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I studied AM before studying universal properties. When I solved the following exercise, I had a tedious solution that involved dealing with elements.

Let $ A $ be a ring with multiplicatively closed subsets $ S $ and $ T $. Define $ U$ to be the image of $ T$ in $S^{− 1}A $. Show that $(ST)^{−1}A$ and $U ^{−1}(S^{−1}A)$ are isomorphic rings.

I think I have a clean solution that uses universal properties. To my surprise, none of the solution manuals does this.

Let $A \rightarrow R$ be a map that sends $ST$ to units. Since this sends each element of $S$ to a unit, it factors as $A \rightarrow S^{-1}A \rightarrow R$. Again, the second map sends each element of $U$ to a unit, thus it factors as $A \rightarrow S^{-1}A \rightarrow U^{-1}(S^{-1}A) \rightarrow R$. The last map is uniquely determined by $A\rightarrow R$. I conclude that $U^{-1}(S^{-1}A)$ satisfies the universal property of $(ST)^{-1}A$.

Is this correct?

user26857
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PeterM
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1 Answers1

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It looks correct to me. A more precise statement would be that the morphism $A \rightarrow U^{-1}(S^{-1}A)$ is universal, instead of just saying that $U^{-1}(S^{-1}A)$ is universal. The underlying category is that whose objects are ring homomorphisms $f:A \rightarrow R$, where $A$ is fixed and $R$ is some ring, such that every element of $f(S)$ is a unit. A morphism from an object $f: A \rightarrow R$ to another object $g:A \rightarrow C$ is a ring homomorphism $h: R \rightarrow C$ such that $h \circ f = g$.

One reason why this solution may not appear in a solution manual, is that it is based on categorical arguments, i.e. uniqueness up to isomorphism of universal objects. On the other hand, this exercise usually appears in early stages of commutative algebra study, so that the reader is not expected to have any knowledge of category theory. As far as know, AM does not really go into categorical arguments, as e.g. Lang does.

Manos
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