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 tth 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.

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  • 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.

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