Showing irreducibility in $\mathbb Z[X]$ for a quartic polynomial.

Question

Let $f(X) = X^4 - 10X^2 + 11 \in \mathbb Z[X]$ and let $\bar f(X) \in \mathbb Z_7[X]$ be the polynomial obtained from $f(X)$ by reduction modulo 7. By using the factorisation of $\bar f(X)$ or otherwise, show that $f(X)$ is irreducible in $\mathbb Z[X]$.

What I have done so far is to factorise $\bar f(X)$ as a product of monic irreducible polynomials in $\mathbb Z_7[X]$, getting that $\bar f(X) = (X^2 + \bar 2)^2$.

What I am struggling with is how to use this to show that $f(X)$ is irreducible in $\mathbb Z[X]$.

My idea is to show this by contradiction, assuming that $f(X)$ is reducible in $\mathbb Z[X]$, but I'm unsure where to go from there.

Answer 1

$f$ is irreducible in $\Bbb Q[X]$, because $f(X+1)=X^4+4X^3-4X^2-16X+2$ is Eisenstein with $p=2$.

Also, $f$ is irreducible modulo $3$ (but not modulo $7$), and hence irreducible in $\Bbb Z[X]$ and $\Bbb Q[X]$.

Answer 2

If you want a continuation of your method, notice that the reduction into irreducibles in $\mathbb Z_7$ as you've done demonstrates that if the polynomial is reducible in $\mathbb Z[X]$, we have

$f(X)=(X^2+aX+b)(X^2+cX+d)$ where both of these factors are $X^2+\bar{2}$ after reduction mod $7$. That means that $b,d\equiv 2 \bmod 7$, but $bd=11$ causes a contradiction.