I'm reading the paper "Quillen's solution of Serre's Problem" in which i am unable to follow one statement:

Suppose $P$ be a finitely generated projective $k[x_1,...,x_n]$ module.We first note that $P$ obviously becomes free after we invert all non zero polynomial. Then the author writes this implies:

By a classical lemma of Noether there exists a polynomial $f$ which is monic in $x_n$ such that $P$ becomes free after we invert $f$.

I'm unable to see this.Could someone explain how does one see the above claim?

  • $\begingroup$ Can you prove that there is a polynomial $f$ (not necessarily monic) such that $P$ becomes free after inverting $f$? $\endgroup$ – user26857 Apr 12 '17 at 7:32
  • $\begingroup$ @user26857: Sorry i'm unable to see the existence of such $f$ $\endgroup$ – Math Lover Apr 12 '17 at 7:49
  • $\begingroup$ Then please read this thread: math.stackexchange.com/questions/16814/… $\endgroup$ – user26857 Apr 12 '17 at 7:51
  • $\begingroup$ @user26857: Sorry i can't see how does one uses this to prove the required result.( I know that projective modules are locally free) $\endgroup$ – Math Lover Apr 12 '17 at 7:57
  • $\begingroup$ I've asked if you know to prove that $P_f$ is $A_f$-free (here $A$ denotes the polynomial ring) for some polynomial $f$ and you said "Sorry I'm unable to see the existence of such $f$". As far as I can see the linked thread deals with a similar question requiring a little more: "There are elements $f_1,\dots,f_n \in A$ such that $(f_1,\dots,f_n) = 1$, and such that $M_{f_i}$ is a free $A_{f_i}$-module for all $1\le i \le n$." $\endgroup$ – user26857 Apr 12 '17 at 8:00

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