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If $V$ is a $K$-vector space and $L$ is a field extension of $K$ then why is the map $v \to v\otimes 1$ an embedding of $V$ into $V\otimes_K L?$

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Note that if $L$ is a field extension of $K$, then $L$ is a $K$-vector space.

$1$ is linearly independent in $L$, so we can find a $K$-basis of $L$ including $1$.

But there is a lemma which states that if $M$ and $N$ are free $R$-modules with basis $\lbrace m_{i} \rbrace _{i \in I} $ and $\lbrace n_{j} \rbrace _{j \in J} $ then $M \otimes_R N$ is free with basis $$\lbrace m_i \otimes n_j \rbrace _{i \in I , \ j \in J}$$ So if $\lbrace v_h \rbrace _{h \in H}$ is a $K$-basis for V and $$v = \sum_h a_h v_h$$ is such that $$0 = v \otimes 1 = \biggl( \sum_h a_h v_h \biggr) \otimes 1 = \sum_h a_h \cdot ( v_h \otimes 1) $$ we can conclude that $a_h = 0 $ for all $h$ because $\lbrace v_h \otimes 1 \rbrace _h$ are part of a $K$-basis of $V \otimes_K L$. Therefore $v = 0$.

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