Let $R$ and $S$ be commutative $k$-algebras, for some commutative ring $k$. All rings are assumed to have the identity.
Let $U(R)$ denote the group of invertible elements of $R$. What can we say about $U(R\otimes_k S)$?
Is something like $$U(R\otimes_k S)\cong U(R)\times U(S)/U(k)$$ true?
I.e. can we express it for example in terms of the pushout?
Cheers.
https://www.jstor.org/stable/2373506 shows that under rather strict circumstances the units are as suggested. For a small example where the result is different, let R = S = k[X]/<X^2> where k is the field with 2 elements, i.e. GF(2). U(k)={1}. Let Y be the image of X in k[X]/<X^2>. The units in R and S are 1 and 1+Y. The other 2 elements 0, Y are not invertible. So the cardinality of U(R)×U(S) is 4.
The 5 units in the tensor product are: 1(x)1 + Y(x)Y and the 4 expected units 1(x)1, (1+Y)(x)1, 1(x)(1+Y), (1+Y)(x)(1+Y)
So the cardinality of U(R)(×)U(S)/U(k) is 5.