Exercise 3.8

Question

Let $S,T$ be multiplicatively closed subsets of $A$, such that $S \subseteq T$. Let $\phi : S^{-1}A 	o T^{-1}A$ be the homomorphism which maps each $a/s \in S^{-1}A$ to $a/s$ considered as an element of $T^{-1}A$. Show that the following statements are equivalent:
  \begin{enumerate}[label=oman*)]
    \item $\phi$ is bijective.
    \item For each $t \in T$, $t/1$ is a unit in $S^{-1}A$.
    \item For each $t \in T$ there exists $x \in A$ such that $xt \in S$.
    \item $T$ is contained in the saturation of $S$ (Exercise $7$).
    \item Every prime ideal which meets $T$ also meets $S$.
  \end{enumerate}

Amit Awasthi · Accepted Answer

\begin{proof}
  $i) \Rightarrow ii)$. $(t/1)(1/t) = 1$ in $T^{-1}A$. $\phi$ has an inverse $\phi^{-1}$ so $1 = \phi^{-1}(1) = \phi^{-1}(t/1)\phi^{-1}(1/t) = t/1\phi^{-1}(1/t)$ in $S^{-1}A$, i.e., $t/1$ is a unit.
  \par $ii) \Rightarrow iii)$. Let $a/b = (t/1)^{-1} \in S^{-1}A$; then, $(a/b)(t/1) = 1$, i.e., $c(at-b) = 0$ for some $c \in S$. Then, $act = bc \in S$, and $ac \in A$; thus, $x = ac$ works.
  \par $iii) \Rightarrow iv)$. $t \in T \implies \exists x \in A$ such that $xt \in S \subseteq \overline{S} \implies t \in \overline{S}$.
  \par $iv) \Rightarrow v)$. Suppose $\mathfrak{p} \cap S 
e \emptyset$; then, $\mathfrak{p} \cap \overline{S} = \emptyset$ by AM Exercise $3.7ii)$, and so $\mathfrak{p} \cap T = \emptyset$ by $iv)$. This proves the contrapositive.
  \par $v) \Rightarrow iii)$. Suppose $iii)$ does not hold. Then, $(t) \cap S = \emptyset$ for some $t \in T$; note that $t$ is not a unit otherwise $(t) = A$. We then see that $(t)^e$ is contained in some maximal ideal $\mathfrak{m}$ of $S^{-1}A$ by AM Corollary $1.4$, and so $\mathfrak{m}^c \subseteq A$ is a prime ideal that does not intersect $S$. But then, $\mathfrak{m}^c \cap S = \emptyset$ while $t \in \mathfrak{m}^c \cap T$, a contradiction.
  \par $iii) \Rightarrow ii)$. Consider $t/1 \in S^{-1}A$. Since there exists $x \in A$ such that $xt \in S$, we see that $t/1 = xt/x$, and $x/xt \in S^{-1}A$ is the inverse of $t/1$.
  \par $ii) \Rightarrow i)$. Suppose $\phi(a/s) = \phi(a^{\prime}/s^{\prime})$ in $T^{-1}A$, i.e., there exists $t \in T$ such that $t(as^{\prime}-a^{\prime}s) = 0$. Choosing $x \in A$ such that $xt \in S$, which we have since we have already shown $ii) \Rightarrow iii)$, we see that $xt(as^{\prime}-a^{\prime}s) = 0$, and so $a/s = a^{\prime}/s^{\prime}$ in $S^{-1}A$, i.e., $\phi$ is injective.
  \par Now let $t \in T$ and let $a/s \in S^{-1}A$ such that $(t/1)(a/s) = 1$, i.e., $x(at-s) = 0$ for some $x \in S$. But since $S \subseteq T$, wesee that this implies $1/t = a/s$ in $T^{-1}A$. This implies $\phi$ is surjective since all elements of the form $1/t$ are in the image of $\phi$.
\end{proof}

Exercise 3.8

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Exercise 3.8

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