Proposition 6

a) Every ring of fractions of a presentable ring is presentable. Every algebra of finite type over a presentable ring is a presentable ring.

b) Let A be a presentable ring and J an ideal of A. The separated completion $\widehat{A}$ of A for the J-adic topology is presentable.

c) Every complete local noetherian ring is presentable.

d) Let A be a presentable local ring. There exists a regular local ring R and a surjective local homomorphism of R onto A.

Let A be a presentable ring; choose a regular ring R and a surjective homomorphism $\rho : R \to A$.

a) Let S be a multiplicative subset of A; put T = $\rho^{-1}(S)$. The homomorphism $T^{-1}R \to S^{-1}A$ deduced from $\rho$ is surjective and the ring $T^{-1}R$ is regular (No. 3, Example 1), hence $S^{-1}A$ is presentable.

Let B be an A-algebra of finite type; there exist a finite set I and a surjective homomorphism $A[(T_i)_{i \in I}] \to B$, hence a surjective homomorphism $R[(T_i)_{i \in I}] \to B$. Since the ring $R[(T_i)_{i \in I}]$ is regular (No. 2, Cor. 5 of Prop. 4), the ring B is presentable.

b) Put $I = \rho^{-1}(J)$ and denote by $\hat{R}$ the separated completion of R for the I-adic topology; for each integer $n \geq 0$, the canonical mapping $I^n/I^{n+1} \to J^n/J^{n+1}$ is surjective. Consequently, the homomorphism $\hat{R} \to \hat{A}$ deduced from $\rho$ is surjective (III, § 2, No. 8, Cor. 2 of Th. 1); since $\hat{R}$ is regular (No. 2, Cor. 3 of Prop. 4), the ring $\hat{A}$ is presentable.

c) This follows from IX, § 2, No. 5, Th. 3 a) and IX, § 3, No. 3, Th. 2 a).

d) Let $m_A$ be the maximal ideal of A; then $p = \rho^{-1}(m_A)$ is a prime ideal of R, the local ring $R_p$ is regular and the homomorphism $R_p \to A$ deduced from $\rho$ is local and surjective.

Since fields and Dedekind rings are regular, hence presentable, Proposition 6 implies that most rings ordinarily encountered in algebraic geometry are presentable.