By Keerthi Madapusi
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So Nil R = 0, and R is reduced. 5. If R is reduced, and P ∈ Spec R, then K(RP ) ∼ = K(R)P Recall that K(R) is the total quotient ring of R. Proof. 4). Since all the primes in K(R) are the primes contained in the associated primes of R, we see that all the primes in K(R) are minimal. 3), it is a product of fields. To be more specific, it’s isomorphic to Q∈Ass R K(R)Q . Now, observe that since RQ is a field, and K(R)Q is a localization of RQ , K(R)Q ∼ = RQ . Moreover, if Q P , for some prime P ⊂ R, then (RQ )P = 0.
Both statements are easy. For the second, note that if s/u ∈ U −1 S, and s satisfies the monic equation sn + an−1 sn−1 + . . + a0 = 0 over R, then s/u satisfies the monic equation (s/u)n + (an−1 /u)(s/u)n−1 + . . + a0 /un = 0 over U −1 R. 3. 1. Reduced Rings. 1. We will say that a ring R is of dimension 0 if every maximal ideal of R is also minimal. Reduced rings of dimension 0 are easy to describe. 2. The only reduced local rings of dimension 0 are fields. 58 4. INTEGRALITY: THE COHEN-SEIDENBERG THEOREMS Proof.
6). (2) ⇒ (3) is trivial, so we’ll finish by proving (3) ⇒ (1). So suppose N → N is a monomorphism of R-modules; then we’ll be done if we show that M ⊗R N → M ⊗R N is a monomorphism of S-modules. It suffices to show this after localizing 2. HOMOLOGICAL CRITERION FOR FLATNESS at a maximal ideal Q ⊂ S; but that this is true follows immediately from our hypothesis in (3). The next Proposition will be used often in the remainder of these notes. 11. Let S be an R-algebra, let M and N be R-modules, with M finitely presented and let P be an S-module flat over R.