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I do not fully understand the proof of Lemma 5.6 in the book A Course in Commutative Algebra of Gregor Kemper (you can find it here)

The lemma states that : If $A$ is an algebra over a field $k$, $S\subset A$ be a generating set of $A$ as a $k$-algebra. Then : dim$A \le \sup\lbrace |T||T\subseteq S$ is finite and algebraically indenpendent$\rbrace$

I post here the paragraph that I do not understand(it will be written in incline text), if you need a full proof, please comment.

Assume $n>0$ and let $P_{0}\subset P_{1}\cdots \subset P_{m}$ be a chain in Spec$(A)$. Factoring by $P_{1}$ yields a chain in Spec$(A/P_1)$ of length $m-1$. If we can show that all algebraically independent subsets $T\subseteq \lbrace a+P_1|a\in S\rbrace \subseteq A/P_1$ have size $|T|<n$, then we can use induction on $n$

By way of contradiction , assume that there exist $a_1,\cdots, a_n \in S$ such that $\lbrace a_1+P_1, \cdots. a_n+P_1 \rbrace \subseteq A/P_1$ is algebraically independent of size $n$ then $\lbrace a_1,\cdots, a_n\rbrace \in S$ is algebraically independent. By the definition of $n$, all $a\in S$ are algebraic over $L:=$Quot$(k[a_1,\cdots,a_n])$, so Quot$(A)$ is algebraic over $L$, too. There exist a non-zero element $a$ of $P_1$. We have a non-zero polynomial $G=\sum^{k}_{i=0}g_{i}x^{i}\in L[x]$ with $G(a)=0$

My two question are :

  • What kind of induction that Kemper mention? Can anyone point it out explicitly?
  • Why we can find a non-zero polynomial $G$ above? Kemper just mentioned only the algebraic property of Quot$(A)$ over $L$.

Thank for reading my question. I hope you will fell freely helping me. Thank you very much!

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1 Answer 1

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1) The induction is ordinary induction on $n$, the transcendence degree of $A$, defined as $\sup\lbrace |T||T\subseteq S$ is finite and algebraically independent$\rbrace$.
This $n$ is assumed finite, else the inequality is obvious.

The initial case $n=0$ of the induction , corresponding to the hypothesis that $A$ is algebraic over $k$, is not trival. Here is how to proceed :
Let $\mathfrak p \subset A$ be a minimal ideal. The domain $A/\mathfrak p$ is also algebraic over the field $k$ hence is a field (Atiyah-Macdonald, Prop.5.7), so that $\mathfrak m\subset A$ is maximal.
Since every minimal ideal is maximal in $A$, we have $dim(A)=0$.

2)(Edited)
We can assume that $A$ is a domain.
Since $A$ has transcendence degree $n$ over $k$ (induction hypothesis) and since $a_1,...,a_n$ are $n$ algebraically independant elements , these elements are a transcendency basis of $\text {Quot}(A)$ over $k$. But then every element $a\in \text {Quot}(A)$ is algebraic over $L$, in particular that nonzero $a\in P_1$.
This proves the existence of the required polynomial $G$.

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  • $\begingroup$ Dear Georges, As I posted, there are two $a$. The second $a$ is a nonzero element in $P_{1}$ which may not be in $S$. That's why I ask this question. $\endgroup$
    – Arsenaler
    Commented Mar 20, 2012 at 14:41
  • $\begingroup$ Dear @msnaber: I have edited my answer in order to answer your question.I must say that the proof, although correct, is not written in an optimal way in my opinion. In particular I find the use of the letters $n$, $m$ and especially $a$ somewhat confusing. $\endgroup$ Commented Mar 20, 2012 at 21:30

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