Let $i : A \to A_S$ and $j : A \to A_T$ be the localization homomorphisms. === 1 implies 2 === Let $h : A_S \to A_T$ be an $A$-algebra homomorphism. Then by definition, $j = h \circ i$: :$\xymatrix{ A \ar[d]_i \ar[r]^{j} & A_T\\ A_S \ar[ru]_{h} }$ Let $s \in S$. By definition of localization, $i(s)$ is a unit of $A_S$. By Ring Homomorphism Preserves Invertible Elements, $j(s) = h(i(s))$ is a unit of $A_T$. Thus $s$ is an element of the saturation of $T$. {{qed|lemma}} === 2 implies 1 === Let $S$ be a subset of the saturation of $T$. Then its image $j(S) \subseteq A_T^\times$ consists of units of $A_T$. By definition of localization at $S$, there exists a unique $A$-algebra homomorphism $h : A_S \to A_T$. {{qed|lemma}} === 2 implies 3 === Let $S$ be a subset of the saturation of $T$. By definition, its saturation is the smallest saturated multiplicatively closed subset of $A$ containing $S$. Thus the saturation of $S$ is a subset of the saturation of $T$. {{qed|lemma}} === 3 implies 2 === By definition, $S$ is a subset of its saturation. {{qed|lemma}} === 3 iff 4 === By definition, the saturation of $S$ is the complement of the union of prime ideals that are disjoint from $S$: :$\operatorname{Sat}(S) = A - \displaystyle \bigcup \left\{ \mathfrak p \in \operatorname{Spec} A : \mathfrak p \cap S = \varnothing \right\}$ Thus: {{begin-eqn}} {{eqn | l = \operatorname{Sat}(S) \subseteq \operatorname{Sat}(T)       | r = A - \bigcup \left\{ \mathfrak p \in \operatorname{Spec} A : \mathfrak p \cap S = \varnothing \right\} \subseteq A -  \bigcup \left\{ \mathfrak p \in \operatorname{Spec} A : \mathfrak p \cap T = \varnothing \right\}       | o = \iff }} {{eqn | r = \bigcup \left\{ \mathfrak p \in \operatorname{Spec} A : \mathfrak p \cap S = \varnothing \right\} \supseteq \bigcup \left\{ \mathfrak p \in \operatorname{Spec} A : \mathfrak p \cap T = \varnothing \right\}       | o = \iff       | c = Subset iff Complement is Superset of Complement }} {{eqn | r = \forall \mathfrak p \in \operatorname{Spec} A : \left( \mathfrak p \cap T = \varnothing \implies \mathfrak p \subseteq \bigcup \left\{ \mathfrak q \in \operatorname{Spec} A : \mathfrak q \cap S = \varnothing \right\} \right)       | o = \iff       | c = Sets are Subset iff Union is Subset }} {{end-eqn}} To finish, we show that the last statement is equivalent to: :$\forall \mathfrak p \in \operatorname{Spec} A : \mathfrak p \cap T = \varnothing \implies \mathfrak p \cap S = \varnothing$ We show that, for $\mathfrak p \in \operatorname{Spec} A$: :$\mathfrak p \subseteq \displaystyle \bigcup \left\{ \mathfrak q \in \operatorname{Spec} A : \mathfrak q \cap S = \varnothing \right\} \iff \mathfrak p \cap S = \varnothing$ Let $\mathfrak p \in \operatorname{Spec} A$. If $\mathfrak p \cap S = \varnothing$, then by Set is Subset of Union: :$\mathfrak p \subseteq \displaystyle \bigcup \left\{ \mathfrak q \in \operatorname{Spec} A : \mathfrak q \cap S = \varnothing \right\}$ Conversely, let $\mathfrak p \subseteq \displaystyle \bigcup \left\{ \mathfrak q \in \operatorname{Spec} A : \mathfrak q \cap S = \varnothing \right\}$. By: :Union of Sets Disjoint with Set :Subset of Disjoint Set we have $\mathfrak p \cap S = \varnothing$. We conclude that $ \operatorname{Sat}(S) \subseteq \operatorname{Sat}(T)$ {{iff}} :$\forall \mathfrak p \in \operatorname{Spec} A : \mathfrak p \cap T = \varnothing \implies \mathfrak p \cap S = \varnothing$ That is: :$\forall \mathfrak p \in \operatorname{Spec} A : \mathfrak p \cap S \neq \varnothing \implies \mathfrak p \cap T \neq \varnothing$ {{qed}}