§ Tensor is right exact
Consider an exact sequence
We wish to consider the operation of tensoring with some ring .
For a given ring morphism this induces a new
So we wish to contemplate the sequence:
To see if it is left exact, right exact, or both. Consider the classic sequence of
modules over :
§ A detailed example
Where is for inclusion, is for projection. This is an exact sequence, since it's of the form
kernel-ring-quotient. We have three natural choices to tensor with: .
By analogy with fields, tensoring with the base ring is unlikely to produce anything of interest.
maybe more interesting, but see that the map gives us an
isomorphism between the two rings. That leaves us with the final and most interesting element (the one with torsion),
. So let's tensor by this element:
- See that has elements of the form , We might imagine that the full ring collapses since (since in ). But this in fact incorrect! Think of the element . We cannot factorize this as since . So we have the element .
- See that : Factorize .
- Similarly, see that . Elements and .
- In general, Let's investigate elements . We can write this as . The gives us a "machine" to reduce the number by and by . So if we first reduce by , we are left with (for remained) for some . We can then reduce by to get . So if , then we get . But see that all elements of the form is divisible by . Hence, all multiples of are sent to zero, and the rest of the action follows from this. So we effectively map into
- In fact, we can use the above along with (1) write finitely generated abelian groups as direct sum of cyclic groups, (2) tensor distributes over direct sum. This lets us decompose tensor products of all finitely generated abelian groups into cyclics.
- This gives us another heuristic argument for why . We should think of as , since we have "no torsion" or "torsion at infinity". So we get the tensor product should have .
- Now see that the first two components of the tensor give us a map from which sends:
- This map is not injective, since this map kills everything! Intuitively, the "doubling" that is latent in is "freed" when injecting into . This latent energy explodes on contant with giving zero. So, the sequence is no longer left-exact, since the map is not injective!
So finally, we have the exact sequence:
- So the induced map is identically zero! Great, let's continue, and inspect the tail end . Here, we sent the element . This clearly gives us all the elements: For example, we get as the preimage of and we get as the preimage of (predictably) . Hence, the map is surjective.
We do NOT have the initial since is no longer injective.
It fails injectivity as badly as possible, since . Thus, tensoring is
RIGHT EXACT. It takes right exact sequences to right exact sequences!
§ The general proof
Given the sequence:
We need to show that the following sequence is exact:
- First, to see that is surjective, consider the basis element . Since is surjective, there is some element such that . So the element maps to by ; (by definition of , and choice of ). This proves that is exact.
- Next, we need to show that .
- To show that , consider an arbitrary . Now compute:
So we have that any element in is in the kernel of .
Next, let's show . This is the "hard part" of the proof. So let's
try a different route. I claim that if iff . This
Since each line was an equality, if I show that , then I have that .
So let's prove this:
I claim that the (informally, "take common").
Define the quotient map . This is a legal quotient map because
is a submodule of .
Let's now study . It contains all those elements such that .
But this is only possible if . This means that .
Also see that for every element , there is an inverse
element . So, the map is surjective . Hence, .
Combining the two facts, we get: