Tor and Ext, Absolutely Flat

Absolutely Flat

A ring R is absolutely flat if every R module M is flat. This assumes R is commutative, so we can talk about flat modules.

Several conditions are equivalent to absolutely flat.

  1. Every R module is flat.

  2. Every ideal is idempotent.

  3. Every finitely generated ideal is generated by an idempotent.

Assume (1), whence any quotient ring R/H in R is a flat R module. Tensor this with 0 → H → R → R/H → 0. Use the quotient ring formula. Naturally R becomes R/H, but R/H also becomes R/H. Descend into the algebra of the tensor product, and imagine x,y in R cross R/H becoming x,y in R/H cross R/H. For this to be 0, some factor of x or y must lie in H, but then we could pass this factor of x over to y, and x,y would be 0 in R×(R/H). In other words, R×(R/H) → (R/H)×(R/H) has no kernel, and is an isomorphism. Since the sequence is exact, H×(R/H), which is H/H2, is equal to 0. Therefore H2 = H, and H is an idempotent ideal.

Now assume H is generated by x. Thus xxy = x for some y in R. Set e = xy to find an idempotent that generates H.

If H has two generators, replace them with their idempotent equivalents e and f. Now e+f-ef generates H and is idempotent. By induction, every finitely generated ideal is generated by an idempotent. Equivalently, any proper finitely generated ideal is a summand of R. Let e generate the ideal, and let f = 1-e generate the "other" summand.

Finally go from (3) back to (1). Let H be a finitely generated ideal of R, with G as the "other" summand. Thus 0 → H → R → G → 0 is split exact. Tensor with any module M and the result is split exact. Thus tor(R/H,M) = 0 for every finitely generated ideal H. This implies M is flat, hence R is absolutely flat.

Quotient Ring

If a homomorphism f takes an absolutely flat ring R onto a ring S, then S is absolutely flat. every S module is also an R module. Tensor any short exact sequence of S modules with another S module, and the result is exact as R modules, and hence as S modules.

Boolean Rings

Let x be an arbitrary member of the commutative ring R. If xn = x, for some n > 1, which may depend on x, then R is absolutely flat. Verify that xn-1 is the idempotent that generates the principal ideal {x}. Reason as above to show every finitely generated ideal is generated by an idempotent, which is condition (3).

As a special case, every boolean ring is absolutely flat.

Interestingly, these rings consist of units and zero divisors. The ideal generated by x equals the ideal generated by x2, so for some y, xxy = x, whence x*(1-xy) = 0. If x is not 0 and is not a 0 divisor then x is invertible.

Local Rings

Let R be a local absolutely flat ring. Let H be a principal ideal in the maximal ideal M. Assuming H is nontrivial, let the idempotent e generate H. Set f = 1-e, giving the other idempotent. A maximal ideal containing f cannot contain e, hence R has at least two maximal ideals. This is a contradiction, hence the maximal ideal is 0, and R is a field.

conversely, every field is absolutely flat, since the only nontrivial ideal is generated by 1, which is idempotent. Another proof: every module over a field is free, hence flat.