Lemma 10.108.1 (Schanuel's lemma). Let $R$ be a ring. Let $M$ be an $R$-module. Suppose that

\[ 0 \to K \xrightarrow {c_1} P_1 \xrightarrow {p_1} M \to 0 \quad \text{and}\quad 0 \to L \xrightarrow {c_2} P_2 \xrightarrow {p_2} M \to 0 \]

are two short exact sequences, with $P_ i$ projective. Then $K \oplus P_2 \cong L \oplus P_1$. More precisely, there exist a commutative diagram

\[ \xymatrix{ 0 \ar[r] & K \oplus P_2 \ar[r]_{(c_1, \text{id})} \ar[d] & P_1 \oplus P_2 \ar[r]_{(p_1, 0)} \ar[d] & M \ar[r] \ar@{=}[d] & 0 \\ 0 \ar[r] & P_1 \oplus L \ar[r]^{(\text{id}, c_2)} & P_1 \oplus P_2 \ar[r]^{(0, p_2)} & M \ar[r] & 0 } \]

whose vertical arrows are isomorphisms.

**Proof.**
Consider the module $N$ defined by the short exact sequence $0 \to N \to P_1 \oplus P_2 \to M \to 0$, where the last map is the sum of the two maps $P_ i \to M$. It is easy to see that the projection $N \to P_1$ is surjective with kernel $L$, and that $N \to P_2$ is surjective with kernel $K$. Since $P_ i$ are projective we have $N \cong K \oplus P_2 \cong L \oplus P_1$. This proves the first statement.

To prove the second statement (and to reprove the first), choose $a : P_1 \to P_2$ and $b : P_2 \to P_1$ such that $p_1 = p_2 \circ a$ and $p_2 = p_1 \circ b$. This is possible because $P_1$ and $P_2$ are projective. Then we get a commutative diagram

\[ \xymatrix{ 0 \ar[r] & K \oplus P_2 \ar[r]_{(c_1, \text{id})} & P_1 \oplus P_2 \ar[r]_{(p_1, 0)} & M \ar[r] & 0 \\ 0 \ar[r] & N \ar[r] \ar[d] \ar[u] & P_1 \oplus P_2 \ar[r]_{(p_1, p_2)} \ar[d]_ S \ar[u]^ T & M \ar[r] \ar@{=}[d] \ar@{=}[u] & 0 \\ 0 \ar[r] & P_1 \oplus L \ar[r]^{(\text{id}, c_2)} & P_1 \oplus P_2 \ar[r]^{(0, p_2)} & M \ar[r] & 0 } \]

with $T$ and $S$ given by the matrices

\[ S = \left( \begin{matrix} \text{id}
& 0
\\ a
& \text{id}
\end{matrix} \right) \quad \text{and}\quad T = \left( \begin{matrix} \text{id}
& b
\\ 0
& \text{id}
\end{matrix} \right) \]

Then $S$, $T$ and the maps $N \to P_1 \oplus L$ and $N \to K \oplus P_2$ are isomorphisms as desired.
$\square$

## Comments (3)

Comment #4019 by MatthÃ© van der Lee on

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