Proposition 4

Let $k$ be a field, $A$ a $k$-algebra essentially of finite type and $B$ a $k$-algebra.

a) If $A$ and $B$ are Macaulay rings, the same is true of $A \otimes_k B$.

b) If $A$ and $B$ are Gorenstein rings, the same is true of $A \otimes_k B$.

Suppose that $A$ and $B$ are Macaulay rings (resp. Gorenstein rings), and let us prove that the same is true of $A \otimes_k B$. The ring $A \otimes_k B$ is noetherian (No. 1, cor. of Prop. 2). The $A$-module $A \otimes_k B$ is free, hence flat. By Prop. 10 of § 2, No. 7 (resp. Cor. 1 of Prop. 12 of § 3, No. 8), it is enough for us to prove that $\kappa(p) \otimes_k B$ is a Macaulay ring (resp. a Gorenstein ring) for every prime ideal $p$ of $A$. The extension $\kappa(p)$ of $k$ is of finite type (No. 1, Prop. 2 and Example 1); we are therefore reduced to proving the statement in the case where the $k$-algebra $A$ is a finite-type extension $K$ of $k$.

The ring $K \otimes_k B$ is then a free $B$-module of finite rank, hence a Macaulay $B$-module. By § 2, No. 6, Cor. 1 of Prop. 8, it is a Macaulay ring, which proves a). Suppose now that $B$ is a Gorenstein ring. There exist intermediate extensions $K_i,\ 0 \leq i \leq m$, of $K$ such that

and such that $K_i$ is a monogenic $K_{i-1}$-algebra for $i = 1, \ldots, m$; this enables us to reduce to the case where the extension $K$ of $k$ is monogenic (of finite degree). The canonical homomorphism $B \to K \otimes_k B$ makes $K \otimes_k B$ into a flat $B$-algebra and, for every $q \in \mathrm{Spec}(B)$, the ring $(K \otimes_k B) \otimes_B \kappa(q)$, isomorphic to $K \otimes_k \kappa(q)$, is a monogenic $\kappa(q)$-algebra, hence a Gorenstein ring ($§ 3$, n° 9, example 5). Thus $K \otimes_k B$ is a Gorenstein ring ($§ 3$, n° 8, cor. 1 of Prop. 12), which completes the proof of b).