## A flatness criterion for modules

Let $R$ be a ring with 1, and let $A$ be a right unital $R$-module.

1. Prove that if $A$ is flat then for every ideal $I \subseteq R$ the mapping $A \otimes_R I \rightarrow A \otimes_R R$ induced by the inclusion $I \rightarrow R$ is injective.
2. Suppose that for every finitely generated ideal $I \subseteq R$, the mapping $A \otimes_R I \rightarrow A \otimes_R R$ induced by the inclusion map is injective. Prove that in fact the induced map $A \otimes_R I \rightarrow A \otimes_R R$ is injective for every ideal $I$. Show further that if $K$ is any submodule of a finitely generated free module $F$ then $A \otimes_R K \rightarrow A \otimes_R F$ is injective. Show that the same is true for any free module (not necessarily finitely generated).
3. Under the hypothesis in part (2), suppose $\psi : L \rightarrow M$ is an injective left $R$-module homomorphism. Prove that the induced map $1 \otimes \psi : A \otimes_R L \rightarrow A \otimes_R M$ is injective. Conclude that $A$ is flat.
4. Let $F$ be a flat right unital $R$-module and $K \subseteq F$ an $R$-submodule of $F$. Prove that $F/K$ is flat if and only if $FI \cap K = KI$ for every finitely generated ideal $I \subseteq R$.

1. This is true by our definition of flat module.
2. Suppose the induced map $1 \otimes \iota : A \otimes_R I \rightarrow A \otimes_R R$ is injective for all finitely generated ideals $I \subseteq R$. Let $I \subseteq R$ be an arbitrary ideal and consider the induced map $1 \otimes \iota : A \otimes_R I \rightarrow A \otimes_R R$. Suppose $(1 \otimes \iota)(\sum a_i \otimes r_i) = 0$. Now there is a finitely generated ideal $J \subseteq R$ which contains all of the $r_i$. Moreover, $1 \otimes \iota$ restricted to $A \otimes J$ is injective by our hypothesis. Thus $\sum a_i \otimes r_i = 0$, and so $1 \otimes \iota$ is injective.

Now let $F = \oplus_T R$ be a finitely generated free module, and let $K \subseteq F$ be a submodule. Now $K \subseteq \oplus_T I_t$, where $I_t \subseteq R$ is a submodule (i.e. a left ideal). (Specifically, $I_t$ is the set of all $t$th coordinates of elements of $K$. This set is an $R$-submodule of $R$ (an ideal) since $K$ is an $R$-module.) Now $A \otimes_R (\oplus_T I_t)$ $\cong \oplus_T (A \otimes_R I_t)$ $\hookrightarrow \oplus_T A \otimes_R R$ $\cong A \otimes_R (\oplus_T R)$ $= A \otimes_R F$. Thus the induced map $1 \otimes \iota : A \otimes_R K \rightarrow A \otimes_R F$ is injective.

Finally, let $F$ be an arbitrary free module and let $K \subseteq F$ be an arbitrary submodule. Consider the induced map $1 \otimes \iota : A \otimes_R K \rightarrow A \otimes_R F$. If $(1 \otimes \iota)(\sum a_i \otimes k_i) = 0$, then there is a finitely generated submodule of $F$ containing all the $k_i$. Moreover, $1 \otimes \iota$ restricted to this submodule is injective, so that $\sum a_i \otimes k_i = 0$. Thus $1 \otimes \iota$ is injective.

