Invertible elements in $A \otimes B$ Let $A$ and $B$ be commutative, associative $k-$algebras with $1$, where $k$ is a field. Is there an easy way to describe the invertible elements in the $k-$algebra $A \otimes_k B$? Clearly the tensor of two invertible elements is invertible, but I suspect that there might be more, since not all elements are pure tensors.
Edit: Simpler version of the question: let $k=\mathbb{R}$ and $B=\mathbb{C}$. So the question asks for a nice description of the units of the complexification of an $\mathbb{R}-$algebra $A$ (in terms of $A^\times$). Since each element of $A \otimes \mathbb{C}$ can be written uniquely as $a \otimes 1 + b \otimes i$, the question reduces to finding a nice description for those tuples $(a,b)$ of elements of $A$ such that there exists a tuple $(c,d)$ (the inverse) so that
$$ \begin{cases} ac-bd=1 \\ ad+bc=0 \end{cases}$$
Clearly, $(a,0)$ with $a \in A^\times$ is such a tuple with inverse $(a^{-1},0)$, and so is $(0,b)$ with inverse $(0,-b^{-1})$. But as Andrew Hubery points out in his comment, there are also other examples, and I would like to know if they can be described in a simple way.
 A: When $A$ is a commutative $\Bbb{R}$-algebra, there are two cases we have to consider, namely when $A$ contains a square root of $-1$, $u$ and when $A$ doesn't already contain a square root of $-1$. $\newcommand\RR{\Bbb{R}}\newcommand\CC{\mathbb{C}}$
This is because $A\otimes_\RR\CC\simeq A\otimes_\RR \RR[x]/(x^2+1)\simeq A[x]/(x^2+1)$.
Case 1, $A$ contains $u$ with $u^2=-1$
When $A$ contains a square root of $-1$, $u$, then $x^2+1$ factors in $A[x]$ as $x+u$, $x-u$, giving comaximal ideals, since $(x+u)-(x-u)=2u$, which is a unit in $A$. Thus in this case, by the Chinese Remainder Theorem, we have $A\otimes_\RR\CC \simeq A[x]/(x-u)\times A[x]/(x+u)\simeq A\times A.$
Then the units of $A\otimes_\RR\CC$ are naturally isomorphic to $(A^\times)^2$. This isomorphism is via $a+bi \mapsto (a+bu,a-bu)$ and $(x,y)\mapsto \left(\frac{x+y}{2},\frac{x-y}{2u}\right)$.
Case 2, The general case, somewhat helpful when $A$ doesn't contain $u$ with $u^2=-1$
This case is less helpful, because I don't see any way to get the concrete structure of the unit group.
If $(a+bi)$ is a unit, then $(a+bi)(c+di)=1$ for some $c$ and $d$. Complex conjugation induces an automorphism of $A\otimes_\RR \CC$, so we get $(a-bi)(c-di)=1$ as well.
Multiplying these together, we get $(a+bi)(a-bi)(c+di)(c-di)=1$, or $(a^2+b^2)(c^2+d^2)=1$. Thus if $a+bi$ is a unit, then $a^2+b^2$ is a unit in $A$.
Conversely, if $a^2+b^2$ is a unit in $A$, with inverse $v$, then $(a+bi)(a-bi)v=1$, so $a+bi$ is a unit.
Thus $a+bi$ is a unit in $A\otimes_\RR \CC$ if and only if $a^2+b^2$ is a unit in $A$.
