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In Atiyah and Macdonald, Chapter 5, Exercise 10, there defines the so called "going-down property" (GDP). Then in Chapter 7, Exercise 24, the hint says, the ring map $f: A\rightarrow B$ has GDP implies, for prime ideals $a\subset b$, if there is a prime ideal $b'$ of $B$ such that $b=f^{-1}(b')$, then there is a prime ideal $a'$ of $B$ such that $a=f^{-1}(a')$. Of course this is true iff $\ker (f) \subset a$. Then if $\ker (f) \notin a$, then what can we say by using GDP? My goal is to remedy the hint of Chapter 7, Exercise 24.

Here is a related problem (it's in fact part of Chapter 5, Exercise 10 in Atiyah and Macdonald), and I know this problem is general not true.

Now I have known the right form of GDP: this is a related problem that I think is the right notion of GDP, which I call the "strong going-down property" (see the last comment by me there, and one can easily show that strong going-down property implies GDP).

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  • $\begingroup$ Why is the going down property of $f:A \rightarrow B$ true if and only if $kerf \subset a$? $\endgroup$ – Manos Feb 13 '14 at 1:30
  • $\begingroup$ Since the kernel is the inverse image of 0. I didn't say the going down property of f:A→B true if and only if kerf⊂a, I only said "for $a$"...The other direction is true by thinking about quotient ring. $\endgroup$ – Lao-tzu Feb 13 '14 at 1:30
  • $\begingroup$ I think you should formulate your question a little bit more clearly, i still don't understand what it is you are asking. $\endgroup$ – Manos Feb 13 '14 at 2:03
  • $\begingroup$ @Lao-tzu I agree with Manos. Can you rewrite? Just citing problems in A&M is bad form--we don't know what you're talking about. $\endgroup$ – Alex Youcis Feb 14 '14 at 7:11
  • $\begingroup$ @Alex Youcis Oh, I think there is no need, [this][1] is a related problem that I think is the right notion of GDP, which I call the "strong going-down property". [1]: math.stackexchange.com/questions/674574/… $\endgroup$ – Lao-tzu Feb 14 '14 at 7:16

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