This is an Exercise on Kemper, A Course in Commutative Algebra:

Exercise 5.6 (G. Kemper) If $A$ is an affine $K$-domain, then the transcendence degree of $A$ is the size of a maximal algebraically independent subset of $A$.

The transcendence degree of an algebra $A$ over a field $K$ is defined by: $$ trdeg(A):=\sup \{|T|\ \mid T\subset A \text{ is finite and algebraically independent} \}$$

The solution Kemper wrote used Noether Normalization (in particular the transcendence basis), which is far after Chapter 5 in his book. I wonder if this question can be solved without using any of that since this is an exercise of Chapter 5. I tried using Krull Dimension, which was just introduced by Chapter 5, but got stuck by a gap:

Let $a_1,\cdots,a_n$ be a maximal algebraically independent subset of $A$, consider $R=K[a_1,\cdots,a_n]$. Clearly $\dim A=trdeg(A)\geq trdeg(R)=n$, so it remains only to show that $n=\dim R\geq \dim A$. It suffices to show that for any chain of prime ideals $P_0\subsetneq P_1\subsetneq\cdots\subsetneq P_m$ in $A$, $R\cap P_i\in Spec(R)$ form a chain $R\cap P_0\subset\cdots \subset R\cap P_m$ of length $m$. The only thing needs to prove is that $R\cap P_i\neq R\cap P_{i+1}$ for each $i$.

It seems that this is not impossible to be fulfilled if one uses the properties of $A$ and $R$ smartly. My attempt is as below:

Assume not, so $R\cap P_i=R\cap P_{i+1}$, then for any $p\in P_{i+1}\setminus P_i$ we have $p\not\in R$. Since $R$ is generated by a maximal algebraically independent subset of $A$, this means that there exists $r_k\in R$ such that $\sum_{k=0}^N r_kp^k=0$. Thus $$-r_0=p\left(\sum_{k=1}^N r_kp^{k-1}\right)\in R\cap P_{i+1}=R\cap P_i,$$ which implies that $r_0\in P_i$ and $r_1+\sum_{k=2}^N r_k > p^{k-1}\in P_i\subset P_{i+1}$. Since $\sum_{k=2}^N r_k p^{k-1}\in P_{i+1}$, we have $r_1\in R\cap P_{i+1}=R\cap P_i$, so $r_1\in P_i$, which implies that $p\left(\sum_{k=2}^N r_kp^{k-2}\right)\in P_i$, so $r_2+\sum_{k=3}^N r_kp^{k-2}\in P_i\subset P_{i+1}$. Repeat this argument and we see that $r_0,\cdots,r_N\in P_i$.

So we conclude that for any $p\in P_{i+1}$ there exists $r_k\in R\cap P_i$ such that $\sum_{k=0}^N r_kp^k=0$ (For $p\in P_i\setminus R$ a similar argument applies).

Notice that since at least one of the $r_k$'s must be non-zero, we know that if $R\cap P_i=\{0\}$ then there must be $R\cap P_i\neq R\cap P_{i+1}$. So far I haven't used the property that $A$ is an integral domain, and I cannot see a clue how to use it. If this property can be used to restrict the choice of the $r_k$'s, for example if one of the $r_k$'s is $1$, then a contradiction can be drawn. Or if this property can be used to draw a contradiction directly from the conclusion below? I'd be grateful if anyone could tell me the solution or give me some hints that would work.

Thanks in advance.


1 Answer 1


Let $R=\mathbb Z[X]$ and $A=\mathbb Z[X,Y]/(XY-2)$. The ring extension $R\subset A$ is algebraic, but does not have the incomparable (INC) property. Let $P=(2,X)/(XY-2)$ and $P'=(2,X,Y)/(XY-2)$. We have $P\cap R=P'\cap R=(2,X)R$.

If one wants $R$ and $A$ to be finitely generated algebras over a field I suggest to consider $R=K[X,Y]$ and $A=K[X,Y][\frac{X}{Y}]$. (In fact, $A\simeq K[X,Y,Z]/(YZ-X)$.) The ring extension $R\subset A$ is algebraic, and does not have the incomparable property since $P=(X,Y)A$ and $P'=(X,Y,X/Y)A$ lie over the same prime ideal, that is, $(X,Y)R$.

  • $\begingroup$ This answer shows that the OP's try can not lead to a proof. $\endgroup$
    – user26857
    Mar 31, 2021 at 22:28
  • $\begingroup$ Very nice examples. Thanks. $\endgroup$
    – Shana
    Apr 1, 2021 at 5:19

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