Hurkyl said:For a dimension 1 regular semi-local ring, does the Picard group vanish?
What if it is not regular? (and what if I ask for the ideal class group?)
What if it's dimension greater than 1?
micromass said:I don't know very much about this. But I think that the Picard group does vanish. A dimension 1 regular ring is a Dedekind domain. And it's known that a semi-local Dedekind domain is a PID. And the Picard group of a PID vanishes.
A semilocal ring is a commutative ring with identity in which every element that is not a unit belongs to exactly one maximal ideal. This means that the ring has a finite number of maximal ideals, but not necessarily a unique maximal ideal.
The Picard group of a ring is the group of isomorphism classes of invertible modules over that ring. In other words, it is the set of all possible ways that a given ring can be "twisted" or "deformed" by invertible modules.
In a semilocal ring, every element that is not a unit belongs to exactly one maximal ideal. This means that the non-unit elements can be "twisted" or "deformed" by a unique maximal ideal. This relationship between the non-unit elements and maximal ideals is what connects semilocal rings to the Picard group.
The Picard group is an important tool in algebraic geometry for studying the behavior of algebraic varieties under deformations. It helps to classify the different ways that a given variety can be deformed, and also provides information about the geometry of the variety such as its singularities and divisors.
Yes, the Picard group can be used to distinguish between non-isomorphic semilocal rings. If two semilocal rings have different Picard groups, then they are not isomorphic. However, the converse is not always true as there are examples of non-isomorphic rings with the same Picard group.