The Stacks project

Proof. Choose a surjection $p : F \to P$ where $F$ is a free $R$-module. Since $P/IP$ is a projective $R/I$-module, we can choose a map $f : P/IP \to F/IF$ which is a right inverse to $p \bmod I$. Consider the map $q : F/JF \to F/(I + J)F \times _{P/(I + J)P} P/JP$ induced by the quotient map $F/JF \to F/(I + J)F$ and $p$. Note that $q$ is a surjective map of $R/J$-modules (small detail omitted). Consider the map $f' : P/JP \to F/(I + J)F \times _{P/(I + J)P} P/JP$ induced by $f$ and the identity map. Since $P/JP$ is a projective $R/J$-module and $q$ surjective, we can choose a map $g : P/JP \to F/JF$ such that $q \circ g = f'$. Then $f \bmod I + J = g \bmod I + J$. Since $F = F/JF \times _{F/(I + J)F} F/IF$ because $I \cap J = 0$, we conclude that we obtain a well defined homomorphism $h = (f, g) : P \to F$ of $R$-modules. The map $a = p \circ h : P \to P$ reduces to the identity modulo $I$ and modulo $J$. Then $a$ also induces the identity map on the submodule $IP$: if $x = \sum i_\alpha x_\alpha $ with $i_\alpha \in I$ and $x_\alpha \in P$, then $a(x) = \sum i_\alpha a(x_\alpha ) = \sum i_\alpha x_\alpha = x$ because $a(x_\alpha ) = x_\alpha + j_\alpha $ with $j_\alpha \in J$ and $i_\alpha j_\alpha = 0$. Thus $a : P \to P$ acts as the identity on $IP$ and on $P/IP$ and we conclude that $a$ is an automorphism of $P$ (small detail omitted). Thus $P$ is a summand of $F$, whence projective. $\square$