Clearly, none of the roots are in $\mathbb{Q}$ so $f(x) = x^4 + 1$ does not have any linear factors. Thus, the only thing left to check is to show that $f(x)$ cannot reduce to two quadratic factors.
My proposed solution was to state that $f(x) = x^4 + 1 = (x^2 + i)(x^2 – i)$ but $\pm i \not\in \mathbb{Q}$ so $f(x)$ is irreducible.
However, I stumbled across this post $x^4 + 1$ reducible over $\mathbb{R}$… is this possible? with a comment suggesting that $x^4 + 1 = (x^2 + \sqrt{2}x + 1)(x^2 – \sqrt{2}x + 1)$ which turns out to be a case that I did not fully consider. It made me realize that $\mathbb{Q}[x]$ being a UFD only guarantees a unique factorization of irreducible elements in $\mathbb{Q}[x]$ (which $x^2 \pm i$ nor $x^2 \pm \sqrt{2} x + 1$ aren't in $\mathbb{Q}[x]$) so checking a single combination of quadratic products is not sufficient.
Therefore, what is the ideal method for checking that $x^4 + 1$ cannot be reduced to a product of two quadratic polynomials in $\mathbb{Q}[x]$? Am I forced to just brute force check solutions of $x^4 + 1 = (x^2 + ax + b)(x^2 + cx + d)$ don't have rational solutions $(a,b,c,d) \in \mathbb{Q}^4$?
Best Answer
I'd say it's easiest to use Eisenstein's criterion after a shift, although this might be a sledgehammer. $$(x+k)^4 + 1 = x^4 + 4k x^3+6k^2 x^2 + 4k^3 x + k^4 +1$$ So we want a prime $p$ dividing each of $\{4k, 6k^2, 4k^3, k^4 + 1\}$ but $p^2$ not dividing $k^4 + 1$. That looks easy enough: let $p = 2$ and $k = 1$.
(If you want to use a lot more theory and make the test a little simpler, note that the discriminant of $x^4+1$ is $256$, whose only prime factor is $2$, so in fact $p=2$ is the only prime that could possibly work. I used to understand why this was true, but I no longer do; it's something to do with theorem 1.3 in https://kconrad.math.uconn.edu/blurbs/gradnumthy/disc.pdf . I believe it may be possible to use this theory to show that $p=2$ does work without finding $k$, but that's far beyond my pay grade.)