Let $A$ be a $K$-algebra and $x$ an indeterminate. Show that $ A \otimes K[x] \cong A[x]$ as $K$-algebras
I'm stuck. I thought the map $f: A \times K[x] \to A[x]$ given by $(a,p(x)) \mapsto ap(x)$ would work but I can't find an inverse for it. Neither can I think of another way to show that it's an isomorphism.
I think I need to add that $(a,p(x)) \mapsto ap(x)$ induces a map $a \otimes p(x) \mapsto ap(x)$ by the universal property, but then any map $A[x] \to A \otimes K[x]$ has to send a polynomial with coefficients in $A$ to some polynomial in $K[x]$ and it's not obvious how that should be done because the tensor is over $K$ and not over $A$.
Define $ A[x] \to A \otimes K[x] $ by $ a\cdot x \to a \otimes x $, and extending by the universal property of the polynomial ring. (Note that $ A \otimes K[x] $ is an $ A $-algebra in the obvious way.) Now, verify that this map is an inverse to the map you defined, and you're done.