Prove that $x^4+15x^3+7$ is irreducible over $ \mathbb Q$?

Question

I am reasonably comfortable with groups and have just started with fields. This question popped up in an assignment sheet.

Answer 1

Since the polynomial has integer coefficients and is monic, it is irreducible in $\mathbb{Q}[x]$ if and only if it is irreducible in $\mathbb{Z}[x]$.

Now, if it were reducible, it would be so modulo $p$, for every $p$.

If we try modulo $2$, the polynomial is $x^4+x^3+1$, which has no roots over $\mathbb{Z}/2\mathbb{Z}$. Let's check for a factorization \begin{align} x^4+x^3+1 &=(x^2+ax+1)(x^2+bx+1)\\ &=x^4+(a+b)x^3+abx^2+(a+b)x+1 \end{align} that requires $a+b=1$ and $a+b=0$, which is contradictory.

Answer 2

Equivalently, we may ask for irreducibility over $\Bbb Z$. As none of $\pm1, \pm7$ is a root, there is no linear factor, hence we are left with $$ x^4+15x^3+7=(x^2+ax+b)(x^2+cx+d)$$ Then wlog $b=\pm1$, $d=7b$, so
$$ x^4+15x^3+7=(x^2+ax\pm1)(x^2+cx\pm7)$$ which leads to $$c=-7a,\qquad a+c=15$$ which is impossible in $\Bbb Z$.