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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.

Community
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Divya
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2 Answers2

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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
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P is finitely generated projective R[t] module $\Rightarrow$ there is a R[t] module Q such that P$\oplus$Q = R$[t]^s$

Now take P $\otimes$R[t]/($t^n$)$\oplus$ Q $\otimes$R[t]/($t^n$)

this is $\cong$ to $(\textbf{R[t]}/(t^n))^s$

Now $\textbf{R[t]}/(t^n) $ has a basis set {$1$, $t$, $t^2$,.....$t^{n-1}$} thus it is a free R module.

Divya
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