Lemma 90.6.4. Let $A \to B$ and $A \to C$ be ring maps. Then the map $L_{B \times C/A} \to L_{B/A} \oplus L_{C/A}$ is an isomorphism in $D(B \times C)$.

**Proof.**
Although this lemma can be deduced from the fundamental triangle we will give a direct and elementary proof of this now. Factor the ring map $A \to B \times C$ as $A \to A[x] \to B \times C$ where $x \mapsto (1, 0)$. By Lemma 90.5.8 we have a distinguished triangle

in $D(B \times C)$. Similarly we have the distinguished triangles

Thus it suffices to prove the result for $B \times C$ over $A[x]$. Note that $A[x] \to A[x, x^{-1}]$ is flat, that $(B \times C) \otimes _{A[x]} A[x, x^{-1}] = B \otimes _{A[x]} A[x, x^{-1}]$, and that $C \otimes _{A[x]} A[x, x^{-1}] = 0$. Thus by base change (Lemma 90.6.2) the map $L_{B \times C/A[x]} \to L_{B/A[x]} \oplus L_{C/A[x]}$ becomes an isomorphism after inverting $x$. In the same way one shows that the map becomes an isomorphism after inverting $x - 1$. This proves the lemma. $\square$

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