Vector space/Change of base field/Properties/Fact/Proof

Proof

(1). The multiplication

$L\times L\longrightarrow L,(r,s)\longmapsto rs,$

is ${}L$-bilinear, and, in particular, ${}K$-bilinear; therefore, it yields, according to fact, to a ${}K$-linear mapping

$L\otimes _{K}L\longrightarrow L.$

This induces, due to fact  (2) and fact, a ${}K$-linear mapping

$L\otimes _{K}{\left(L\otimes _{K}V\right)}\cong {\left(L\otimes _{K}L\right)}\otimes _{K}V\longrightarrow L\otimes _{K}V.$

This gives a well-defined scalar multiplication

$L\times {\left(L\otimes _{K}V\right)}\longrightarrow {\left(L\otimes _{K}V\right)},$

which is explicitly given by

${}s\cdot {\left(\sum _{j=1}^{n}r_{j}\otimes m_{j}\right)}=\sum _{j=1}^{n}{\left(sr_{j}\right)}\otimes m_{j}\,.$

From this description, we can deduce directly the properties of a scalar multiplication.
(2). The ${}K$-homomorphism follows directly from the bilinearity of the tensor product. For ${}L=K$, the mapping is surjective. The scalar multiplication ${}K\times V\rightarrow V$ induces a ${}K$-linear mapping

$K\otimes _{K}V\longrightarrow V.$

The composition of the canonical mapping ${}V\rightarrow K\otimes _{K}V$ with this mapping is the identity on ${}V$, so that the first mapping is also injective.
(3) follows from the explicit description in (1).
(4) follows from fact.
(5) follows from (4).
(6). Because of part (2), we have, one one hand, a ${}K$-linear mapping ${}V\rightarrow L\otimes _{K}V$. This yields a ${}K$-multilinear mapping

$M\times V\longrightarrow M\times {\left(L\otimes _{K}V\right)}\longrightarrow M\otimes _{L}{\left(L\otimes _{K}V\right)},$

which induces a ${}K$-linear mapping

$M\otimes _{K}V\longrightarrow M\otimes _{L}{\left(L\otimes _{K}V\right)}.$

On the other hand, we have a ${}L$-linear mapping

$L\otimes _{K}V\longrightarrow M\otimes _{K}V.$

On the right-hand side, we have an ${}M$-vector space; therefore, the scalar multiplication may be considered as an ${}L$-multilinear mapping

$M\times {\left(L\otimes _{K}V\right)}\longrightarrow L\otimes _{K}V.$

This induces an ${}L$-linear mapping

$M\otimes _{L}{\left(L\otimes _{K}V\right)}\longrightarrow L\otimes _{K}V.$

The two mappings are inverse to each other, as this can be checked on the decomposable tensors. This shows also that the ${}M$-multiplications are the same.

To fact