# Exercise 6.3 from Miles Reid's Commutative Algebra

I'm trying to do all the exercises from Chapter 6 of M. Reid CA, but atm I'm having difficulties understanding what's going on with localizations and everything. I would appreciate some help please!

The problems says:

a) Let $A = A' \times A''$; prove that $A'$ and $A''$ are rings of fractions of $A$.

b) If $A'$ and $A''$ are integral domains and $A \subset A' \times A''$ is a subring that maps onto each factor, then what are the necessary and sufficient conditions that a multiplicative set $S$ of $A$ must satisfy in order for $S^{-1}A$ to be a ring of fractions of $A'$?

The hint they give is to look at $k[X,Y]/(XY) \subset K[X] \times K[Y]$, but I don't really get it... Thanks.

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 Surely there is some hypothesis on $A$ for part a. – Michael Albanese Oct 10 '12 at 9:05

a) Consider $S=\{ (1, 0), (1, 1) \} \subset A$ and compute $S^{-1}A$.
b) What you need is (1) the image of $S$ in $A'$ doesn't contain zero (I guess in his definition, a multiplicative set never contains $0$, otherwiser forget about (1)) and (2) that $S^{-1}A\to {S'}^{-1}A'$, where $S'$ is the image of $S$ in $A'$, is injective (it is already surjective). Now try to make explicit these conditions on $S$.
b') Let $a=(a', a'')\in A$ and $s=(s', s'')\in S$. Then the image of $a/s$ in $S'^{-1}A'$ is $a'/s'$. It is zero if and only if $a'=0$ because $A'$ is a domain. For the injectivity, we then must require that $(0, a'')/s=0$ for all $a''\in A''$. So there must be a $t=(t',t'')\in S$ such that $t.(0, a'')=0$. So an element of the form $t''=0$. The conclusion for b) is then $S$ contains an element of the form $(t', 0)$ (check that this condition is sufficient for the injectivity).
 Thank you! I got everything except I'm not sure however how to translate (2) into something about S... – Dquik Oct 4 '12 at 1:03 @Dquik: Ok I now put a complete answer. – QiL'8 Oct 4 '12 at 7:01 The definition of multiplicative systems requires the unity of the ring to be there. This is not the case for answer a) above. I suggest you to consider $S=\{1\}\times A''$ and try to prove that $S^{-1}A\cong A'$. – YACP Oct 4 '12 at 18:59 @navigetor23: you are right, thanks ! – QiL'8 Oct 4 '12 at 19:26