# Basis of extension of scalars

Let $$A$$ be a $$k$$-algebra of finite dimension, where $$k$$ is a field, with $$k$$-basis $$e_1,\cdots,e_n$$. Let $$K$$ be some field extension of $$k$$. The extension of scalars is defined as $$A\otimes_kK$$.

What's a $$K$$-basis of $$A\otimes_kK$$? Would $$e_1\otimes1,\cdots,e_n\otimes1$$ work?

It clearly generates $$A\otimes_kK$$. However, I am having trouble showing linear independence.

Let $$\lambda_1,\cdots,\lambda_n \in K$$, suppose $$\sum_{i=1}^ne_i\otimes\lambda_i=0$$, why does this imply $$\lambda_1=\cdots=\lambda_n=0$$?

If $$\lambda_i\in k$$ then I can exploit bilinearity to get $$\sum_{i=1}^ne_i\otimes\lambda_i=0=(\sum_{i=1}^ne_i\lambda_i)\otimes1.$$

So $$(\sum_{i=1}^ne_i\lambda_i)=0$$ and $$\lambda_1=\cdots=\lambda_n=0.$$

But I don't know what to do in the case where $$\lambda_i\in K.$$

Any hints are appreciated.

Suppose $$\sum_{i=1}^ne_i\otimes_k\lambda_i=0$$ for some $$\lambda_1,\ldots,\lambda_n\in K$$. If we define a $$k$$-bilinear map $$\beta_i:A\times K\to K$$ as in Lord shark's answer, that is, $$\beta_i\left(\sum_{j=1}^n a_j e_j,b\right)=a_ib,$$ we have a $$k$$-linear map $$f_i: A\otimes_k K\to K$$ which takes $$\left(\sum_{j=1}^n a_j e_j\right)\otimes_k b$$ to $$a_ib$$.
Note that $$f_i(e_i\otimes_k b)=b$$ and $$f_i(e_j\otimes_k b)=0$$ if $$i\neq j$$.
So $$f_i\left(\sum_{j=1}^ne_j\otimes_k\lambda_j\right)=\lambda_i.$$ But $$f_i\left(\sum_{j=1}^ne_j\otimes_k\lambda_j\right)=f_i(0)=0.$$ So $$\lambda_i=0$$. Since $$i$$ is arbitrary, the claim follows.
For each $$i$$, there's a $$k$$-bilinear map $$\beta_i:A\otimes_k K\to K$$ defined by $$\beta_i\left(\sum_{j=1}^n a_j e_j,b\right)=a_ib.$$ This corresponds to a $$k$$-linear map taking $$\sum_{j=1}^n e_j\otimes b_j$$ to $$b_i$$. So if $$\sum_{j=1}^n e_j\otimes b_j=0$$ then $$b_i=0$$ for all $$i$$.