# 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 $$\textbf{P/t^nP}$$ 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.

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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
September 11, 2019 19:16 PM

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