## What is a quantum vector space?September 24, 2010

Posted by Sarah in Uncategorized.
Tags:

I went to the Friday grad seminar — this one by Hyun Kyu on the quantum Teichmuller space. (Here’s his paper, which I haven’t read as of now.) I thought it might be helpful to learn some background about this whole “quantum” business.

One way of thinking about the ordinary plane is to consider it the algebra freely generated by the elements x and y subject to the commutation relationship yx = xy. Now, what if we alter this description to instead have yx = qxy? This defines something known as the quantum plane. Here, q is an element of the ground field. Obviously, except when q = 1, this is a non-commutative algebra. For any pair of integers i and j, we have
$y^i x^i = q^{ij} x^i y^j.$
We can define quantum versions of lots of things — for example, $SL_q(2, \mathbb{C})$ is the group of 2-by-2 matrices with determinant 1, satisfying the relations

ab = 2ba, bc = cb, cd = q dc, ac = q ca, bd = q db, and ad-da = (q – 1/q)bc.

The “quantum” here is in the mathematical sense of a non-commutative deformation of a commutative algebra.
More on the subject: “What is a Quantum Group?”

## Atiyah-Macdonald: Modules and Exact SequencesJune 23, 2010

Posted by Sarah in Uncategorized.
Tags:

A module is a generalization of the notion of ideals. Rings, ideals, quotient rings, and vector spaces are all modules. If A is a ring, then an A-module is an abelian group M on which A acts linearly:
$a(x+y) = ax + ay$
$(a+ b) x = a x + bx$
$(ab) x = a(bx)$
$1x = x$
Submodules are subgroups of M closed under multiplication by elements of A; the sum and intersection of submodules is again a submodule.

The annihilator of M is the set of all a in A such that aM = 0. A module is called faithful if its annihilator is zero.

Direct sum of two modules is just defined as an ordered pair (x, y), with componentwise addition and scalar multiplication. An A-module is called free if it’s isomorphic to the direct sum of modules isomorphic to A. (A finitely generated A-module is isomorphic to a quotient of $A^n$ for some n.)

Proposition.
Let M be a finitely generated A-module. Let a be an ideal of A and let $\phi$ be an A-module automorphism of M st $\phi(M) \subset aM$. Then $\phi$ satisfies an equation of the form

$\phi^n + a_1 \phi^{n+1} \dots + a_n = 0$.

Proof.
Let $x_1 \dots x_n$ be a set of generators of M. Then $\phi(x_i) = \sum a_{ij} x_j$ for some $a_{ij}$. That is,
$\sum (\delta_{ij} \phi - a_{ij}) x_j = 0$
Left-multiplying by the adjoint of the matrix $(\delta_{ij} \phi - a_{ij}$, we see that the determinant annihilates each $x_i$, hence is the zero endomorphism of M. Expanding out the determinant gives us an equation of the desired form.

Exact sequences. Suppose we have a sequence of homomorphisms between modules,
$M_{i-1} \to M_i \to M_{i+1} \dots$
This is exact at $m_i$ if $Im(f_i) = Ker(f_{i+1})$. The sequence is exact if it is exact at every module.
$0 \to M' \to M$ is exact iff f is injective;
$M \to M^* \to 0$ is exact iff g is surjective.
A short exact sequence:
$0 \to M' \to M \to M^* \to 0$.
Every long exact sequence can be split into these.
If the above is an exact sequence, then
$0 \to Hom(M^*, N) \to Hom(M, N) \to Hom(M', N)$ is also exact; note that the order reverses when we go to the dual space.

Next time I’ll introduce the boundary map and the Snake Lemma.

## Atiyah-MacDonald: IdealsJune 18, 2010

Posted by Sarah in Uncategorized.
Tags:

An ideal, a, of a ring, A is a subset which is an additive subgroup and such that $A a \subset a$. The cosets of this ideal, $A / a$, are known as a quotient ring.

A basic, though special, example here is the ring of integers. Every ideal is the set of multiples of a single integer, n. The quotient rings are the integers mod n. The ideal generated by n is denoted (n), and it is called a principal ideal. Because all the ideals in the integers are multiples of a single element, the integers are known as a principal ideal domain.

Notice that $(mn) \subset (n)$. Multiples of six are also multiples of three. An ideal generated by a prime number is not contained in any other ideal (except the whole ring of integers.) This brings us to the notion of prime and maximal ideals.

A prime ideal p in a ring A is an ideal that is not equal to (1) and if $xy \in p$ then $x \in p$ or $y \in p$. Equivalently, a prime ideal is one where $A / p$ is an integral domain.

A maximal ideal m in a ring A is an ideal that is not equal to (1) and there is no ideal a such that $m \subset a \subset (1)$. Equivalently, a maximal ideal is one where $A / m$ is a field. (This last is true because A is a field iff the only ideals in A are 0 and (1).)

Maximal ideals are always prime, because a field is always an integral domain.

Every ring (not equal to the zero ring) has at least one maximal ideal. This is a result of Zorn’s lemma.

As a result, every non-unit of A is contained in a maximal ideal.

In a principal ideal domain, every non-zero prime ideal is maximal.
For if $(x) \neq 0$ is a prime ideal and $(x) \subset (y)$, then we have some $x = yz$ so that $yz \in (x),$ hence $z \in (x)$, say $z = tx$. Then $x = yz = ytx$, so $yt = 1$ and so $(y) = (1)$.

The set of nilpotent elements in a ring is called the nilradical and it is an ideal. The product of anything with a nilpotent element is also nilpotent; the sum of nilpotent elements is nilpotent by the binomial theorem.

The nilradical is the intersection of all the prime ideals; if f is nilpotent, then $f^n = 0$ for some n, so f itself is in all the prime ideals. If f is not nilpotent, you can construct a prime ideal — the largest in the set of ideals a such that $f^n$ is not in a. (The product of two elements not in this ideal is not itself in this ideal.)

The Jacobson Radical is the intersection of all the maximal ideals in A. It (obviously) contains the nilradical.

You can get new ideals from old by summing them (that gives you a new ideal), intersecting them (also an ideal) or finding the product (the set of all finite sums $\sum x_i y_i$ of products of elements in X and Y). In the integers, $a \cup b$ is the ideal generated by their lcm, and $ab = (mn)$, the ideal generated by the product of their generators. If the generators are relatively prime, then the intersection and the product are the same ideal.

This is true for general rings, as can be proven by induction. The generalization of relatively primality is that two ideals are coprime if a + b = (1).