# Torsion and Flatness

Definition.Let A be a ring and M an A-module; let .

- If satisfies , we call it an –
torsionelement.- If is an -torsion for some non-zero-divisor we call it a
torsionelement.- M is said to be
torsion-freeif it has no torsion elements other than 0.Note that a torsion element in A is the same as a zero-divisor.

**Exercise A**

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:

Lemma 1.A flat module M over a ring A is torsion-free.

**Proof**

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. ♦

**Note**

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*.

**Exercise B**

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”.

# Examples

We will work through some concrete examples now.

## Example 1

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.

## Example 2

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.

Lemma 2.Let be an ideal of ring A. If is A-flat, then .

**Proof**

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 .

**Exercise C**

Find an ideal of a ring *A* which satisfies . For further challenge, give two different constructions.

## Example 3

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.

**Direct Proof **

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 .

**Indirect Proof **

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.

## Example 4

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.*

### Exercise D

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 ?