작은숲:공책/가환대수/Zariski continous function between two rings

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This is about the explanation of Zariski continuity of Algebraic function $f:R\to S$

Ring homomorphism

Referred to [ICA] Q.21 of section 1:
$\phi :A\to B$ be a ring homomorphism. Let $X=Spec(A)$ and $Y=Spec(B)$ . If $q\in Y$ , then ${\phi }^{-1}(q)=\phi ^{\ast }(q)$ is a prime ideal of A. The map ${\phi }^{\ast }:Y\to X$ satisfies the property :

Define $X_{f}\in Spec(A)$ be an open set $X_{f}=\{x\in Spec(A)|f\notin x\}$

1

i) If

f\in A

then

\phi ^{\ast -1}(x_{f})=Y_{\phi (f)}

and hence that

\phi ^{\ast }

is continuous.

Solution
$p\in \phi ^{\ast -1}(X_{f})\Leftrightarrow \phi ^{\ast }(p)\ {=}\phi ^{-1}(p)\in X_{f}\Leftrightarrow f\notin \phi ^{-1}(p)\Leftrightarrow \phi (f)\notin p\Leftrightarrow p\in Y_{\phi (p)}$

2

ii) If

a

is an ideal of A, then

\phi ^{\ast -1}(V(a))=v({a}^{c})

Solution
$p\in \phi ^{\ast -1}(V(a))\Leftrightarrow \phi ^{\ast }(p)=\phi ^{-1}(p)\in V(a)\Leftrightarrow \phi ^{-1}(p)\supset a\Leftrightarrow p\supset B\phi (a)={a}^{e}\Leftrightarrow p\in V({a}^{e})$

3

iii) If

b

is an ideal of B, then

{\bar {\phi ^{\ast }(V(b))}}=V(b^{c})

.

Solution

Let  $p\triangleleft _{p}rA$  Then  $p\in \phi ^{\ast }(V(b))\Leftrightarrow p\supset \phi ^{-1}(b)\Leftrightarrow p\in V(b^{c})$  and  $\phi ^{\ast }(V(b))=V(b^{c})$

Meanwhile, $p\in V(b^{c})\Leftrightarrow p\supset \phi ^{-1}(b).$ and take a set ${\bar {\phi ^{/ast}(V(b))}}{=}V(\phi ^{ast}(b))$ and $p\in V(\phi ^{\ast }(b))=V(b^{c})$ . So ${\bar {\phi ^{\ast }(V(b))}}{=}V(b^{c}).$ is a prime ideal of A satisfying $\phi ^{/}\ker(\phi )=q$

4

iv) If

\phi

is surjective, then

\phi *

is a homeomorphism of Y onto the closed subset

V(Ker(\phi ))

of X. In particular, Spec(A) and

{\rm {{Spec}({\it {{A/\Re })}}}}

Solution

Suppose  $\phi$  is surjective, then by lattice isomorphism theorem, there is a bijective relation between ideal p of A containing  $\ker \phi$  and the ideal  $\phi (p)$  of  $B=A/\ker(\phi )$ . Especially, take  $q{\triangleleft }_{\rm {pr}}{\it {B}}$ . Then  $\phi *(q){\vartriangleleft }_{\rm {pr}}{\it {A}}$  satisfies  $\phi (q)^{\ast }/\ker(\phi )\cong q$ . Also,  $\phi ^{\ast }$  is a homeomorphism because for any ideal  $I\triangleleft B$ ,  ${\phi }^{\ast }(I)$  is also an ideal and there is a one-two-one correspondence between  $V({\text{p}})$  and  $V(\phi ^{-1}({\text{p}})$  for any  $p\supset I$ . That is,  $\phi ^{\ast }$  sends V(I) to  $V(\phi ^{-1}(I))$ . Thus,  $\phi ^{\ast }$  is a homeomorphism between  ${\rm {{Spec}({\it {{B})}}}}$  and  $V(\ker(\phi ))$ .