Is it an element of $\mathbb{Q}[a]$?

Question

We have that $f(x)=x^4-2x^2-1$ is irreducible in $\mathbb{Q}[x]$.

Let $a\in \mathbb{C}$ be a root of $f$.

How could we check if $\sqrt{3}\in \mathbb{Q}[a]$ without using Galois theory?

Answer 1

Check if $f$ splits over $\Bbb Q[\sqrt 3]$.

We can exclude any linear factor $(x-u-v\sqrt 3)$ right away, as then also $(x-u+v\sqrt 3)$ would be a linear factor, resulting in a quadratic factor in $\Bbb Q[x]$. Remains the case of two quadratic factors $$(x^4-2x-1)=(x^2+\alpha x+\beta)(x^2+\gamma x+\delta) $$ with $\alpha,\beta,\gamma,\delta\in\Bbb Q[\sqrt 3]$. Again, we see that ew obtain another such factorization by replacing $\sqrt 3$ with $-\sqrt 3$ everywhere. Thus either both factors are rational (which is impossible) or they are conjugate of each other. Additionally, with $\beta=v+w\sqrt 3$, we have $-1=\beta\delta=(v+w\sqrt 3)(v-w\sqrt 3)=v^2-3w^2$, which is impossible

Answer 2

Not saying that this is an easy way, but it's the first thing to try in any such problem in my opinion. It is an approach that is perhaps better suited for lower degree extensions. Anyways, here goes:

The elemtents of $\Bbb Q[a]$ can all be written as degree-$3$ (or less) polynomials in $a$ with coefficients in $\Bbb Q$. If $\sqrt3 \in \Bbb Q[a]$, then the same should be true there. In that case, we have $$ \sqrt3 = pa^3 + qa^2 + ra + s,\quad p, q, r, s \in \Bbb Q $$ Squaring both sides yield $$\begin{align} 3 ={}& (pa^3 + qa^2 + ra + s)^2 \\ ={} &(2a^3+ a^2) p^2+2 (2a^2+a) p q+2 (2a+1) p r+(2a+1)q^2\\ &+2 a^3 p s+2 a^3 q r+2 a^2 q s+a^2 r^2+2 a r s+s^2 \end{align} $$ which implies $$ \begin{align} {}& (2p^2 + 2ps + 2qr)a^3 + (p^2+4pq + 2qs + r^2)a^2 \\ &+ (2pq+4pr+2q^2+2rs)a +(2pr + q^2 + s^2-3)={}0\end{align} $$ Such a number can only be $0$ if each of the coefficients of the different powers of $a$ is $0$ ($1, a, a^2$ and $a^3$ are linearly independent over $\Bbb Q$). Thus we get a system of four equations which in theory can be solved. Note that the powers of $a$ are linearly dependent over $\Bbb R$ and $\Bbb C$, so you specifically have to use information about roots known to be irrational or similar in order to solve this.

Answer 3

$ X^4 - 2X - 1 $ is separable modulo $ 17 $ and has the root $ X = 10 $, therefore Hensel's lemma gives us a root $ \beta $ of $ X^4 - 2X - 1 $ in the field $ \mathbf Q_{17} $ of $ 17 $-adic numbers. This gives us an embedding $ \mathbf Q(\alpha) \to \mathbf Q_{17} $. However, $ 3 $ is not a quadratic residue modulo $ 17 $, therefore $ \sqrt{3} \notin \mathbf Q_{17} $, and therefore $ \sqrt{3} \notin \mathbf Q(\alpha) $, as it is isomorphic to a subfield of $ \mathbf Q_{17} $.

If this feels like an unnatural proof, it shouldn't: a common trick to show that $ \sqrt{-1} $ is not in a number field is to identify it with a subfield of $ \mathbb R $, and thus conclude that the polynomial $ X^2 + 1 $ cannot have a root in this field. What we are doing is essentially equivalent to this, except we need a local structure which can distinguish between numbers with real conjugates. This calls for examining the field locally at other places, and some simple congruence checks give the above argument.