Commutative Algebra 32

Torsion and Flatness


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

  • If m\in M satisfies am=0, we call it an atorsion element.
  • If m\in M is an a-torsion for some non-zero-divisor a\in A 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.

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.


Let a\in A be a non-zero-divisor. The injection 0 \to A \stackrel {a}\to A (where a here means multiplication by a) gives an injection upon tensoring with M : 0 \to M \stackrel {a}\to 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

A = k[X] \longrightarrow k[X, Y, Z]/(XZ, YZ) = B.

As an A-module, B has torsion since X \cdot \overline Z = 0 so it is not A-flat. The homomorphism corresponds to a morphism f from the variety W = \{(x, y, z): xz = yz= 0\} to V = \mathbb A^1 by projecting onto the x-coordinate. We treat W as a collection of varieties parametrized by points on V, with W_x = f^{-1}(x).


Note that when x\ne 0, we have W_x = \{(x, y, 0) : y \in k\} \cong \mathbb A^1. And when x = 0, we have W_0 = \{(0, y, z) : yz = 0\}, a union of two lines. On an intuitive level, this means \lim_{x\to 0} W_x \ne W_0, although such a statement makes no sense here.

The idea is: if k[V] \to k[W] is flat, then W\to V 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 U = V - \{0\}; we get the restriction f : f^{-1}(U) \to U where f^{-1}(U) = \{(x, y, z) : x\ne 0, z=0, y\in k\}. Taking the Zariski closure in their respective spaces gives us

W^\circ := \overline{f^{-1}(U)}= \{(x,y,0) : z=0\}\longrightarrow \overline U = V, \quad (x,y,0) \mapsto x

which is now flat since k[W^\circ] \cong k[X, Y] and the homomorphism of coordinate rings is k[X]\hookrightarrow k[X, Y]. This extension is flat since it is free.

In a nutshell, the problem is that f^{-1}(U) is not dense in W even though U is dense in V.

Exercise B

Let f:W\to V be a morphism of k-varieties with corresponding

f^* : A := k[V] \longrightarrow k[W] =: B.

Pick an open subset D(g) \subseteq V for g\in k[V]; recall D(g) = \{v\in V: g(v) \ne 0\}.

  • Prove that D(g) is dense in V if and only if g is not a zero-divisor.
  • Prove that f^{-1}(D(g)) is exactly D(f^*(g)).

Hence D(g) and f^{-1}(D(g)) have coordinate rings k[V][\frac 1 g] and k[W][\frac 1 g] respectively. The geometric condition “if D(g) is dense in V then f^{-1}(D(g)) is dense in W” then translates to “if g\in k[V] is not a zero-divisor then k[W] has no g-torsion”.



We will work through some concrete examples now.

Example 1

We have a sequence of flat extensions A \subset A[X] \subset A[X, \frac 1 X]. The first inclusion is flat because A[X] is free over A; the second inclusion is flat because it is a localization. By transitivity of flatness (proposition 3 here) we see that A[X, \frac 1 X] is a flat A-algebra.

Example 2

The injection k[X] \subset k[X, Y]/(XY) is not flat because X becomes a zero-divisor in k[X, Y]/(XY).

But what about the map k[X, Y]/(XY) \to k[X] taking Y\mapsto 0? This is also not A-flat. For that we note that k[X] is the quotient of k[X, Y]/(XY) by the ideal (Y). Now apply the following lemma.

Lemma 2.

Let \mathfrak a be an ideal of ring A. If A/\mathfrak a is A-flat, then \mathfrak a = \mathfrak a^2.


Take the exact sequence of A-modules 0 \to \mathfrak a \to A \to A/\mathfrak a \to 0. Taking (A/\mathfrak a) \otimes_A - gives us, together with (A/\mathfrak a) \otimes_A (A/\mathfrak a) \cong A/\mathfrak a, we get the exact sequence

0 \longrightarrow \mathfrak a/\mathfrak a^2 \longrightarrow A/\mathfrak a \stackrel {\text{id.}} \longrightarrow A/\mathfrak a \longrightarrow 0.

