Schanuel’s Lemma

Let R be a ring with 1.

  1. If 0 \rightarrow K_1 \stackrel{\alpha_1}{\rightarrow} P_1 \stackrel{\beta_1}{\rightarrow} M \rightarrow 0 and 0 \rightarrow K_2 \stackrel{\alpha_2}{\rightarrow} P_2 \stackrel{\beta_2}{\rightarrow} M \rightarrow 0 are short exact sequences of (left unital) modules and P_1 and P_2 are projective, show that P_1 \oplus K_2 \cong_R P_2 \oplus K_1.
  2. If 0 \rightarrow M \stackrel{\alpha_1}{\rightarrow} Q_1 \stackrel{\beta_1}{\rightarrow} L_1 \rightarrow 0 and 0 \rightarrow M \stackrel{\alpha_2}{\rightarrow} Q_2 \stackrel{\beta_2}{\rightarrow} L_2 \rightarrow 0 are short exact sequences of (left unital) modules and Q_1 and Q_2 are injective, show that Q_1 \oplus L_2 \cong_R Q_2 \oplus L_1.

(Side note: two modules A and B are called projectively equivalent if there exist projective modules P_1 and P_2 such that A \oplus P_1 \cong_R B \oplus P_2. Likewise, A and B are injectively equivalent if there are injective modules Q_1 and Q_2 such that A \oplus Q_1 \cong_R B \oplus Q_2.)


  1. Let X denote the pullback of \beta_1 and \beta_2. (See this previous exercise about pullbacks.) We claim that the pullback morphisms \lambda_1 and \lambda_2 are surjective. To see this, suppose a \in P_1 and consider \beta_1(a). Since \beta_2 is surjective, there exists b \in P_2 such that \beta_2(b) = \beta_1(a). Thus (a,b) \in X, and so \lambda_1(a,b) = a. Thus \lambda_1 is surjective; a similar argument shows that \lambda_2 is surjective. Thus 0 \rightarrow \mathsf{ker}\ \lambda_1 \stackrel{\iota}{\rightarrow} X \stackrel{\lambda_1}{\rightarrow} P_1 \rightarrow 0 and 0 \rightarrow \mathsf{ker}\ \lambda_2 \stackrel{\iota}{\rightarrow} X \stackrel{\lambda_2}{\rightarrow} P_2 \rightarrow 0 are short exact sequences. We now claim that there exist module homomorphisms \theta_1 : \mathsf{ker}\ \lambda_2 \rightarrow K_1 and \theta_2 : \mathsf{ker}\ \lambda_1 \rightarrow K_2 such that the following diagram of modules commutes.

    A diagram

    To this end, we prove a lemma.

    Lemma 1: Let f : A \rightarrow C and g : B \rightarrow C be injective module homomorphisms. If \mathsf{im} f \subseteq \mathsf{im}\ g, then there exists a unique module homomorphism h : A \rightarrow B such that f = g \circ h. Moreover, h is injective. Proof: Let a \in A. Since \mathsf{im}\ f \subseteq \mathsf{im}\ g, there exists b \in B such that f(a) = g(b). Since g is injective, this b is unique. Define h(a) = b. Certainly h is well defined. Moreover, if h(a_1) = b_1 and h(a_2) = b_2 and r \in R, then we have f(a_1+ra_2) = f(a_1)+rf(a_2) = g(b_1) + rg(b_2) = g(b_1+rb_2), so that h(a_1+ra_2) = b_1+rb_2 = h(a_1) + rh(a_2). So h is a module homomorphism. Moreover, (g \circ h)(a) = g(h(a)) = f(a), so that f = g \circ h. To see uniqueness, if k : A \rightarrow B is a module homomorphism such that f = g \circ k, then if k(a) = b, we have g(b) = f(a), so that b = h(a). Thus h = k. Finally, if h(x) = 0, then f(x) = 0, so that x = 0. Thus h is injective. \square

