I've been reading some commutative algebra, but have been struggling with this idea for a while.

Let $k$ be a field, and let $A=k[x_1,\dots,x_n]$ be a finitely generated integral domain, such that $\operatorname{tr. deg}_k(k(x_1,\dots,x_n))=r$. I want to know why for any maximal chain of (nonempty) irreducible closed sets $P_1\subset P_2\subset\cdots\subset P_m=\operatorname{Spec}(A)$, with $P_i\neq P_j$ when $i\neq j$, then $m=r+1$.

I know that since the $P_i$ are closed and irreducible, then each $P_i=Z(p_i)$, the set of zeroes for some prime ideal $p_i$. So I tried looking at a maximal chain $$ Z(p_1)\subset Z(p_2)\subset\cdots\subset Z(p_m)=\operatorname{Spec}(A). $$ I also know that $Z(a)\subset Z(b)\iff\text{rad }a\supset\text{rad }b$, so this gives a maximal chain of prime ideals $$ p_1\supset p_2\supset\cdots\supset p_m=(0). $$

I've not seen a way to relate this back to the transcendence degree to conclude that $m=r+1$. I thought about assuming $m<r+1$ or $m>r+1$ to get a contradiction, but didn't see a way to proceed. Is there a nice, relatively self contained proof that the length of every maximal chain of ideals is equal to the transcendence degree of the field of fractions in this case?

I've been looking around, but haven't found a very digestible proof.

  • 1
    $\begingroup$ Note that if the commutative ring $R$ has Krull dimension $d$, then there exists a maximal chain $p_1\subset p_2\subset\ldots\subset p_{d+1}$ of primes consisting of $d+1$ members. The prime ideal $p_1$ necessarily is a minimal prime of $R$; thus if $R$ is a domain $P_1=0$. To avoid or at least minimize the risk of confusion many/most authors prefer to start the numbering of the prime ideals in a maximal chain with the index $0$ instead of $1$. $\endgroup$
    – Hagen Knaf
    Feb 21, 2012 at 8:40

1 Answer 1


This is a consequence of the Noether normalization theorem, and some inductions. (It is not trivial.) First of all, there is a weak interpretation of the statement, which says all chains of primes in $A$ of maximal length have length $r$. The stronger interpretation is that any chain of primes in $A$ in which no further prime insertions are possible has length $r$. The stronger statement is true, but somewhat harder to prove. [The length of $P_0 \subsetneq \cdots \subsetneq P_r$ is declared to be $r$ here.]

The normalization theorem says there is a polynomial subring $B = k[y_1, \ldots , y_r]$ of $A$ so that $A$ is integral over $B$. Then the Going Up Theorem will tell you chains of primes in $B$ lift to chains in $A$, and chains of length $s$ in $A$ contract to chains of length $s$ in $B$. There are obviously chains of length $r$ in $B$, hence also in $A$. I'll prove by induction on $r$ that there are no chains of length greater than $r$ in any $A$ of trans deg $r$. If $r = 1$, this holds because $B = k[y]$ is a PID, and chains in $B$ and $A$ correspond. In general, the first non-zero prime in a maximal length chain in $B$ must be principal, generated by an irreducible polynomial $f$. But now it is easy that $B/(f)$ has transcendence degree $r-1$, so by induction chains going on up from $(f)$ in $B$ have length bounded by $r-1$.

This only proves the weak theorem. The stronger theorem uses the Going Down Theorem for integral extensions of an integrally closed domain. The polynomial ring $B$ is integrally closed, so Going Down applies. The point now is, if we have a chain in $A$ for which no insertions are possible, then by Going Down the smallest non-zero prime $P$ in that chain must intersect $B$ in a minimal prime $(f)$ in $B$. Now induction for the stronger statement applies to $B/(f) \subseteq A/P$.

I guess this isn't a 'self-contained' proof, so perhaps doesn't provide what you hoped for. But I don't believe there is any easier method.

  • $\begingroup$ Do you mean by weak interpretation "all chains of primes in $A$ has length $\le r$" (and at least one has length equal to $r$) ? $\endgroup$
    – user18119
    Mar 10, 2012 at 22:19

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