## 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 $L_1 \oplus Q_2 \cong_R L_2 \oplus Q_1$.

• 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!