The author wrote the following remark without a proof:
Let $f(x)=c_0+c_1x+c_2x^2+\dots+c_lx^l$ be a polynomial such that $c_i\in\mathbb{Z}$ and $p^k\mid c_0,p^{k+1}\nmid c_0, p\mid c_1,\dots,p\mid c_{l-1},p\nmid c_l$ for some prime number $p$.
Then the number of the factors of $f(x)$ is less than or equal to $k$.
My proof of this proposition is here:
If $k=1$, then the number of the factors of $f(x)$ is $1$ by Eisenstein's criterion.
So, the above proposition holds when $k=1$.
Suppose that the above proposition holds when $k=1,\dots,k^{'}$.
Let $k:=k^{'}+1$.
If $f(x)$ is irreducible in $\mathbb{Q}[x]$, then the above proposition holds when $k=k^{'}+1$.
If $f(x)$ is reducible in $\mathbb{Q}[x]$, we can write $f(x)=\phi(x)\psi(x)$, where $\phi(x),\psi(x)\in\mathbb{Z}[x]$ by a famous theorem.
Let $\phi(x)=a_0+a_1x+a_2x^2+\dots+a_mx^m$.
Let $\psi(x)=b_0+b_1x+b_2x^2+\dots+b_nx^n$.
Then, $p\nmid a_i$ for some $i\in\{0,\dots,m\}$ and $p\nmid b_j$ for some $j\in\{0,\dots,n\}$.
Let $h:=\min\{i\in\{0,\dots,m\}\mid p\nmid a_i\}$.
Let $k:=\min\{j\in\{0,\dots,n\}\mid p\nmid b_j\}$.
Then, $c_{h+k}=a_hb_k+a_{h-1}b_{k+1}+a_{h-2}b_{k+2}+\dots+a_{h+1}b{k-1}+a_{h+2}b_{k-2}+\dots$.
Since $a_hb_k$ is not divisible by $p$ and the other terms is divisible by $p$, $p\nmid c_{h+k}$.
So, $c_{h+k}=c_l$.
Therefore, $h=m$ and $k=n$.
So, $p\mid a_0$ and $p\mid b_0$.
Let $p^{k_1}\mid a_0$ and $p^{k_1+1}\nmid a_0$.
Let $p^{k_2}\mid b_0$ and $p^{k_2+1}\nmid b_0$.
Then, $k_1+k_2=k$.
Since $k_1<k$, we can say the number of the factors of $\phi(x)$ is less than or equal to $k_1$.
Since $k_2<k$, we can say the number of the factors of $\psi(x)$ is less than or equal to $k_2$.
So, the number of the factors of $f(x)$ is less than or equal to $k=k_1+k_2$.
Is my proof ok or not?
Best Answer
The argument is correct. Neither $a_m$ nor $b_n$ can be divisible by $p$, so neither set can be empty.
This is also correct, with a minor edit added in $\color{red}{\small\text{red}}$. At this point, it has been established that $p \mid a_0$ so the exponent $k_1 \ge 1$, and the same goes for $p \mid b_0$ and $k_2 \ge 1$. This is worth (re)stating explicitly, since applying the criterion requires that the constant term be divisible by (at least) $p^1$.