Is it a Projective R-module?

by ZOne   Last Updated September 11, 2019 19:20 PM

R is a commutative ring with unity.

Let $\textbf{P}$ be a finitely generated projective $\textbf{R[t]}$ module, then is $\textbf{P/$t^n$P}$ a finitely generated projective $\textbf{R}$ module?

Now in my attempt I have desperately tried using the idempotent lifting of projective module but I have failed miserably, it seems that I am missing something very simple.

$\textbf{Idempotent Lifting}$ : If I is a nilpotent ideal or a complete ideal then there is a bijection between isomorphism classes of finitely generated projective mdules over R and finitely generated projective modules over R/I.

So here in the ring R[t]/($t^n$) I have $(t)+(t^n)$ as a nilpotent ideal and R[t]/(t) $\cong$ R. I was thinking along this line. If you could help me I would be grateful.



Answers 1


Hint:

If $P$ is a finitely generated projective $R[t]$-module, it is (isomorphic to) a direct summand of some finitely generated free $R[t]$-module, say $R[t]^s$, so , tensoring with $R[t]/(t^n)$ over $R[t]$, we obtain that $P/t^nP$ is a direct summand of $$R[t]^s\otimes_{R[t]}R[t]/(t^n)\simeq \bigl(R[t]/(t^n)\bigr)^{\!s}$$ and each factor of the latter is a finitely generated .free $R$-module

Bernard
Bernard
September 11, 2019 19:16 PM

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