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Commutative rings, modules, ideals, homological algebra, computational aspects, invariant theory, connections to algebraic geometry and combinatorics.
19
votes
What is the "intuition" behind "brave new algebra"?
This is a general phrase that refers to the direction of
higher category theory, per Lurie (you know references)
scheme homotopy theory, per Voevodsky
derived spaces, per Ben-Zvi and Nadler (0706.032 …
14
votes
Accepted
What representative examples of modules should I keep in mind?
Yes, there is a big class of modules that have an intuition different from the abstract algebra, namely the ones that come from an algebraic geometry. If $R$ is a (say, Noetherian) commutative ring, t …
14
votes
Why is an elliptic curve a group?
Short answer: because it's a complex torus. Explanation below would take as through many topics.
Topological covers
The curve should be considered over complex numbers, where it can be seen as a Rie …
5
votes
What is interesting/useful about Castelnuovo-Mumford regularity?
Here's an example paper: 0905.2212
It uses some bound on Castelnuovo–Mumford regularity to prove that cohomology of smooth complex projective variety can be computed in parallel polynomial time.
4
votes
Class number measuring the failure of unique factorization
Class number $h(K)$ is exactly the quantitative measure of the failure of unique factorization: by its definition it measures "how many more ideas are there compared to numbers".
To clarify: decompo …
3
votes
Rings of integers of function fields
Congratulations, you've just discovered the definition of normal variety — precisely the one for which locally all rings are integral closures in their fields of fractions.
This definition leads to s …
1
vote
What does a projective resolution mean geometrically?
Ah, great question!
I'm not a big expert, but one thing it obviously does is constructing the sheaf M from the locally free bundles (locally free = projective). For example, consider a skyscraper she …
0
votes
Homology of koszul complex is finitely generated?
At least for the definition I know, it seems to be because all the groups in the Koszul complex are finitely generated as $A$-modules, being explicitely constructed as sums of tensor products of $M$!
0
votes
Solving polynomial equations when you know in which number field the solutions live
Well, it's certainly easier to intersect lines and quadrics. I would simply go on intersecting all lines, and then select some nicer quadrics to intersect until I'm having a finite set of points, from …
-1
votes
Dense section of sheaves of modules
Here's a typical example of $M$ with the property that $O_x = M_x$ for $x$ in open subset.
Take $U = \mathop{\mathrm{Spec}}A-\{f=0\}$. Note that $U$ is $\mathop{\mathrm{Spec}} A_f$ where $A_f$ is a l …