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.


I believe the the following conjecture is still open:

For any commutative noetherian ring $A$ and any projective module $P$ over $A[T]$, if $P$ is stably extended then $P$ is extended.

Since "$P$ stably extended'' implies "$det(P)$ extended'', a positive answer to your question would settle this notoriously difficult problem.


You basically are asking if a family (over a line) of vector bundles (on an affine scheme) with constant determinant is constant. The answer is no. Here is an example:

Let $A$ be the coordinate ring of a cuspidal curve: $A = \mathbb{C}[x, y]/(x^2 - y^3)$. Then the class group of $A$ is equal to $\mathbb{C}$. Denote by $L_a$ the line bundle (projective module over $A$ of rank one) corresponding to $a\in\mathbb{C}$. Take $P$ to be the naturally defined $A[T]$-module such that $P/(T-a) = L_a\oplus L_{-a}$. Then $P$ is not extended, but its determinant is trivial.

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    $\begingroup$ I believe that such a $P$ is in fact extended, actually free. If $P$ is a projective module over $A[T]$, with $A$ Noetherian of dimension one and $P$ has determinant trivial, then $P$ is trivial. This is because, $A(T)$ ($A[T]$, localized with respect to the set of all monic polynomials) is one dimensional and so any projective module with trivial determinant is trivial. Thus, Quillen-Suslin imples $P$ is trivial. $\endgroup$ – Mohan Jul 14 '12 at 13:32
  • $\begingroup$ @Mohan In the above example, because the determinant is extended from A, the Quillen ideal, $J(P,A)$, is must have height $\geq 2$, correct? Which would mean that $J(P,A) = A$ because dim$(A) = 1$, implying $P$ must be extended? $\endgroup$ – Andrew Parker Jul 15 '12 at 16:09
  • $\begingroup$ @Andrew Parker, Yes, $J(P,A)$ has height at least two. $\endgroup$ – Mohan Jul 15 '12 at 19:15
  • $\begingroup$ @Mohan: Sorry for being sketchy and giving the wrong answer. Unfortunately, I don't understand your argument. Is my "vector bundle" intuition wrong? Are you saying that $L_a \oplus L_{-a}$ does not depend on $a$? $\endgroup$ – Piotr Achinger Aug 1 '12 at 21:04

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