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I am trying to get used to $\operatorname{Spec}$ of a ring. I know an example, when one prime ideal is contained in another for $\mathbb{C}[x,y]$. $(f) \subset (x-a,y-b)$, where $f(a,b) = 0$.

Is there, say, an example of ascending chain of prime ideals exactly $5$ terms long?

Could you give an example of an infinite ascending chain of prime ideals?

The same question for an infinite descending chain of prime ideals?

I think that I want to see an explicit example of usage of going up/down theorem, something useful is discussed here.

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  • $\begingroup$ $(0)\subset(X_1)\subset(X_1,X_2)\subset(X_1,X_2,X_3)\subset(X_1,X_2,X_3,X_4)$ has five terms. $\endgroup$
    – user26857
    Apr 17, 2019 at 13:10

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The answer can be found googling "Krull Dimension" and "Nagata example of an Krull infinitely dimensional Noetherian ring." The non-intuitive though conceptual answer is found.

Furthermore, the question about the descending chain of ideals is still to answer.

UPD: the ideal $I$ is called prime if $ab \in I$ implies $a \in I$ or $b \in I$

Lemma ideal $I = (x_1, x_2 ... x_n) $ , where $x_i$ are distinct variables of $\mathbb{C}[x_1, x_2 ... ]$ is prime.

Justification: if $f*g$ is in $I$, then it should be divisible by a polynomial of first $n$ variables. If neither $f$ nor $g$ of them are, then $fg$ is not divisible also.

That's because complex polynomial ring is a UFD

This gives us the way to construct an ascending chain of prime ideals of any length.

Lemma 2 $(x_2 ... x_n ...)$ is a prime ideal for the same reason. This gives us a descending chain of prime ideals.

The ring of polynomials with infinitely many variables is not noetherian by definition, though.

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    $\begingroup$ In $K[X_1,\dots,X_n,\dots]$ consider $P_i=(X_i,\dots)$. These form a descending chain. And consider $Q_i=(X_,\dots,X_i)$. These form an ascending chain. $\endgroup$
    – user26857
    Apr 17, 2019 at 13:09
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    $\begingroup$ And btw, Nagata's example doesn't (and can't) contain an infinite chain of ascending prime ideals. And any noetherian ring doesn't. $\endgroup$
    – user26857
    Apr 17, 2019 at 13:15
  • $\begingroup$ @user26857 damn. you are right, I will make a clear explaining soon. Feeling sorry for making poor quality answers. $\endgroup$ Apr 17, 2019 at 15:56

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