Tensor product of short exact sequences

Let $k$ be a field and $R=k[x_j:j\in J]/I$ be a smooth $k$-Algebra. Then $R \otimes R$ is also a smooth $k$-Algebra and by the universal property of the tensor product, we obtain a (surjective) map $k[x_i:i\in J] \otimes k[x_j :j\in J] \rightarrow R\otimes R$.

Can we say anything about the kernel? If yes, how Is it related to $I \otimes k[x_j :j\in J] +k[x_j:i\in J] \otimes I$?

• The kernel should look more like $I\otimes 1+1\otimes I$. – Pedro Tamaroff Aug 27 '18 at 12:30
• Indeed you are right. This is also what I wanted to express, that was a bit careless off me. – slinshady Aug 27 '18 at 12:43