3. Let $\psi : L \rightarrow M$ be an injective left module homomorphism. Now every module (for instance $M$) is isomorphic to a quotient of a free module. Suppose $F$ is a free module and $K \subseteq F$ a submodule such that $M = F/K$. In particular, the sequence $0 \rightarrow K \rightarrow F \rightarrow M \rightarrow 0$ is short exact. Let $J \subseteq F$ be the subset consisting of precisely those elements $x \in F$ such that $\pi(x) \in \mathsf{im}\ \psi$, where $\pi : F \rightarrow M$ is the natural projection. That is, $J = \{ x \in F \ |\ \pi(x) \in \mathsf{im}\ \psi \}$. Define $\theta : J \rightarrow L$ by $\theta(x) = y$, where $\pi(x) = \psi(y)$. Now $\theta$ is total by definition and is well-defined because $\psi$ is injective. Suppose now that $x,y \in J$ and $r \in R$. There exist $a,b \in L$ such that $\theta(x) = a$ and $\theta(y) = b$. So $\pi(x) = \psi(a)$ and $\pi(y) = \psi(b)$. Now $\pi(x) + r\pi(y) = \psi(a) + r\psi(b)$, so that $\pi(x+ry) = \psi(a+rb)$. Thus we have $\theta(x+ry) = a+rb = \theta(x) + r\theta(y)$, and so $\theta$ is a left $R$-module homomorphism. We claim also that $K \subseteq J$. It is certainly the case that $K \subseteq J$, since $K = \mathsf{ker}\ \pi$. This gives the following diagram of module homomorphisms.

A diagram of modules

We claim that this diagram commutes and that the rows are exact. First we show exactness; the bottom row is exact by definition. For the top row, certainly $\mathsf{im}\ \iota \subseteq \mathsf{ker}\ \theta$, since $\psi$ is injective. Now suppose $x \in \mathsf{ker}\ \theta$. Then $\pi(x) = 0$, so that $x \in \mathsf{ker}\ \pi = K = \mathsf{im}\ \iota$. Thus the rows are exact. Now we show that the diagram commutes. The left square certainly commutes, since the maps are all simply inclusions. Now let $x \in J$. Note that $(\psi \circ \theta)(x) = \psi(\theta(x))$ $= \psi(y)$, where $\pi(x) = \psi(y)$. Thus $\psi \circ \theta = \pi$, and so the diagram commutes. In fact, the triple $(\mathsf{id}, \iota, \psi)$ is a homomorphism of short exact sequences.

Recall that the functor $A \otimes_R -$ is right exact. Thus the induced diagram of modules

Another diagram of modules

commutes and has exact rows. By part (2) above, $1 \otimes \iota : A \otimes_R J \rightarrow A \otimes_R F$ is injective, and certainly $\mathsf{id} : A \otimes_R K \rightarrow A \otimes_R K$ and $1 \otimes \theta$ are surjective. By part 4 of this previous exercise, $1 \otimes \psi$ is injective.

Thus $A$ is flat as a right $R$-module.

4. Suppose $F$ is flat, $K \subseteq F$ a submodule, and that for all finitely generated ideals $I$ we have $FI \cap K = KI$. We wish to show that $F/K$ is flat. Let $I \subseteq R$ be an arbitrary finitely generated ideal. By the flatness criterion proved above, it suffices to show that $1 \otimes \iota : F/K \otimes_R I \rightarrow F/K \otimes_R R$ is injective. To that end, consider the exact sequences shown in the rows of the following diagram.

Yet another diagram of modules

With $t_K$ given by $k \otimes a \mapsto ka$ (and likewise $t_F$) and $\iota_K$ by $k \mapsto k \otimes 1$ (and likewise $\iota_F$), we claim that there exist suitable injective maps $\alpha$ and $\beta$ so that this diagram commutes.

Define $\alpha^\prime : F/K \times I \rightarrow FI/KI$ by $\alpha^\prime(x+K, a) = xa+KI$. Certainly if $x-y \in K$, then $xa-ya \in KI$, so that $\alpha^\prime$ is well defined. Moreover, this map is clearly bilinear. Thus we have an $R$-module homomorphism $\alpha : F/K \otimes_R I \rightarrow FI/KI$ given by $(x+K) \otimes a \mapsto xa+KI$. Clearly the upper right hand square commutes.