Hence \mathfrak a / \mathfrak a^2 = 0. ♦

Now applying this to our case, we see that (Y) \ne (Y^2) since Y \in (Y) - (Y^2).

Exercise C

Find an ideal \mathfrak a \ne 0, (1) of a ring A which satisfies \mathfrak a = \mathfrak a^2. For further challenge, give two different constructions.

Example 3

Take the affine k-varieties (k algebraically closed) V = \{ (x,y) \in \mathbb A^2 : y^2 = x^3\} and W = \mathbb A^1 with W\to V, t \mapsto (t^2, t^3), which corresponds to

A = k[X, Y]/(Y^2 - X^3) \longrightarrow k[T] = B, \quad X \mapsto T^2, Y \mapsto T^3.

We claim that B is not a flat A-algebra. There are a few ways to see this.

Direct Proof 

Let \mathfrak m = (X, Y) \subset A be a maximal ideal. It suffices to show that the product map \mathfrak m \otimes_A B \to \mathfrak m B not injective. Indeed X \otimes T^2, Y\otimes T both map to T^2 \cdot T^2 = T^3 \cdot T = T^4. On the other hand, they are distinct elements of \mathfrak m\otimes_A B. To see why we take

(A/ \mathfrak m) \otimes_A \mathfrak m \otimes_A B \cong \mathfrak m /\mathfrak m^2 \otimes_A B,

and the action of A on \mathfrak m/\mathfrak m^2 factors through A/\mathfrak m. Since \mathfrak m/\mathfrak m^2 has basis XY (see the calculations from earlier) the above module is a direct sum of (k\cdot X \otimes_A B) and (k\cdot Y \otimes_A B); in particular X\otimes T^2 \ne Y\otimes T in \mathfrak m \otimes_A B.

Indirect Proof 

If A\to B is flat, then so is A/(X) \to B/(X)B. But X\mapsto T^2 \in B so the resulting map is

k[X, Y]/(Y^2 - X^3, X) \cong k[Y]/(Y^2) \longrightarrow k[T]/(T^2), \quad Y \mapsto 0.

This map factors through k[Y]/(Y^2) \to k \hookrightarrow k[T]/(T^2). The first map is not flat by applying lemma 2 to the ideal \mathfrak a = (Y) of ring A' = k[Y]/(Y^2). Specifically the injective map \mathfrak a\to A', upon applying the functor (A'/\mathfrak a)\otimes_{A'} -, gives us \mathfrak a \to k taking Y\mapsto 0. Since k[T]/(T^2) is free over k, upon applying k[T]/(T^2) \otimes_k -, the map is still not injective.

Example 4

We consider a family of ellipses \{x,y\in \mathbb A^2 : ax^2 + by^2 = c\} parametrized by abc from earlier. This corresponds to the homomorphism

f : k[A, B, C] \hookrightarrow k[A, B, C, X, Y]/(AX^2 + BY^2 - C).

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

f' : k[A] \hookrightarrow k[A, X, Y]/(AX^2)

is not flat. But X \in \text{ RHS} 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 k[T] \hookrightarrow k[X, Y, T]/(Y - TX^2) a flat extension? What if we restrict to T\ne 0?

2. Apply the same analysis to k[T] \hookrightarrow k[X,Y,T]/(XY - T).

3. For example 4, prove that the extension is flat if we restrict to the open affine subset B \ne 0, i.e. we get the extension

k[A, B, C, \frac 1 B] \hookrightarrow k[A, B, C, \frac 1 B, X, Y]/(AX^2 + BY^2 - C).

Intuitively, this means if b\ne 0, the family of ellipses ax^2 + by^2 = c is well-behaved. What about the open subset C\ne 0?


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