Let $A$ be a commutative noetherian ring, and let $P$ be a projective $A[T]$-module with constant rank $n$. Let $L$ be the determinant of $P$, $\wedge^n(P)$. We say that $P$ (resp. $L$) is *extended* when $P = P/(T) \otimes A[T]$. It seems natural to me that if $P$ is extended, then so is $L$.

**Question:**
Does $L$ being extended necessarily imply that $P$ is also extended? In other words, can we have a non-extended $P$ with an extended determinant?

**EDIT:**
By a result of Bhatwadekar and Roy, if the determinant of $P$ is extended from $A$, then $\text{ht}(J(P,A)) \geq 2$ - forcing any counter-example to have $\dim(A) \geq 2$.

*Definition:* The Quillen ideal, $J(P,A)$ is defined to be the set of $a \in A$ such that $P_a$ is extended from $A_a$. Quillen proved that this set is a radical ideal.