$A\otimes B\cong A$? I am struggling for quite a while now with tensor products, since I have limeted time to learn to understand them. Now on several occasions I have read that we have $A\otimes_KK\cong A$ for $A$ an $K$-algebras. I convinced myself of this by "showing" the following function is an isomorphism: 

\begin{align}f:A\otimes_KK&\to A\\a\otimes_Kk&\mapsto a\end{align}
  Surjectivity: obvious
Injectivity: $\ker(f)=\{a\otimes_Kk\in A\otimes_KK\mid f(a\otimes_Kb)=a=0\}=\{0\otimes_Kk\mid 0\in A, k\in K\}=\{0\}$ hence $f$ is injective.

I suspect something is terribly wrong with this proof, because we can do the same thing if we replace $K$ by any $K$-algebra $B$. In wich case $A\otimes B\cong A$, wich doesn't appear right...
My question is: What would be the correct way to show relations like this? (And implicitly: what am I not getting when using tensor products?)
 A: This map is actually not even well defined. This is because $f(a \otimes 2) = a$, but $f(2a \otimes 1) = 2a$, even though $a \otimes 2 = 2a \otimes 1$ in $A \otimes_K K$. The correct map is $a \otimes k \mapsto ka$.
A: One of your errors is that you think every element of a tensor product $A\otimes_K B$ is an elementary tensor $a\otimes b$. I say this because from the way you are defining $f$ and trying to compute the kernel it seems you think every tensor is elementary. Elements of a tensor product are sums of elementary tensors $a_1\otimes b_1 + \cdots + a_r \otimes b_r$. Now it is true in $A \otimes_K K$ that all tensors are elementary, but it is not true in general tensor products. In particular, to show a linear map out of a tensor product is zero it is insufficient in general to show the only elementary tensor that goes to $0$ is the zero tensor. For example, there is an $\mathbf R$-linear map $\mathbf C \otimes_{\mathbf R} \mathbf C \rightarrow \mathbf C$ by $z \otimes w \mapsto zw$ on elementary tensors and the only elementary tensor that goes to $0$ is $0 \otimes 0$ (do you understand that $z \otimes 0 = 0 \otimes w = 0 \otimes 0$ in this tensor product?), but lots of non-elementary tensors also go to $0$ so the map is not injective. For instance, $1\otimes i - i \otimes 1$ is mapped to $0$ and $1\otimes i - i \otimes 1$ is not elementary in $\mathbf C \otimes_{\mathbf R} \mathbf C$. Being able to prove that tensor is not elementary is one indicator that you can work properly with tensors.
A: Your isomorphism should be
\begin{align}
f: A \otimes_K K& \to A\\
a \otimes_K k& \mapsto k \cdot a
\end{align}
This is a well-defined isomorphism and is $K$-linear, which your function wasn't. Because the scalar multiplication is only defined for scalars, this isomorphism can only be defined when one of the domain vector spaces is $K$ itself.
