Irreducible polynomial over integers : sufficient conditions?

Question

Why is $9x^{2}-3$ reducible over integers? I am not able to understand what are the necessary and sufficient condition for the irreducibility of a polynomial.

Answer 1

In fact, there are different definitions of irreducibility. The confusion comes mainly from the context.

Some literature has defined irreducible polynomials (over some ring, in this case, $\mathbb{Z}$) as the polynomials that cannot be factored into the product of two NON-CONSTANT polynomials. In this way $9x^2-3$ is irreducible.

However if we consider the irreducibility in the domain $\mathbb{Z}[x]$, the definition of irreducibility is that when $f$ is factored into $g\dot\ h$, either $g$ or $h$ is a unit (has an inverse). In this way, $3$ is definitely not a unit, so $9x^2-3$ is not irreducible.

I have encountered both definitions, and they have also brought confusion. Hope my explanation helps.

Answer 2

The polynomial is irreducible in $\mathbb{Q}[x]$, but reducible in $\mathbb{Z}[x]$.

Definition. Let $R$ be an integral domain. An element $a\in R$, $a\ne0$ and not a unit, is reducible if (and only if) there exist non units $b,c\in R$ such that $a=bc$. An element $a\in R$, $a\ne0$ and not a unit, is irreducible if and only if it is not reducible.

Equivalently, $a$ ($a\ne0$ and not a unit) is irreducible if and only if, for every $x,y\in R$, if $a=bc$ then either $b$ is a unit or $c$ is a unit.

In your case $$ 9x^2-3=3(3x^2-1) $$ is a factorization where neither factor is a unit in $\mathbb{Z}[x]$.

On the other hand, the polynomial is irreducible in $\mathbb{Q}[x]$ because it has no roots (which is a sufficient condition for polynomials of degree $1$ or $2$ over a field).

What’s a necessary and sufficient condition for a nonzero polynomial $f(x)$ in $\mathbb{Z}[x]$ to be irreducible?

First, define the content $c_f$ of $f(x)$ to be the positive greatest common divisor of its coefficients. Any nonzero polynomial can be written as $$ f(x)=c_f \hat{f}(x) $$ where the content of $\hat{f}(x)$ is $1$ (such a polynomial is called primitive).

If the polynomial is constant (and not $\pm1$), then it is irreducible if and only if $c_f$ is prime.

Assume $f$ is not constant. Then, in order to be irreducible in $\mathbb{Z}[x]$, it is necessary that $c_f=1$ and that $\hat{f}(x)$ is irreducible.

Now (a consequence of) Gauss’ lemma tells you that a primitive polynomial is irreducible in $\mathbb{Z}[x]$ if and only if it is irreducible in $\mathbb{Q}[x]$.