Topics in algebra Chapter IV: Commutative rings and modules I - 1

Let M be an R-module; M is said to be flat if ⊗RM preserve short
exact sequences,i.e., if 0 −→ N1
φ
−→ N2
ψ
−→ N3 −→ 0 is a s.e.s., then
0 −→ N1 ⊗R M φ⊗idM
−→ N2 ⊗R M ψ⊗idM
−→ N3 ⊗R M −→ 0 stays exact.
Lemma 6.11 Let M be an R-module; then R⊗R M and M ⊗R R are canonical
isomorphic to M.
Proof. Let i : R ⊗R M −→ M be the map given by i(r ⊗ x) = rx; then i
is well-defined and is an R-homomorphism as i is induced by the R-bilinear
map ĩ : R × M −→ M with ĩ(r, x) = rx. It is easy to see that i is an
isomorphism and R ⊗R M ∼ = M. Similarly, M ⊗R R ∼ = M.
Corollary 6.12 R is a flat R-module.
φ
Proof. Let 0 −→ N1 −→ N2 be an exact sequence of R-modules; then ⊗RR
induces the sequence 0 −→ N1 ⊗R R φ⊗idR
−→ N2 ⊗R R. Moreover,
0 −→ N1 ⊗R R φ⊗idR
−→ N2 ⊗R R
↓ ↓
0 −→ N1
φ
−→ N2
is a commutative diagram of R-modules and R-homomorphisms such that
the second row is exact and the vertical homomorphisms are isomorphisms.
Therefore we conclude that φ ⊗ idR is 1-1.
Let R be a ring and S be a multiplicative subset of R; then the construction
of RS can be carried through with an R-module M in place of the ring R.
Definition 6.13 Let R be a ring, S be a multiplicative subset of R and M be
an R-module. Define a relation on the set M × S = {(x, s) | x ∈ M, s ∈ S}
as follows: (x, s) ∼ (y, t) if and only if there is an element u ∈ S such that
u(xt − ys) = 0. It is easy to show that ∼ is an equivalent relation. As
the above (M × S)/ ∼ can be made into an RS-module. We use x
for the
s
equivalent class of the element (x, s), and MS for the module (M × S)/ ∼.
As before, if S = R−P for some prime P , then we use MP instead of MS.i.e.,
MP = { x
| x ∈ M, s /∈ P }.
s
Lemma 6.14 Let M be an R-module and S be a multiplicative subset of R;
then M ⊗R RS is canonically isomorphic to MS, so is RS ⊗R M.
142
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