Proposition 3

— Let $K$ be a field. a) Every subalgebra of a regular $K$-algebra is itself regular. b) The tensor product of two regular $K$-algebras is again regular. c) Let $A$ be a $K$-algebra and $K'$ an extension of $K$. For $A$ to be regular it is necessary and sufficient that the $K'$-algebra $A_{(K')}$ derived from $A$ by extension of scalars should be regular.

The proof of a) (resp. c)) is identical to that of part a) (resp. d)) of Prop. 3, V, p. 119, after replacing everywhere « reduced ring » by « integral domain » and « separable algebra » by « regular algebra ». Let us prove b).

Let $A$ and $B$ be two regular $K$-algebras. Let $L$ be an extension of $K$. Since $A$ is regular, the ring $L \otimes_K A$ is an integral domain. By Prop. 2, the ring $(L \otimes_K A) \otimes_K B$ is an integral domain, because $B$ is regular. Finally, the ring $L \otimes_K (A \otimes_K B)$ is isomorphic to $(L \otimes_K A) \otimes_K B$, and hence is an integral domain. This shows $A \otimes_K B$ to be a regular $K$-algebra.