Torsion and Flatness
Let A be a ring and M an A-module; let .
- If satisfies , we call it an –torsion element.
- If is an -torsion for some non-zero-divisor we call it a torsion element.
- M is said to be torsion-free if it has no torsion elements other than 0.
Note that a torsion element in A is the same as a zero-divisor.
Decide if each of the following is true or false.
- A submodule of a torsion-free module is torsion-free.
- A quotient module of a torsion-free module is torsion-free.
- An intersection of torsion-free submodules of M is torsion-free.
- A sum of torsion-free submodules of M is torsion-free.
- A direct sum of torsion-free modules is torsion-free.
- A direct sum of a torsion-free module and any module is torsion-free.
Now we wish to prove:
A flat module M over a ring A is torsion-free.
Let be a non-zero-divisor. The injection (where here means multiplication by ) gives an injection upon tensoring with M : . Hence M has no a-torsion elements other than 0. ♦
Eventually, we will see that for a principal ideal domain, the converse is true.
A Little Geometry
Let us give a geometric interpretation to lemma 1. Take the concrete example
As an A-module, B has torsion since so it is not A-flat. The homomorphism corresponds to a morphism f from the variety to by projecting onto the x-coordinate. We treat W as a collection of varieties parametrized by points on V, with .
Note that when , we have . And when , we have , a union of two lines. On an intuitive level, this means , although such a statement makes no sense here.
The idea is: if is flat, then gives a family of varieties “smoothly parametrized” by elements of V.
For our case, let us rephrase our statement so that it is more algebra-friendly. Let ; we get the restriction where . Taking the Zariski closure in their respective spaces gives us
which is now flat since and the homomorphism of coordinate rings is . This extension is flat since it is free.
In a nutshell, the problem is that is not dense in W even though U is dense in V.
Let be a morphism of k-varieties with corresponding
Pick an open subset for ; recall .
- Prove that is dense in V if and only if g is not a zero-divisor.
- Prove that is exactly .
Hence and have coordinate rings and respectively. The geometric condition “if is dense in V then is dense in W” then translates to “if is not a zero-divisor then has no g-torsion”.
We will work through some concrete examples now.
We have a sequence of flat extensions . The first inclusion is flat because is free over A; the second inclusion is flat because it is a localization. By transitivity of flatness (proposition 3 here) we see that is a flat A-algebra.
The injection is not flat because X becomes a zero-divisor in .
But what about the map taking ? This is also not A-flat. For that we note that is the quotient of by the ideal (Y). Now apply the following lemma.
Let be an ideal of ring A. If is A-flat, then .
Take the exact sequence of A-modules . Taking gives us, together with , we get the exact sequence
Hence . ♦
Now applying this to our case, we see that since .
Find an ideal of a ring A which satisfies . For further challenge, give two different constructions.
Take the affine k-varieties (k algebraically closed) and with , , which corresponds to
We claim that B is not a flat A-algebra. There are a few ways to see this.
Let be a maximal ideal. It suffices to show that the product map not injective. Indeed both map to . On the other hand, they are distinct elements of . To see why we take
and the action of A on factors through . Since has basis X, Y (see the calculations from earlier) the above module is a direct sum of and ; in particular in .
If is flat, then so is . But so the resulting map is
This map factors through . The first map is not flat by applying lemma 2 to the ideal of ring . Specifically the injective map , upon applying the functor , gives us taking . Since is free over k, upon applying , the map is still not injective.
We consider a family of ellipses parametrized by a, b, c from earlier. This corresponds to the homomorphism
We claim that this is not flat. Using the trick in the indirect proof above, we factor by B = C = 0. It suffices to show that
is not flat. But is an A-torsion element, so we are done.
Geometrically, the ellipse degenerates at the point A = B = C = 0, so it becomes the whole plane.
1. Is the extension a flat extension? What if we restrict to ?
2. Apply the same analysis to .
3. For example 4, prove that the extension is flat if we restrict to the open affine subset , i.e. we get the extension
Intuitively, this means if , the family of ellipses is well-behaved. What about the open subset ?