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:
$\varphi :A\to B$ be a ring homomorphism. Let $X=Spec(A)$ and $Y=Spec(B)$. If $q\in Y$ , then ${\varphi }^{-1}(q)={\varphi }^{\ast }(q)$ is a prime ideal of A. The map ${\varphi }^{\ast }:Y\to X$ 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 ${\varphi }^{\ast -1}(x_f)=Y_{\varphi (f)}$ and hence that ${\varphi }^{\ast }$ is continuous.

Solution
$p\in {\varphi }^{\ast -1}(X_f)\iff {\varphi }^{\ast }(p)={\varphi }^{-1}(p)\in X_f\iff f\notin {\varphi }^{-1}(p)\iff \varphi (f)\notin p\iff p\in Y_{\varphi (p)}$

2

ii) If $a$ is an ideal of A, then ${\varphi }^{\ast -1}(V(a))=v(a^c)$

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

3

iii) If $b$ is an ideal of B, then $\overline{{\varphi }^{\ast }(V(b))}=V(b^c)$.

Solution

Let $p{◃}_{p}rA$ Then $p\in {\varphi }^{\ast }(V(b))\iff p\supset {\varphi }^{-1}(b)\iff p\in V(b^c)$ and ${\varphi }^{\ast }(V(b))=V(b^c)$ Meanwhile, $p\in V(b^c)\iff p\supset {\varphi }^{-1}(b).$ and take a set $\overline{{\varphi }^{/ast}(V(b))}=V({\varphi }^{ast}(b))$ and $p\in V({\varphi }^{\ast }(b))=V(b^c)$. So $\overline{{\varphi }^{\ast }(V(b))}=V(b^c).$ is a prime ideal of A satisfying ${\varphi }^{/}\ker (\varphi )=q$

4

iv) If $\varphi$ is surjective, then $\varphi *$ is a homeomorphism of Y onto the closed subset $V(Ker(\varphi ))$ of X. In particular, Spec(A) and $\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}\mathrm{(}𝐴/\mathrm{\Re }\mathit{)}$

Solution

Suppose $\varphi$ is surjective, then by lattice isomorphism theorem, there is a bijective relation between ideal p of A containing $\ker \varphi$ and the ideal $\varphi (p)$ of $B=A/\ker (\varphi )$. Especially, take $q{◃}_{\mathrm{p}\mathrm{r}}𝐵$. Then $\varphi *(q){⊲}_{\mathrm{p}\mathrm{r}}𝐴$ satisfies $\varphi (q{)}^{\ast }/\ker (\varphi )\cong q$. Also, ${\varphi }^{\ast }$ is a homeomorphism because for any ideal $I◃B$, ${\varphi }^{\ast }(I)$ is also an ideal and there is a one-two-one correspondence between $V(\text{p})$ and $V({\varphi }^{-1}(\text{p})$ for any $p\supset I$. That is, ${\varphi }^{\ast }$ sends V(I) to $V({\varphi }^{-1}(I))$. Thus, ${\varphi }^{\ast }$ is a homeomorphism between $\mathrm{S}\mathrm{p}\mathrm{e}\mathrm{c}\mathrm{(}𝐵\mathit{)}$ and $V(\ker (\varphi ))$.

5

v) If φ is injective, then ${\varphi }^{\ast }(Y)$ is dense in X. More precisely, ${\varphi }^{\ast }(Y)$ is dense in $X\iff \ker (\varphi )\subseteq \mathrm{N}\mathrm{i}\mathrm{l}\mathrm{(}𝐴\mathit{)}$

Solution
Suppose ${\varphi }^{\ast }(Y)$ is dense, then the set

References

• M.F. Atiyah and I.G.MacDonald, 《Introduction to Commutative Algebra》, Westview Press, 1969. ISBN 0-201-00361-9 [ICA]