Note that $t_K$ is surjective (obviously). Moreover, since $F$ is flat, the map $1 \otimes \iota : F \otimes_R I \rightarrow F \otimes_R R$ is injective (by part (1) above). Now $t_F : F \otimes_R R \rightarrow F$ is an isomorphism and we have $t_F = t_F \circ (1 \otimes \iota)$, so that $t_F$ in our diagram is injective. By part 1 of this previous exercise, $\alpha$ is injective.

Now define $\beta : FI/KI \rightarrow F/K \otimes_R R$ by $x+KI \mapsto x \otimes 1$. First, if $x-y \in KI \subseteq K$, then $(x+K) \otimes 1 = (y+K) \otimes 1$, so that $\beta$ is well defined. It is clear that $\beta$ is a homomorphism of modules, and that the lower right hand square commutes. We claim also that $\beta$ is injective. To see this, suppose $x+KI \in \mathsf{ker}\ \beta$. Now $(x+K) \otimes 1 = 0$. Recall that $F/K \otimes_R R \cong F/K$ via the “multiplication” homomorphism $(x+K) \otimes r \mapsto xr+K$; so in fact we have $x+K = 0$, so that $x \in K$. Now $x \in FI$ and $x \in K$, so that (by our hypothesis) $x \in KI$. So $x+KI = 0$, and in fact $\beta$ is injective.

Finally, we claim that $\beta \circ \alpha = 1 \otimes \iota$. To see this, note that $(\beta \circ \alpha)((x+k) \otimes a) = \beta(xa+KI) = (xa+K) \otimes 1$ $= (x+K) \otimes a$.

Thus $1 \otimes \iota : F/K \otimes_R I \rightarrow F/K \otimes_R R$ is injective for all finitely generated ideals $I \subseteq R$. By the flatness criterion, $F/K$ is flat as a right $R$-module.

Conversely, suppose $F$ and $F/K$ are flat and let $I \subseteq R$ be a finitely generated ideal. Note that $KI \subseteq FI$ and $KI \subseteq K$, so that the inclusion $KI \subseteq FI \cap K$ always holds. Now consider the following commutative diagram of modules.

Last one, I promise!

Again, $t_K$ (and $t_F$ and $t_{F/K}$ denotes the isomorphism $K \otimes_R R \rightarrow K$ given by $k \otimes r \mapsto kr$. Let $\sum \zeta = x_ia_i \in FI \cap K$. (Letting $x_i \in F$ and $a_i \in I$.) We will chase $\zeta$ around the diagram to see that $\zeta \in KI$. Note that $0 = \pi(\zeta) = \sum x_ia_i + K$, so that $\sum (x_i+K) \otimes a_i = 0$ in $F/K \otimes_R R$. Note that $(1 \otimes \iota)(\sum (x_i+K) \otimes a_i) = \sum (x_i+K) \otimes a_i$, and because $1 \otimes \iota$ is injective (by our hypothesis that $F/K$ is flat), $\sum (x_i+K) \otimes a_i = 0$ in $F/K \otimes_R I$. In particular, $\sum x_i \otimes a_i \in \mathsf{ker}\ \pi \otimes 1$. Since the top row is exact, we have $\sum x_i \otimes a_i = \sum y_i \otimes b_i$ for some $y_i \in K$ and $b_i \in I$. Following these elements down to $F$, we see that $\zeta = \sum y_ib_i \in KI$ as desired.

Thus if $F$ and $F/K$ are flat, then $FI \cap K = KI$ for all finitely generated ideals $I \subseteq R$.

Post a comment or leave a trackback: Trackback URL.

### Comments

• mcoulont  On September 24, 2011 at 2:29 pm

In the second paragraph of 2, why K = direct sum of the I_t ?

• nbloomf  On September 24, 2011 at 6:05 pm

Good question. I’m about to go out, but I’ll get to this later today.

• nbloomf  On September 25, 2011 at 12:10 pm

On further reflection, that line is clearly not true. (I confused this with the analogous situation for ideals in a product of rings.)

It will take more time than I thought to fix this, so I’ll have to mark this ‘incomplete’ and come back later.