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Facts before question:

$\textbf{Fact 1:}$ Let $F(X) = X^n + a_{n-1}X^{n-1} + \cdots + a_1X+a_0\in \mathbb{Z}[X]$, with $a_0\neq 0$.

If $|a_{n-1}|>1+|a_{n-2}| + \cdots +|a_1| + |a_0|$, then $F$ irrreducible in $\mathbb{Z}[X]$.

$\textbf{Fact 2:}$ Let $K$ be a field, $F(X,Y)=a_n(X)Y^n + \cdots + a_1(X)Y + a_0\in K[X,Y]$, with $a_o,\ldots,a_{n-1}\in K[X]$, $a_n \in K$ and $a_0a_n\neq 0$.

If $\deg(a_{n-1})>\max\bigl(\{\deg(a_0), \deg(a_1), \ldots, \deg(a_{n-2})\}\bigr)$, then $F$ is irreducible over $K[X]$.

Facts end here.

$\textbf{Conjecture:}$ Let $F(X) = a_nX^n + a_{n-1}X^{n-1} + \cdots + a_1X+a_0\in \mathbb{Z}[X]$, with $a_0\neq 0$.

If $|a_{n-1}|>|a_n|+|a_{n-2}| + \cdots +|a_1| + |a_0|$, then $F$ irrreducible in $\mathbb{Z}[X]$.

Is the above conjecture true? I know nothing about multivariable polynomials, so I'm just looking for a yes or no answer, and, in case it is false, can you please provide a counterexample?

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Comment, down voter? – Git Gud Mar 16 '14 at 16:40
up vote 4 down vote accepted

Try this for the polynomial $2X^2 + 5X + 2$.

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$2X^2 + 5X + 2 = (X+2)(2X+1)$. Thanks! – Git Gud Jan 5 '13 at 11:16

There shouldn't be an absolute value of $a_n$ in there. The polynomial must also be monic.

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