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In this note proposition 1.3 claimed that :

If $k$ is a field, $R$ is a finitely generated $k$-algebra which is a domain then $\dim R$ is finite and any saturated chain of prime ideals has length equal to $\dim R$.

I understand mostly the proof, however, I do not understand the following sentences:

Let $P_0\subset P_1\subset...\subset P_m$ be a saturated chain of prime ideals in $R$.

So, what is the $m$ here ? Or it is just a typo mistake, it should be $n$ ?

After proving $\dim B=n-1$, the author wrote :

By induction, we are done.

So, what did he want to prove ? If he want to prove any saturated chain of prime ideals has length equal to $\dim R$ by induction on $\dim R$, what is the induction hypothesis ? I do not understand how can we get any saturated chain of prime ideals in $R$ has length equal to $\dim R$ from any any saturated chain of prime ideals in $R$ has length equal to $\dim R-1$.

Please help me point it out. Thanks.

Arsenaler
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  • Dear Arsenaler, You are misunderstanding the inductive hypothesis. It is not just about saturated chains of prime ideals in the particular ring $R$. (After all, the theorem tells you that, when all is said and done, these will all be of a fixed length, namely $\dim R$.) Rather, the variable in the induction is $\dim R$, and the inductive hypothesis is that the theorem is true for all f.g. $k$-algebras of dimension $< \dim R$. (And I guess that $B$ is such an algebra.) Regards, – Matt E Jun 04 '13 at 03:54
  • @MattE : May be I have misunderstood the inductive method applied here. But I still can not guess or imagine how the proposition was proven by induction. Could you please write it more precisely ? If you feel my comment is stupid, and loosing your time, simply delete it. Thanks. – Arsenaler Jun 04 '13 at 04:43
  • Dear Arsenaler, In my previous comment I already sketched the idea; I wrote an answer below with slightly more details. Regards, – Matt E Jun 04 '13 at 23:15

1 Answers1

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I haven't looked at the text in question, but here is a sketch of how such arguments go:

We want to prove that for all f.g. $k$-algebras, some quantity $q(R)$ attached to $R$ is equal to $\dim R$. (In this case, the quantity in question is the length of a saturated chain of prime ideals.)

One proves it for all $R$ at once, by induction on dimension of $R$, i.e. the inductive hypothesis is that for every f.g. $k$-algebra of dim'n $< n$, the statement is true, and one goes on to prove the statement for every f.g. $k$-algebra of dimension $n$.

Thus we give ourselves an arbitrary f.g. $R$ algebra of dimension $n$, and try to prove that $q(R) = n$. One typical approach is to find a quotient $B$ of $R$ of dimension $n -1$, and prove that the quantity in question for $B$, i.e. $q(B)$, is $1$ less than the quantity in question for $R$, i.e. $q(R)$. Assuming we've done this: then by the inductive hypothesis one has $q(B) = \dim B = n - 1$ (since $B$ is a f.g. $k$-algebra of dimension $n -1$, which is $< n$), and so $q(R) = q(B) + 1$ (as I already wrote, I'm assuming that we've proved this in some way; this is where the heart of the proof lies) $ = \dim B + 1 = (n-1) + 1 = n,$ and we're done.

Based on what is written by the OP, I think the above proof schema is the one being used.

My impression is that the OP's confusion comes from imagining the the ring $R$ is fixed (and that the induction somehow involves the length of chains of prime ideals in this fixed ring $R$ --- which it can't, because all saturated chains of primes ideals will be of the same length, and so there is no variable $n$ to induct on in this setting), whereas it is not: the statement we are proving by induction involves all $R$ at once.

Matt E
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