USAMO 1973 #4 August 19, 2009
Posted by lumixedia in algebra, Problem-solving.Tags: algebra, contest math, olympiad math, USAMO, USAMO 1973
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A fairly straightforward algebra problem. Could appear on a modern AMC-12, though the decoy answers would have to be carefully written.
USAMO 1973 #4. Determine all the roots, real or complex, of the system of simultaneous equations
Integrality, invariant theory for finite groups, and more tools for Noetherian testing August 11, 2009
Posted by Akhil Mathew in algebra, commutative algebra.Tags: algebra, integrality, invariant theory, Noetherian rings
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There are quite a few more tools to tell whether a ring is Noetherian. In this post, I’ll discuss another basic tool: integrality. I’ll discuss the application to invariant theory for finite groups.
Subrings
In general, it is not true that a subring of a Noetherian ring is Noetherian. For instance, let be the polynomial ring in infinitely many variables over a field . Then is not Noetherian because of the ascending chain
However, the quotient field of is Noetherian. This applies to any non-Noetherian integral domain.
There are special cases where we can conclude a subring of a Noetherian ring is Noetherian.
USAMO 1973 #2 August 11, 2009
Posted by lumixedia in Problem-solving.Tags: algebra, contest math, number theory, olympiad math, USAMO, USAMO 1973
3 comments
USAMO 1973 #2. Let and denote two sequences of integers defined as follows:
Thus, the first few terms of the sequence are:
Prove that, except for “1”, there is no term which occurs in both sequences. (more…)
“Undergraduate Algebra”: Or How I Relearned Algebra in a Week August 11, 2009
Posted by Martin Camacho in General.Tags: algebra, books
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A few weeks ago I vowed to relearn all of my forgotten algebra – advanced group theory, rings, modules, and fields especially. The main problem,at least for me, was finding a viable resource to tutor me. Wikipedia proved futile as there was no use in clicking links in an unsystematic manner, and Wikibooks’ algebra section was simultaneously obtrusive and incomplete.
How to tell if a ring is Noetherian August 9, 2009
Posted by Akhil Mathew in algebra, commutative algebra.Tags: algebra, commutative algebra, Hilbert basis theorem, localization, Noetherian rings
7 comments
I briefly outlined the definition and first properties of Noetherian rings and modules a while back. There are several useful and well-known criteria to tell whether a ring is Noetherian, as I will discuss in this post. Actually, I’ll only get to the first few basic ones here, though these alone give us a lot of tools for, say, algebraic geometry, when we want to show our schemes are relatively well-behaved. But there are plenty more to go.
Hilbert’s basis theorem
It is the following:
Theorem 1 (Hilbert) Let be a Noetherian ring. Then the polynomial ring is also Noetherian.
Generic freeness II July 30, 2009
Posted by Akhil Mathew in algebra, commutative algebra.Tags: algebra, commutative algebra, generic freeness, localization
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Today’s goal is to partially finish the proof of the generic freeness lemma; the more general case, with finitely generated algebras, will have to wait for a later time though.
Recall that our goal was the following:
Theorem 1 Let be a Noetherian integral domain, a finitely generated -module. Then there there exists with a free -module.
USAMO 1972 #4 July 26, 2009
Posted by lumixedia in Problem-solving.Tags: algebra, contest math, olympiad math, USAMO, USAMO 1972
3 comments
USAMO 1972 #4. Let denote a non-negative rational number. Determine a fixed set of integers , , , , , such that, for every choice of ,
The enveloping algebra July 25, 2009
Posted by Akhil Mathew in algebra.Tags: algebra, enveloping algebras, Lie algebras, tensor algebras, universal properties
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As we saw in the first post, a representation of a finite group can be thought of simply as a module over a certain ring: the group ring. The analog for Lie algebras is the enveloping algebra. That’s the topic of this post.
Definition
The basic idea is as follows. Just as a representation of a finite group was a group-homomorphism for a vector space, a representation of a Lie algebra is a Lie-algebra homomorphism . Now, is the Lie algebra constructed from an associative algebra, —just as is the group constructed from taking invertible elements.
Engel’s Theorem and Nilpotent Lie Algebras July 23, 2009
Posted by Akhil Mathew in algebra, representation theory.Tags: algebra, Engel's theorem, Lie algebras, linear algebra, nilpotent
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Now that I’ve discussed some of the basic definitions in the theory of Lie algebras, it’s time to look at specific subclasses: nilpotent, solvable, and eventually semisimple Lie algebras. Today, I want to focus on nilpotence and its applications.
Engel’s Theorem
To start with, choose a Lie algebra for some finite-dimensional -vector space ; recall that is the Lie algebra of linear transformations with the bracket . The previous definition was in terms of matrices, but here it is more natural to think in terms of linear transformations without initially fixing a basis.
Engel’s theorem is somewhat similar in its statement to the fact that commuting diagonalizable operators can be simultaneously diagonalized.
Why simple modules are often finite-dimensional II July 22, 2009
Posted by Akhil Mathew in algebra, representation theory.Tags: algebra, finite-dimensional vector spaces, Nakayama's lemma, representation theory, simple modules
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I had a post a few days back on why simple representations of algebras over a field which are finitely generated over their centers are always finite-dimensional, where I covered some of the basic ideas, without actually finishing the proof; that is the purpose of this post.
So, let’s review the notation: is our ground field, which we no longer assume algebraically closed (thanks to a comment in the previous post), is a -algebra, its center. We assume is a finitely generated ring over , so in particular Noetherian: each ideal of is finitely generated.
Theorem 1 (Dixmier, Quillen) If is a finite -module, then any simple -module is a finite-dimensional -vector space.