# Torsion and Flatness

Definition.

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 $a$torsion 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.

Proof

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. ♦

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

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

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”.

# Examples

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 $\mathfrak a = \mathfrak a^2$.

Proof

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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