Systems of quadratic forms I've been working on a problem in field theory, specifically showing that $\mathbb{Q}(\alpha)$ has no intermediate fields when where $\alpha$ is a root of $x^4-nx-1$ whenever $n \not \in \{-4,0,4\}$. The catch is we're NOT allowed to use Galois theory which would make the problem pretty trivial (the resolvent cubic is irreducible in this case so the Galois group of the splitting field is either $A_4$ or $S_4$ and in either case any subgroup of index $4$ is maximal so that there can be no intermediate fields).
My initial thought was to look at $\mathbb{Q}[x]/(x^4-nx-1)$ and show that there can be no quadratic subfield by trying to show that there is no solution to
$$(ax^3+bx^2+cx+d)^2 \equiv k \mod x^4-nx-1$$
for some $a,b,c,d,k\in\mathbb{Q}$. After squaring, and reducing, this is equivalent to an integer solution (since we can clear denominators from $a,b,c,d$ before-hand) of the system of quadratic forms
\begin{align*}
2ad + 2bc+a^2n &= 0\\
c^2+2bd + a^2 + 2abn &= 0\\
2ab + 2cd +nb^2 + 2nac &= 0
\end{align*}
which have matrix forms
$$\begin{bmatrix}
n & 0 &0 &1\\
0&0&1&0\\
0&1&0&0\\
1&0&0&0
\end{bmatrix} \quad \quad \begin{bmatrix}
1 & n &0 &0\\
n&0&0&1\\
0&0&1&0\\
0&1&0&0
\end{bmatrix}\quad \quad \begin{bmatrix}
0 & 1 &n &0\\
1&n&0&0\\
n&0&0&1\\
0&0&1&0
\end{bmatrix}.$$
Is there any simple criterion to show that this system has no simultaneous solutions? Any thoughts on how I can proceed?
 A: Too long for a comment. I would proceed differently, since intersection of quadrics are nasty...
If $E=\mathbb{Q}(\alpha)$ has a quadratic subfield $F=\mathbb{Q}(\sqrt{d}),$ where is a squarefree non zero integer, let $\mu=X^2-uX+v$ be the minimal polynomial of $\alpha$ over $F$. Denoting by $*$ the non trivial automorphism of $F$, you have $\mu\mid X^4-nX+1$, and applying $*$, $\mu^*\mid X^4-nX+1$.
Since $X^4-nX+1$ is irreducible, $\mu$ and $\mu^*$ are distinct irreducible polynomials of $F[X]$ (if $\mu=\mu^*,$ then $\mu\in\mathbb{Q}[X]$, which is not possible). Hence $\mu \mu^*\mid X^4-nX+1$, and comparing degrees and leading coefficients shows that $X^4-nX+1=\mu \mu^*=X^4-(u+u^*)X^3+(v+v^* +uu^*)X^2-(u^*v+uv^*)X+vv^*$.
Hence $u^*+u=0, v+v^*+uu^*=0, u^*v+uv^*=n,vv^*=1.$
Thus $u=a\sqrt{d}$ by the first equality. Write $v=b+c\sqrt{d}$. Then $2b-da^2=0$, $-2acd=n$ and $b^2-dc^2=1$.
Hence $d^2 a^4-4dc^2=4$ and $d^3a^6-n^2=4da^2$.
Thus the polynomial $d^3 X^3-4dX-n^2$ has an integer root $x$
It follows that $X^3-4X-n^2$ has an integer root (replace $X$ by $dX$).
Hence we are reduced to prove that $X^3-4X-n^2$ has no integer roots, for any $n\neq 0,\pm 4$. This should be easier than studying the intersection of three quadrics.
For example, if $n$ is coprime to $3$, reducing modulo $3$ shows a contradiction.
