Exercise 3.9

Proof of first claim. We first show $S_0$ is saturated. Suppose $\mathit{xy}\in S_0$ but $x\notin S_0\vee y\notin S_0$. Then, there is $z\ne 0$ such that $\mathit{xz}=0\vee \mathit{yz}=0$, and so $\mathit{xyz}=0$, so then $\mathit{xy}\notin S_0$. Now if $\mathit{xy}\notin S_0$, then there exists $z\ne 0$ such that $\mathit{xyz}=0$. If $\mathit{yz}=0$ then $y\notin S_0$, and if $\mathit{yz}\ne 0$ then $x\notin S_0$. $D=A\setminus S_0$ is then a union of prime ideals by AM Exercise $3.7i)$.

We now show that every minimal prime ideal of $A$ is contained in $D$. So suppose not; then, there exists prime and minimal $𝔭$ which contains some $x$ which is not a zero-divisor. Consider $A\setminus 𝔭$, which is a multiplicatively closed subset of $A$. Then, $\{x^{i}y\mid y\in A\setminus 𝔭,i\in \mathbb{N}\}⊋A\setminus 𝔭$, and so we see that $A\setminus 𝔭$ is not maximal, which implies that $𝔭$ is not a minimal prime ideal of $A$ by AM Exercise $3.6$, a contradiction. □

Proof of $i)$. Let $\phi :A\to S_0^{-1}A$, and $S$ be the largest multiplicatively closed subset of $A$ for which $\phi$ is injective. Suppose $a∕1=0∕1$ in $S_0^{-1}A$. Then, $\mathit{sa}=0$ for some $s\in S_0$. But $s\in S_0⟹a=0$, and so $\phi$ is injective. Thus, $S_0\subseteq S$.

Now we show the reverse inclusion by assuming $\phi$ is injective. $a∕1=0∕1$ in $S^{-1}A$ implies that $\mathit{sa}=0$ for some $s\in S$. But then, $a=0$ by the injectivity of $\phi$, and so $s$ is a non-zero-divisor. Thus, $S\subseteq S_0$. □

Proof of $\mathit{ii})$. $a∕s\in S_0^{-1}A⟹a\in S_0\vee a\notin S_0$. $a\in S_0⟹{(a∕s)}^{-1}=s∕a\in S_0^{-1}A$, i.e., $a∕s$ is a unit. $a\notin S_0⟹\exists <!--\; FUNCTION\; APPLICATION\; -->x\in A$ such that $\mathit{ax}=0⟹(a∕s)(x∕1)=0$, i.e., $a∕s$ is a zero-divisor. □

Proof of $\mathit{iii})$. $\phi :A\hookrightarrow S_0^{-1}A$, and so it suffices to show $\phi$ is surjective. $S_0^{-1}A\ni a∕x=a{x}^{-1}∕x{x}^{-1}=a{x}^{-1}∕1=\phi (a{x}^{-1})$, where $x^{-1}$ exists since $S_0$ consists of units. □