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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.

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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.