    Now we claim that \lambda_1 \circ \iota : \mathsf{ker}\ \lambda_2 \rightarrow P_1 is injective. To see this, suppose (a,b) \in \mathsf{ker}\ (\lambda_1 \circ \iota). Then 0 = (\lambda_1 \circ \iota)(a,b) = \lambda_1(a,b) = a. Moreover, since (a,b) \in \mathsf{ker}\ \lambda_2, we have b = 0. Thus \mathsf{ker}\ (\lambda_1 \circ \iota) = 0, and so \lambda_1 \circ \iota is injective. Moreover, suppose a \in \mathsf{im}\ (\lambda_1 \circ \iota) with a = (\lambda_1 \circ \iota)(a,0). Then \beta_1(a) = \beta_2(0) = 0, so that a \in \mathsf{ker}\ \beta_2 = \mathsf{im}\ \alpha_1. Thus we have \mathsf{im}\ (\lambda_1 \circ \iota) \subseteq \mathsf{im}\ \alpha_1. By Lemma 1, there exists a unique module homomorphism \theta_1 : \mathsf{ker}\ \lambda_2 \rightarrow K_1 such that \lambda_1 \circ \iota = \alpha_1 \circ \theta_1. By a similar argument, \theta_2 : \mathsf{ker}\ \lambda_1 \rightarrow K_2 exists such that \lambda_2 \circ \iota = \alpha_2 \circ \theta_2. Moreover, \theta_1 and \theta_2 are injective.

    We claim now that \theta_1 and \theta_2 are surjective. To see this, let k \in K_1. Now k \in \mathsf{ker}\ \beta_1, so that \beta_1(k) = 0 = \beta_2(0). Now (k,0) \in X, and in fact (k,0) \in \mathsf{ker}\ \lambda_2. Clearly \theta_1(k,0) = k, so that \theta_1 is surjective. A similar argument shows that \theta_2 is surjective. In fact, we have \mathsf{ker}\ \lambda_1 \cong_R K_2 and \mathsf{ker}\ \lambda_2 \cong_R K_1. Because P_1 and P_2 are projective, the sequences centered at X are split. Thus we have K_1 \oplus P_2 \cong_R K_2 \oplus P_1.

  2. Let X denote the pushout of \alpha_1 and \alpha_2. (See this previous exercise about pushouts.) We claim that the pushout morphisms \sigma_1 and \sigma_2 are injective. To see this, suppose \sigma_1(x) = 0. Then (x,0) = (\alpha_1(m),-\alpha_2(m) for some m \in M. Since \alpha_2 is injective, we have m = 0, so that x = 0. Thus \sigma_1 is injective. Similarly, \sigma_2 is injective. This yields the two short exact sequences 0 \rightarrow Q_1 \stackrel{\sigma_1}{\rightarrow} X \stackrel{\pi}{\rightarrow} X/\mathsf{im}\ \sigma_1 \rightarrow 0 and 0 \rightarrow Q_2 \stackrel{\sigma_2}{\rightarrow} X \stackrel{\pi}{\rightarrow} X/\mathsf{im}\ \sigma_2 \rightarrow 0. We claim that there exist module homomorphisms \theta_1 : L_1 \rightarrow X/\mathsf{im}\ \sigma_2 and \theta_2 : L_2 \rightarrow X/\mathsf{im}\ \sigma_1 such that the following diagram of modules commutes.

    Another diagram

    To this end, we prove a lemma.

