## USAMO 1973 #4August 19, 2009

Posted by lumixedia in algebra, Problem-solving.
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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

$\displaystyle x+y+z=3$

$\displaystyle x^2+y^2+z^2=3$

$\displaystyle x^3+y^3+z^3=3$

## Integrality, invariant theory for finite groups, and more tools for Noetherian testingAugust 11, 2009

Posted by Akhil Mathew in algebra, commutative algebra.
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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 ${A := k[X_1, X_2, \dots]}$ be the polynomial ring in infinitely many variables over a field ${k}$. Then ${A}$ is not Noetherian because of the ascending chain

$\displaystyle (X_0) \subset (X_0, X_1) \subset (X_0, X_1, X_2) \subset \dots.$

However, the quotient field of ${A}$ 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 #2August 11, 2009

Posted by lumixedia in Problem-solving.
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USAMO 1973 #2. Let ${\{X_n\}}$ and ${\{Y_n\}}$ denote two sequences of integers defined as follows:

$\displaystyle X_0=1,\hspace{0.1cm}X_1=1,\hspace{0.1cm}X_{n+1}=X_n+2X_{n-1}\hspace{0.1cm}(n=1,2,3,...)$

$\displaystyle Y_0=1,\hspace{0.1cm}Y_1=7,\hspace{0.1cm}Y_{n+1}=2Y_n+3Y_{n-1}\hspace{0.1cm}(n=1,2,3,...)$

Thus, the first few terms of the sequence are:

$\displaystyle X:\hspace{0.1cm}1,1,3,5,11,21,...$

$\displaystyle Y:\hspace{0.1cm}1,7,17,55,161,487,...$

Prove that, except for “1”, there is no term which occurs in both sequences. (more…)

## “Undergraduate Algebra”: Or How I Relearned Algebra in a WeekAugust 11, 2009

Posted by Martin Camacho in General.
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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 NoetherianAugust 9, 2009

Posted by Akhil Mathew in algebra, commutative algebra.
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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 ${A}$ be a Noetherian ring. Then the polynomial ring ${A[X]}$ is also Noetherian.

## Generic freeness IIJuly 30, 2009

Posted by Akhil Mathew in algebra, commutative algebra.
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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 ${A}$ be a Noetherian integral domain, ${M}$ a finitely generated ${A}$-module. Then there there exists ${f \in A - \{0\}}$ with ${M_f}$ a free ${A_f}$-module.

## USAMO 1972 #4July 26, 2009

Posted by lumixedia in Problem-solving.
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USAMO 1972 #4. Let ${R}$ denote a non-negative rational number. Determine a fixed set of integers ${a}$, ${b}$, ${c}$, ${d}$, ${e}$, ${f}$ such that, for every choice of ${R}$,

$\displaystyle |\frac{aR^2+bR+c}{dR^2+eR+f}-\sqrt[3]{2}|<|R-\sqrt[3]{2}|.$ (more…)

## The enveloping algebraJuly 25, 2009

Posted by Akhil Mathew in algebra.
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As we saw in the first post, a representation of a finite group ${G}$ 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 ${G}$ was a group-homomorphism ${G \rightarrow Aut(V)}$ for a vector space, a representation of a Lie algebra ${\mathfrak{g}}$ is a Lie-algebra homomorphism ${\mathfrak{g} \rightarrow \mathfrak{g}l(V)}$. Now, ${\mathfrak{g}l(V)}$ is the Lie algebra constructed from an associative algebra, ${End(V)}$—just as ${Aut(V)}$ is the group constructed from ${End(V)}$ taking invertible elements.

## Engel’s Theorem and Nilpotent Lie AlgebrasJuly 23, 2009

Posted by Akhil Mathew in algebra, representation theory.
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1 comment so far

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 ${L \subset \mathfrak{gl} (V)}$ for some finite-dimensional ${k}$-vector space ${V}$; recall that ${\mathfrak{gl} (V)}$ is the Lie algebra of linear transformations ${V \rightarrow V}$ with the bracket ${[A,B] := AB - BA}$. 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 IIJuly 22, 2009

Posted by Akhil Mathew in algebra, representation theory.
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I had a post a few days back on why simple representations of algebras over a field ${k}$ 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: ${k}$ is our ground field, which we no longer assume algebraically closed (thanks to a comment in the previous post), ${A}$ is a ${k}$-algebra, ${Z}$ its center. We assume ${Z}$ is a finitely generated ring over ${k}$, so in particular Noetherian: each ideal of ${Z}$ is finitely generated.
Theorem 1 (Dixmier, Quillen) If ${A}$ is a finite ${Z}$-module, then any simple ${A}$-module is a finite-dimensional ${k}$-vector space.