Rational solution to AX=0

Question

Let $\mathcal{M}_{n,p}(\mathbb{K})$ be the set of matrices $n\times p$ with coefficients in $\mathbb{K}$.

Let $A\in\mathcal{M}_{n,p}(\mathbb{Q})$.

We suppose there exists a non zero solution $X\in\mathcal{M}_{p,1}(\mathbb{R})$ to $AX=0$. ($0$ denotes $[0]_{p,1}$)

Show that there exists a non zero solution $X'\in\mathcal{}_{p,1}(\mathbb{Q})$ to $AX'=0$

PS : This is NOT a duplicate of REVISITED $^2$: Does a solution in $\mathbb{R}^n$ imply a solution in $\mathbb{Q}^n$?, which gives a zero solution in my case.

I've tried building up the system of linear equations linked to $AX=0$.

I got the result for simple small square matrices, but I can't manage to generalize it to all the matrices.

Are determinants something you can use, or are they still officially unknown? — Daniel Fischer, Oct 14 '14 at 17:15
If $p > n$, the existence of $X'$ is clear, so assume $p \leqslant n$, What does the existence of $X$ tell you about the determinants of the $p\times p$ minors? — Daniel Fischer, Oct 14 '14 at 17:45

score 0 · Accepted Answer · answered Oct 14 '14 at 17:15

I don't have time at the moment to write down a full answer, but here's a sketch. Consider the field $K = \mathbb{Q}(X_1, \dots, X_p)$, where $X$ is your nonzero solution. We want to construct a ($\mathbb{Q}$-linear) map $f:K \to \mathbb{Q}$ with some $f(X_i)\not = 0$; since the entries of $A$ lie in $\mathbb{Q}$, we'll then have $Af(X) = 0$ with $f(X)\in \mathbb{Q}^p$ nonzero. Construct a tower of fields $\mathbb{Q} = K_0 \subset \cdots \subset K_r = K$ with each $K_i/K_{i-1}$ either algebraic or isomorphic to $K_{i-1}(t)$ for some purely transcendental element $t$. In the former case, use the nondegeneracy of the trace form. In the latter case, use the fact that the map $f:K_{i-1}(t) \to K_{i-1}$ with $f\vert K_{i-1} = \text{id}$ and $f(t) = \alpha$ for any fixed $\alpha\in K_{i-1}$ is well-defined.

Rational solution to AX=0

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