    Lemma 2: If f : A \rightarrow B and g : A \rightarrow C are surjective module homomorphisms and \mathsf{ker}\ f \subseteq \mathsf{ker}\ g, then there exists a unique module homomorphism h : B \rightarrow C such that g = h \circ f. Proof: Define h as follows: given b \in B, h(b) = g(a) where f(a) = b (a exists since f is surjective). Note that if f(a_1) = f(a_2) = b, then a_1 - a_2 \in \mathsf{ker}\ f \subseteq \mathsf{ker}\ g, so that g(a_1) = g(a_2). Thus h is well defined. Moreover, if b_1,b_2 \in B and r \in R, and f(a_1) = b_1 and f(a_2) = b_2, then f(a_1+ra_2) = b_1+rb_2. Thus h(b_1 + rb_2) = g(a_1+ra_2) = g(a_1) + rg(a_2) = h(b_1) + rh(b_2), so that h is a module homomorphism. That g = h \circ f is clear. To see uniqueness, suppose k : B \rightarrow C is a homomorphism such that g = k \circ f. Given b \in B, there exists a \in A such that b = f(a). Now k(b) = (k \circ f)(a) = g(a) = (h \circ f)(a) = h(b), so that k = h. Finally, to see that h is surjective, note that if c \in C then since g is surjective there exists a \in A with g(a) = c. Let b = f(a); then h(b) = c. \square

    Now we claim that \pi \circ \sigma_1 : Q_1 \rightarrow X/\mathsf{im}\ \sigma_2 is surjective. To see this, let ((a,b)+Y) + \mathsf{im}\ \sigma_2 \in X/\mathsf{im}\ \sigma_2. (Recall that X = (Q_1 \oplus Q_2)/Y, where Y = \{(\alpha_1(m),-\alpha_2(m)) \ |\ m \in M\}.) Note that ((a,b)+Y)+\mathsf{im}\ \sigma_2 = ((a,0)+Y)+\mathsf{im}\ \sigma_2 + ((0,b)+Y)+\mathsf{im}\ \sigma_2 = ((a,0)+Y)+\mathsf{im}\ \sigma_2 = (\pi \circ \sigma_1)(a). So \pi \circ \sigma_1 is surjective; likewise \pi \circ \sigma_2 : Q_2 \rightarrow X/\mathsf{im}\ \sigma_1 is surjective. By Lemma 2, there exist unique surjective module homomorphisms \theta_1 and \theta_2 making the diagram commute.

    Finally, we claim that \theta_1 and \theta_2 are injective. To see this, let x \in \mathsf{ker}\ \theta_1. Since \beta_1 is surjective, there exists an element y \in Q_1 such that \beta_1(y) = x. Now 0 = (\theta_1 \circ \beta_1)(y) = (\pi \circ \sigma_1)(y), so that \pi(\sigma_1(y)) = 0. Thus \sigma_1(y) \in \mathsf{ker}\ \pi = \mathsf{im}\ \sigma_2. That is, \sigma_1(y) = \sigma_2(z) for some z \in Q_2. Now \sigma_1(y) - \sigma_2(z) = (y,-z)+Y = 0 in X, so that (y,-z) = (\alpha_1(m),-\alpha_2(m)) for some m \in M. In particular, y \in \mathsf{im}\ \alpha_1, so that x = \beta_1(y) = 0. Thus \mathsf{ker}\ \theta_1 = 0, and so \theta_1 is injective. Similarly, \theta_2 is injective. So L_1 \cong_R X/\mathsf{im}\ \sigma_2 and L_2 \cong_R X/\mathsf{im}\ \sigma_1. Since Q_1 and Q_2 are injective, the middle row and column split, and we have L_1 \oplus Q_2 \cong_R L_2 \oplus Q_1.

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Comments

  • fanny  On June 3, 2011 at 6:35 pm

    Why you take X/ker(sigma1)? with sigma1 is inyective, so that has no sence, while ker(sigma1)=0, you are saying that the secuence 0 -> Q -> X -> X -> 0 is exact, wich is not true.

    Thanks

    • nbloomf  On June 3, 2011 at 9:05 pm

      You’re right- I think it should be X/\mathsf{im}\ \sigma. These proofs are dual to each other, so I wrote the first and then edited it to work for the second. Apparently I forgot to dualize X/\mathsf{ker}\ \sigma.

      I’ll mark this as incomplete until I have a chance to properly fix it.

      Thanks!

      • nbloomf  On June 7, 2011 at 1:28 pm

        Okay- it should be fixed now. Let me know if there are other errors.

        Thanks!

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