Suppose a finite homogeneous system of linear equations with rational coefficients has a nonzero complex solution. Must it have a nonzero rational solution? Prove your answer or give a counterexample.
Solution
Write the system as \[ Ax=0, \] where \(A\in M_{m\times n}(\QQ)\).
1. The rank of \(A\) over \(\QQ\) equals its rank over \(\CC\).
For each \(k\), every \(k\times k\) minor of \(A\) is a rational number. Such a minor is nonzero when regarded as an element of \(\QQ\) if and only if it is nonzero when regarded as an element of \(\CC\). Since the rank is the largest size of a nonzero minor, the two ranks are equal.
2. One has\[ \rank_{\QQ}A<n. \]
By hypothesis there is a nonzero vector \[ v\in\CC^n \] with \(Av=0\). Hence the complex-linear map defined by \(A\) has nontrivial kernel, so rank-nullity gives \[ \rank_{\CC}A<n. \] Step 1 gives the same inequality over \(\QQ\).
3. The system has a nonzero rational solution, so the answer is\[ \boxed{\text{yes}}. \]
By step 2 and rank-nullity over \(\QQ\), \[ \dim_{\QQ}\ker(A:\QQ^n\to\QQ^m) = n-\rank_{\QQ}A >0. \] Thus this kernel contains a nonzero vector in \(\QQ^n\), which is a nonzero rational solution of the original system.
4. Q.E.D.
Step 3 proves the required assertion.