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In Atiyah-MacDonald, we have the claim that $S^{-1}(I+J) = S^{-1}I + S^{-1}J$ and similarly, $S^{-1}(IJ) = S^{-1}I S^{-1}J$. Here $I,J$ are ideals of a commutative unital ring $R$ and $S$ is a multiplicative subset of $R$.

The proof is omitted with the remark (p.42): "...For sums and products, this follows from (1.18);...."

Where (1.18) is the following exercise on page 10: enter image description here

I showed $S^{-1}(I+J) = S^{-1}I + S^{-1}J$ and $S^{-1}(IJ) = S^{-1}I S^{-1}J$ showing the usual two inclusions $\supset, \subset$ and the proof is one line.

On the other hand, I don't see where exercise 1.18 comes in here. Could someone show me the proof using exercise 1.18? Or if that's not possible, confirm that this is a typo? Thanks.

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If $\iota: R \to S^{-1}R$ is the inclusion morphism, then $S^{-1}(I+J) = (I+J)^e$ for this morphism.