# When can linear independence of vectors over a field be 'carried over' to linear independence its subfield?

Let $k'$ be an field extension of an infinite field $k$. If $\{v_i\}$ is a set of vectors over $k$, and they satisfy $\sum_{i=1}^{n} c_iv_i =0$ for some nontrivial $\{c_i\} \subset k'$, then is the same set of vectors $\{v_i\}$ linearly dependent over $k$? In other words, can the $c_i$'s be chosen in $k$ only?

I see that if $k'$ is a finite extension of $k$, then I can quickly prove why this fact is true. But for infinite extensions I'm less sure. Maybe I'm missing something very obvious? Thanks in advance.

-

Any non trivial linear combination with coefficients from $k'$ will involve only finitely many elements of $k'$. Consider the subfield $k''$ of $k'$ generated by $k$ and by these coefficients. Then $k''$ is a finite extension of $k$, the $v_i$'s satisfy a linear relation with coefficients in $k''$, and you claim to be able to do what you want in that case.
N.B.: In any case, your question, as stated, does not make a lot of sense really: your $v_i$'s are presumably elements of a $k$-vector space, so what sense does it make to compute a linear combination with them using coefficients from $k'$? What I wrote above answers a version of what you meant:
suppose that $V$ is a $k$-vector space, that $(v_i)_{i\in I}$ is a family of elements of $V$, and that the family $(1\otimes v_i)_i$ of elements of the $k'$-vector space $k'\otimes_k V$ is linearly dependent over $k'$: then $(v_i)_{i\in I}$ is linearly dependent in $V$ over $k$.
There is no need, though, to proceed by reducing first to a finite subextension to do this. An alternative is the following: suppose that we have a non-trivial zero linear combination $\sum_i a_i\otimes v_i=0$ of the family $(1\otimes v_i)_{i\in I}$ with coefficients $(a_i)_{i\in I}$ in $k'$. Let $(z_j)_{j\in J}$ be a basis of $k'$ as a $k$-vector space, and suppose for all $i\in I$ we have $a_i=\sum_{j\in J}a_{i,j}z_j$, with all the $a_{i,j}\in k$. Then $$0 = \sum_{i\in I} a_i\otimes v_i = \sum_{j\in J}z_j\otimes\Bigl(\sum_{i\in I}a_{i,j}v_i\Bigr).$$ As $(z_j)_{j\in J}$ is linearly independent, this means that for all $j\in J$ we have $$\sum_{i\in I}a_{i,j}v_i=0,$$ and since we have assumed one of the $a_i$s is not zero, one of these linear combinations, which have coefficients in $k$, is itself non-trivial.