Let $S/R$ be an integral extension. Consider the map $\varphi : \text {Spec} (S) \longrightarrow \text {Spec} (R)$ defined by $$\varphi (\mathfrak B) = \mathfrak B \cap R,\ \mathfrak B \in \text {Spec} (S).$$ Then $\varphi$ is a closed map.
My attempt $:$ Let $A$ be a closed subset of $\text {Spec} (S).$ So $\exists$ an ideal $J$ of $S$ such that $A= V(J),$ where $$V(J) = \{q \in \text {Spec} (S) : q \supseteq J \}.$$ Need to show that $\varphi (A)$ is a closed subset of $\text {Spec} (R).$ Let $I := J \cap R.$ Then $I$ is an ideal of $R.$ It is not very hard to see that $\varphi (A) \subseteq V(I),$ because if $q \in \text {Spec} (S)$ with $q \supseteq J$ then $q \cap R \in \text {Spec} (R)$ with $q \cap R \supseteq J \cap R = I.$ Now if we can show that $V(I) \subseteq \varphi (A)$ then we are through. For that I take $\mathfrak {p} \in V(I).$ Then $\mathfrak {p} \in \text {Spec} (S)$ with $\mathfrak {p} \supseteq I.$ Take $N:= R \setminus \mathfrak {p}.$ Then clearly $N$ is a multiplicatively closed set. Let us consider the ideal $\mathfrak {p} S$ of $S.$ Let $x \in \mathfrak {p} S.$ Then since $S$ is integral over $R$ and $\mathfrak {p} S \subseteq \text {Rad} (\mathfrak {p} S)$ we can say that $x$ is integral over $\mathfrak {p}.$ So $\exists$ $r_i \in \mathfrak {p}$ for $i=1,2 \cdots , n$ with $n \geq 1$ such that $$x^n+r_1 x^{n-1} + \cdots + r_n=0.$$ If $x \in N$ i.e. in particular if $x \in R$ then by the above equation we can conclude that $x^n \in \mathfrak {p}.$ But since $\mathfrak {p}$ is a prime ideal of $R$ it follows that $x \in \mathfrak {p},$ a contradiction to the fact that $x \in N.$ Therefore $\mathfrak {p} S \cap N = \varnothing.$ So by Krull's lemma $\exists$ $\ \mathfrak {B} \in \text {Spec} (S)$ with $ \mathfrak {B} \supseteq \mathfrak {p} S$ and $ \mathfrak {B} \cap N = \varnothing.$ But that implies $ \mathfrak {B} \cap R = \mathfrak {p}.$ So if we can show that $ \mathfrak {B} \in A$ we are through. For that we need only to show that $ \mathfrak {B} \supseteq J.$ Now since $J \cap R = I \subseteq \mathfrak {p} = \mathfrak {B} \cap R.$ So $J \cap R \subseteq \mathfrak {B} \cap R.$
From here can I conclude that $J \subseteq \mathfrak {B}$? Clearly it is not always the case. Two non-comparable sets may have the same intersection with some other set. But why is it happening for this case? I don't find any proper justification to this argument. Please help me in this regard.
Thank you very much for your valuable time.
There is a more elegant way to think that problem. First we note that $S/J$ is integral over $R/I.$ So the map $\psi : \text {Spec} (S/J) \longrightarrow \text {Spec} (R/I)$ defined by $$\psi (\mathfrak {B} / J) = (\mathfrak {B}/J) \cap (R/I),\ \mathfrak {B}/J \in \text {Spec} (S/J)$$ is surjective. Now let us take $\mathfrak {p} \in V(I).$ Then $\mathfrak {p} / I \in \text {Spec} (R).$ Since $\psi$ is surjective so $\exists$ $\mathfrak {B} \in \text {Spec} (S)$ such that $$(\mathfrak {B}/J) \cap (R/I) = \mathfrak {p}/I.$$
Claim $:$ $\mathfrak {B} \cap R = \mathfrak {p}.$
First we observe that $(\mathfrak {B}/J) \cap (R/I) = (\mathfrak {B} \cap R)/I.$ So we have $$(\mathfrak {B} \cap R) / I = \mathfrak {p} / I.$$ From here it is not hard to see that $$\mathfrak {B} \cap R = \mathfrak {p}$$ as claimed.