I'm reading some lecture notes that argue that $P(x) = x^3+3x+1$ is irreducible in $\mathbb{Q}[x]$ by first noting that it's irreducible in $\mathbb{F}_2[x]$. I'm not sure I quite understand this reasoning.
I guess that we can argue roughly like so:
- $P(x)$ is irreducible in $\mathbb{F}_2[x]$
- Since $\mathbb{F}_2[x]$ is a UFD, this means it's prime
- So the ideal it generates is prime
- So the pre-image of this ideal under the quotient map is prime
- Somehow use this to establish the primeness of $P(x)$ in $\mathbb{Q}[x]$.
- Deduce its irreducibility in $\mathbb{Q}[x]$
However:
- I'm not sure about the "somehow" line, and:
- There's probably a better way of thinking about it.
Question. Is there a better way of thinking about this? What's really going on here?
One way to do this is to use the fact that $P$ is monic and of degree$~3$. The degree implies that the only way of being reducible is having a factor of degree$~1$, hence a rational root. Being monic, the rational root must be integer (and it divides the constant coefficient$~1$; be careful to avoid concluding irreducibility even before having had the chance to reduce modulo$~2$); the factor $X-a$ that comes with an integer root$~a$ leads to a similar factor after reduction modulo$~2$, contradicting irreducibility there.
More generally, using just that $P$ has integer coefficients (which is prerequisite to the reduction modulo the prime number $p=2$ being meaningful) and leading coefficient not divisible by$~p$, one can argue as follows. If $P$ were reducible over$~\Bbb Q$, it would be reducible over$~\Bbb Z$ (by the contrapositive of Gauss' lemma, second formulation) into non-constant polynomials with leading coefficients not divisible by$~p$, which after reduction modulo$~p$ imply that the reduction of $P$ modulo$~p$ factors into factors of the same degrees, hence non-constant, contradicting irreducibility there.