Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

a picard group is the set of isomorphism classes of invertible R-modules. I just read that phrase in the CRing project notes without further explanations:

Here are my questions:

1-under which law (I am guessing it's a restriction of a tensor product) how could we restrict it to classes? 3-why an abelian group?

I am not familiar with any algebraic geometry. Thus shall be grateful to you for using a basic vocabulary :)

Many thanks

share|improve this question
    
1. tensor product, 2. tensor product respects isomorphisms by the universal property, 3. tensor product respects switching of the factors by the universal property. –  Qiaochu Yuan May 20 '11 at 11:28
    
@Yuan thanks for the hint –  El Moro May 20 '11 at 12:00
2  
If you are familiar with Number theory, the class group for a number field is in fact the Picard group for the ring of integers. The point is that every invertible module is isomorphic to a fractional ideal, and tensor product is isomorphic to products when you work with fractional ideals. Multiplication by a principal ideal gives you a fractional ideal that's isomorphic to the previous one as $R$-modules. See Theorem 11.6 of Eisenbud's "commutative algebra with a view toward algebraic geometry" for more details. –  Jiangwei Xue May 20 '11 at 12:05
1  
Dear El Moro, The reason considering invertible modules is so important is that, in algebraic geometry, it's important to know what the line bundles on a scheme are, and on an affine these correspond precisely to the invertible modules. In topology, much mathematics has focused on classifying arbitrary vector bundles -- perhaps since line bundles alone on a space X can be classified by the easily computable (usually) $H^2(X,Z)$. –  Akhil Mathew May 20 '11 at 12:35
1  
In algebraic geometry, the analogous characterization (the line bundles are classified by $H^1(X,O_X^*)$) seems much less easy to compute with, since it is the cohomology of a rather complicated (and non-quasicoherent) sheaf. Moreover, on a normal affine scheme, the Picard group measures the deviation from factoriality. –  Akhil Mathew May 20 '11 at 12:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.