# On the irreducibility of a polynomial that almost satisfies Eisenstein's Criterion

Let $f(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0$ be a polynomial in $\mathbb{Z}[x]$ such that for some prime $p$ we have $p$ does not divide $c_n$, $p$ does not divide $c_{n-1}$, $p$ divides each other $c_i$, and $p^2$ does not divide $c_0$. Show that $f$ is irreducible in $\mathbb{Q}$ if and only if there does not exist a rational number $a/b$ such that $f(a/b)=0$.

The $\Rightarrow$ direction is straightforward. I'm not sure how to tackle the $\Leftarrow$ direction. Clearly, the polynomial $g(x)=c_{n-1}x^{n-1}+\cdots+c_1x+c_0$ is irreducible by Eisenstein's criterions using $p$. However, extending this seems to be a challenge. I've played around with plugging in $a/b$ to show that $f(a/b)\neq0$ is irreducible, but didn't get far. Also tried an inductive argument on the degree of the polynomial, but didn't see how to argue for degree above 3. I've considered the contrapositive statement, but I don't know of any necessary conditions for reducibility that imply a rational root. If there are any suggestions on how to approach this, that would be greatly appreciated. Full solution not required.

• Show the contrapositive.(Reducible means there is a linear factor.) Commented Apr 28, 2018 at 5:18
• @daruma how so, why can't it factor into a quadratic and a polynomial of degree $n-2$? Commented Apr 28, 2018 at 5:19
• @Sil No, we are not claiming that $f$ is already irreducible. Eisenstein requires that $p$ does divide $c_{n-1}$, which it doesn't here. Commented Apr 28, 2018 at 18:27
• @Atsina I see, missed that part.
– Sil
Commented Apr 28, 2018 at 19:37
• This follows from the fact the Newton polygon of $f$ consists of two line segments $(0,1)-(n-1,0)$ and $(n-1,0)-(n,0)$, so if it is reducible, it has to have linear factor, so it means it means it would have to have a rational root.
– Sil
Commented May 7, 2018 at 18:22

Suppose that $f$ is reducible.

Then,

$c_n x^n+c_{n-1}x^{n-1}+...+c_1x+c_0=(d_s x^s+...+d_1x+d_0)(e_tx^t+...+e_1x+e_0)$

Without loss of generality, suppose $p\not\mid e_0$. Let $i>0$ be the minimum integer such that $p\not\mid d_i$ such an $i$ must exist as otherwise $p$ divides all of the $d_j$ and then, $p$ divides all of $a_j$.

Claim: $p\not \mid a_i$ and $p \mid a_j$ for every $j<i$.

Proof of Claim: $a_i=d_i e_0+d_{i-1} e_1+...+d_0 e_i$ and $p\mid d_{i-1} e_1+...+d_0 e_i$ but $p\not\mid d_i e_0$. Hence, $p\not\mid a_i$. For $j<i$, $a_j=d_je_0+...d_0e_j$ so $p\mid d_j$.

So we know that $i=n-1$ as $p$ divides $a_j$ for all $j<n-1$. In particular, we learn that $s\geqslant n-1$. So, $e_tx^t+...e_1x_1+e_0$ is a constant factor if $s=n$ or a linear factor if $s=n-1$. (The former case can be ignored as that does not give us a non-trivial factorization.) So

• It should read: Let $i\geq0$... Commented Apr 28, 2018 at 7:24
• Not sure why you renamed the coefficients from $c$ to $a$, and some of the argument could be cleaned up ("such an $i$ must exist" should be its own sentence, the introduction of $s$ is a bit confusing, etc.) but I appreciate the help. Commented Apr 28, 2018 at 18:33