Total quotient ring and its properties

Question

I am not really good in solving problems in localization and I am badly struck on this. I have been self studying algebraic geometry.

(Total Quotient Ring) Let $A\neq 0$ be a commutative ring and $S_0 = Nzd(A) = A\setminus Z(A)$ be the set of all non-zero divisors in $A$. Prove that $S_0$ is a multiplicatively closed subset in $A$. The ring of fractions is $S_0^{-1}A$ is called the total quotient ring of $A$ and is usually denoted by $Q(A)$. The natural ring hom. $i_{S_0} : A\to Q(A)$ is injective and hence $A$ can be identified with a subring of its total quotient ring. In particular, if $A$ is an integral domain, then $Q(A)$ is the field of fractions of $A$.

I have proved $S_0$ to be multiplicatively closed.

(a) Show that $S_0$ is the largest multiplicatevely closed subset of $A$ for which the hom. $i_{S_0}: A \to S_{0}^{-1} A$ is injective.

Work: I assumed that there exists a set $S'$ such that $S\subset S'$ and hom is injective. But what result to use to get a contradiction?

(b) Every element in $Q(A)$ is either a zerodivisor or a unit.

Work: $Q(A) = \frac{A} {Nzd(A)}$. I assumed that let $y \in Q(A)$ which is neither a non zero divisor and non-unit. Again, how to get a contradiction?

(c) Every non-zero ring of fractions $S^{-1} A$ of an integral domain is canonically isomorphic to a subring of the quotient field $Q(A)$ of $A$.

I have no idea how to approach this part.

I am really sorry that I didn't had much work to show.

Kindly guide.

Answer 1

For (a), assume that $A \to S^{-1}A$ is injective. We will show that if $s \in S$, then $s$ is not a zerodivisor. In fact, if there is some $a \in A$ such that $as=0$, then $a/1=0/1$ in $S^{-1}A$, and so $a$ is in the kernel of $A \to S^{-1}A$. It follows that $a =0$ by injectivity, hence $s \in S$ is not a zerodivisor.

For (b), let $p/q \in S^{-1}A$, with $q \in S$ and $p \in A$.We have two options, depending on $p \in S$.

• If $p \notin S$, then $pr=0$ for some $r \neq 0$ (as it will be a zerodivisor), hence $r/1 \neq 0/1$ in $S^{-1}A$ (as $A \to S^{-1}A$ is injective). At last, $p/q \cdot r/1$ is zero. This shows that $p/q$ is a zerodivisor.
• If $p \in S$, then $q/p \in S^{-1}A$ as well, and clearly $p/q \cdot q/p=1/1$, i.e. $p/q$ is a unit.

At last, for (c), let $S \subset A$ be a multiplicative subset which does not contain zero (i.e. $S^{-1}A$ is not the zero ring), and consider the injection $A \to Q(A)$. Note that for every $s \in S$, the image of $s$ in $Q(A)$ is a unit, and so the universal property of localization gives us a map $S^{-1}A \to Q(A)$. The kernel of this map is given by $a/s \in S^{-1}A$ which are zero in $Q(A)$. But $a/s=0/1$ in $Q(A)$ implies that $ab=0$ for some $b \neq 0$, and so $a=0$ (as $A$ is an integral domain). Therefore, the map $S^{-1}A \to Q(A)$ is injective.