# 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).
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 $S^{-1}A\cong A'$. – user26857 Oct 4 '12 at 18:59