작은숲:공책/가환대수/Zariski continous function between two rings 문서 원본 보기

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{{퍼옴|[https://studynotekr.miraheze.org/wiki/Study_Note:Math-CA/Zariski_continous_function_between_two_rings Korean Study Note]}}
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This is about the explanation of Zariski continuity of Algebraic function <math>f:R \to S </math>

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

Define <math>X_f \in Spec(A) </math> be an open set <math>X_f = \{ x \in Spec(A) | f \notin x \} </math>
=== 1 ===
: '''i)''' If <math>f \in A </math> then <math> \phi^{\ast -1}(x_f) = Y_{\phi(f)}</math> and hence that <math>\phi^{\ast}</math> is continuous.
{{숨기기|Solution|
<math> 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)}  </math> }}
=== 2 ===
: '''ii)''' If <math> a </math> is an ideal of A, then <math>\phi^{\ast -1} (V(a)) = v({a}^{c})</math>
{{숨기기|Solution|
<math> 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}) </math> }}
=== 3 ===
: ''' iii)''' If <math>b</math> is an ideal of B, then <math>\bar{\phi^{\ast}(V(b))} =V( b^c )</math>.
{{숨기기|Solution| Let <math>p \triangleleft_pr A</math> Then <math> p \in \phi^{\ast} (V(b)) \Leftrightarrow p \supset \phi^{-1}(b) \Leftrightarrow p \in V( b^c ) </math> and <math> \phi^{\ast} (V(b)) = V( b^c ) </math> <br>
Meanwhile, <math>p \in V(b^c ) \Leftrightarrow p \supset \phi^{-1} (b).</math> and take a set <math>\bar{\phi^{/ast} (V(b))} {{=}} V(\phi^{ast}(b))</math> and <math>p \in V(\phi^{\ast}(b)) =V(b^c )</math>. So <math>\bar{\phi^{\ast} (V(b))} {{=}} V(b^c ).</math> is a prime ideal of ''A'' satisfying <math>\phi^/\ker (\phi)=q </math> }}
=== 4 ===
: ''' iv) ''' If <math> \phi</math> is surjective, then <math>\phi*</math> is a homeomorphism of Y onto the closed subset <math>V(Ker (\phi))</math> of ''X''. In particular, Spec(''A'') and <math>\rm{Spec} (\it{A/ \Re})</math>
{{숨기기|Solution| Suppose <math>\phi</math> is surjective, then by lattice isomorphism theorem, there is a bijective relation between ideal ''p'' of ''A'' containing <math>\ker \phi</math> and the ideal <math>\phi(p)</math> of <math>B=A/\ker(\phi)</math>. Especially, take <math>q {\triangleleft}_{\rm{pr}} \it B</math>. Then <math>\phi*(q) {\vartriangleleft}_{\rm{pr}} \it A</math> satisfies <math>\phi(q)^{\ast}/\ker(\phi) \cong q </math>. Also, <math>\phi^{\ast}</math> is a homeomorphism because for any ideal <math> I \triangleleft B </math>, <math>{\phi}^{\ast}(I)</math> is also an ideal and there is a one-two-one correspondence between <math>V(\p)</math> and <math>V(\phi^{-1} (\p )</math> for any <math>p \supset I </math>. That is, <math>\phi^{\ast}</math> sends V(I) to <math>V(\phi^{-1} (I))</math>. Thus, <math>\phi^{\ast}</math> is a homeomorphism between <math>\rm{Spec} (\it{B})</math> and <math> V(\ker (\phi))</math>.}}
=== 5 ===
: ''' v) ''' If φ is injective, then <math>\phi^{\ast} (Y) </math> is dense in X. More precisely, <math>\phi^{\ast}(Y)</math> is dense in <math>X \Leftrightarrow \ker (\phi) \subseteq \rm{Nil}(\it{A})</math>
{{숨기기|Solution|
Suppose <math>\phi^{\ast} (Y)</math> is dense, then the set 
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== Footnotes ==
<references/>
== References ==
* M.F. Atiyah and I.G.MacDonald, 《Introduction to Commutative Algebra》, Westview Press, 1969. ISBN 0-201-00361-9 [ICA]

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