How many roots does $f=x^4+x^2+2$ have in $\mathbb Q[x]/(f)$?

Question

Given an irreducible $f\in \mathbb Q[x]$, how to determine the number of roots of $f=0$ in $\mathbb Q[x]/(f)$?

For some simple examples, I know how to do. If $f(x)=x^2+x-1$, the we use $x\in\mathbb Q[x]/(f)$ to represent $(-1+\sqrt 5)/2$, and $-x-1$ will be another root.

If $f=x^3-2$, $x\in\mathbb Q[x]/(f)$ will represent $2^{1/3}$, and it is the only root of $f=0$ in $\mathbb Q[x]/(f)$, since the other roots are complex.

But for more complicated $f$, e.g. $f=x^4+x^2+2$, I have no idea how to find the number of roots in the extended field. Thank you.

Answer 1

The more concrete way of thinking about it is, if we take one of the roots such as $r = \sqrt{-\frac{1}{2}+i\frac{\sqrt{7}}{2}}$, which of the other roots can we express in terms of $r$?

In this case, we have $-r = -\sqrt{-\frac{1}{2}+i\frac{\sqrt{7}}{2}}$, another root of $f$. The other two roots are $\pm \sqrt{-1 - r^2}$, so we want to know if there is any element of $\mathbb Q[x]/(f)$ which squares to $-1-x^2$.

The brute force approach is to take an arbitrary element $p = a_0 + a_1 x + a_2 x^2 + a_3 x^3$, square it, and check if it's possible to get $p^2 = -1-x^2$. This is a painful but not really difficult computation; you'll quickly see that either $a_0 = a_2 = 0$ or $a_1 = a_3 = 0$, and then get a quadratic equation to solve with no rational roots.

(So we conclude that $\mathbb Q[x]/(f)$ has only two elements $t$ such that $t^4+t^2+2=0$: $t= \pm x$.)

It might also help to observe that if $p^2 = -1-x^2$, then $(px)^2 = -x^2-x^4 = 2$. I'm pretty sure that this is a contradiction, but I don't actually know what I'm talking about when it comes to field extensions - I just felt really bad when I came across this question and saw everybody interpreting it as "how do I find the roots of $x^4+x^2+2=0$?"

Answer 2

Let $u=x^2$.

Then, $f = u^2+u+2$ whose roots are $\dfrac{-1\pm\sqrt{1-8}}{2}=-\dfrac12\pm\dfrac{\sqrt7}2i$.

Now, you can find all the $4$ roots, but I will not demonstrate it here for obvious reasons.

Answer 3

In the case of $ \mathbf Q(2^{1/3}) $, one examines the situation locally, by embedding this field into a completion of $ \mathbf Q $ (in this case, $ \mathbf R $) and using what we know about $ \mathbf R $ to deduce that there can only be one root of $ X^3 - 2 $ in this number field. We may do something similar for the field $ \mathbf Q(\alpha) $, where $ \alpha $ is a root of $ X^4 + X^2 + 2 $. Indeed, we have the factorization

$$ X^4 + X^2 + 2 = (X+2)(X+9)(X^2 + 5) \pmod{11} $$

which implies, by Hensel's lemma, that $ X^4 + X^2 + 2 $ has exactly $ 2 $ roots in the field $ \mathbf Q_{11} $ of $ 11 $-adic numbers. Now, we have an embedding $ \mathbf Q(\alpha) \to \mathbf Q_{11} $ by sending $ \alpha $ to either of these roots and extending; therefore, since $ \mathbf Q(\alpha) $ is then realized as a subfield of $ \mathbf Q_{11} $, it can also contain at most two roots of $ X^4 + X^2 + 2 $. We know that it contains at least two, since $ \alpha $ and $ -\alpha $ are distinct roots; therefore it contains exactly two such roots.