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Archive for the ‘number theory’ Category

The goal of this post is to collect a list of applications of the following theorem, which is perhaps the simplest example of a fixed point theorem.

Theorem: Let G be a finite p-group acting on a finite set X. Let X^G denote the subset of X consisting of those elements fixed by G. Then |X^G| \equiv |X| \bmod p; in particular, if p \nmid |X| then G has a fixed point.

Although this theorem is an elementary exercise, it has a surprising number of fundamental corollaries.

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Previously I mentioned very briefly Granville’s The Anatomy of Integers and Permutations, which explores an analogy between prime factorizations of integers and cycle decompositions of permutations. Today’s post is a record of the observation that this analogy factors through an analogy to prime factorizations of polynomials over finite fields in the following sense.

Theorem: Let q be a prime power, let n be a positive integer, and consider the distribution of irreducible factors of degree 1, 2, ... k in a random monic polynomial of degree n over \mathbb{F}_q. Then, as q \to \infty, this distribution is asymptotically the distribution of cycles of length 1, 2, ... k in a random permutation of n elements.

One can even name what this random permutation ought to be: namely, it is the Frobenius map x \mapsto x^q acting on the roots of a random polynomial f, whose cycles of length k are precisely the factors of degree k of f.

Combined with our previous result, we conclude that as q, n \to \infty (with q tending to infinity sufficiently quickly relative to n), the distribution of irreducible factors of degree 1, 2, ... k is asymptotically independent Poisson with parameters 1, \frac{1}{2}, ... \frac{1}{k}.

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Today’s post is a record of a very small observation from my time at PROMYS this summer. Below, by \text{Spec } R I mean a commutative ring R regarded as an object in the opposite category \text{CRing}^{op}.

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In the previous post we described the following result characterizing the zeta distribution.

Theorem: Let a_n = \mathbb{P}(X = n) be a probability distribution on \mathbb{N}. Suppose that the exponents in the prime factorization of n are chosen independently and according to a geometric distribution, and further suppose that a_n is monotonically decreasing. Then a_n = \frac{1}{\zeta(s)} \left( \frac{1}{n^s} \right) for some real s > 1.

I have been thinking about the first condition, and I no longer like it. At least, I don’t like how I arrived at it. Here is a better way to conceptualize it: given that n | X, the probability distribution on \frac{X}{n} should be the same as the original distribution on X. By Bayes’ theorem, this is equivalent to the condition that

\displaystyle \frac{a_{mn}}{a_n + a_{2n} + a_{3n} + ...} = \frac{a_m}{a_1 + a_2 + ...}

which in turn is equivalent to the condition that

\displaystyle \frac{a_{mn}}{a_m} = \frac{a_n + a_{2n} + a_{3n} + ...}{a_1 + a_2 + a_3 + ...}.

(I am adopting the natural assumption that a_n > 0 for all n. No sense in excluding a positive integer from any reasonable probability distribution on \mathbb{N}.) In other words, \frac{a_{mn}}{a_m} is independent of m, from which it follows that a_{mn} = c a_m a_n for some constant c. From here it already follows that a_n is determined by a_p for p prime and that the exponents in the prime factorization are chosen geometrically. And now the condition that a_n is monotonically decreasing gives the zeta distribution as before. So I think we should use the following characterization theorem instead.

Theorem: Let a_n = \mathbb{P}(X = n) be a probability distribution on \mathbb{N}. Suppose that a_{nm} = c a_n a_m for all n, m \ge 1 and some c, and further suppose that a_n is monotonically decreasing. Then a_n = \frac{1}{\zeta(s)} \left( \frac{1}{n^s} \right) for some real s > 1.

More generally, the following situation covers all the examples we have used so far. Let M be a free commutative monoid on generators p_1, p_2, ..., and let \phi : M \to \mathbb{R} be a homomorphism. Let a_m = \mathbb{P}(X = m) be a probability distribution on M. Suppose that a_{nm} = c a_n a_m for all n, m \in M and some c, and further suppose that if \phi(n) \ge \phi(m) then a_n \le a_m. Then a_m = \frac{1}{\zeta_M(s)} e^{-\phi(m) s} for some s such that the zeta function

\displaystyle \zeta_M(s) = \sum_{m \in M} e^{-\phi(m) s}

converges. Moreover, \zeta_M(s) has the Euler product

\displaystyle \zeta_M(s) = \prod_{i=1}^{\infty} \frac{1}{1 - e^{- \phi(p_i) s}}.

Recall that in the statistical-mechanical interpretation, we are looking at a system whose states are finite collections of particles of types p_1, p_2, ... and whose energies are given by \phi(p_i); then the above is just the partition function. In the special case of the zeta function of a Dedekind abstract number ring, M = M_R is the commutative monoid of nonzero ideals of R under multiplication, which is free on the prime ideals by unique factorization, and \phi(I) = \log N(I). In the special case of the dynamical zeta function of an invertible map f : X \to X, M = M_X is the free commutative monoid on orbits of f (equivalently, the invariant submonoid of the free commutative monoid on X), and \phi(P) = \log |P|, where |P| is the number of points in P.

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An interesting result that demonstrates, among other things, the ubiquity of \pi in mathematics is that the probability that two random positive integers are relatively prime is \frac{6}{\pi^2}. A more revealing way to write this number is \frac{1}{\zeta(2)}, where

\displaystyle \zeta(s) = \sum_{n \ge 1} \frac{1}{n^s}

is the Riemann zeta function. A few weeks ago this result came up on math.SE in the following form: if you are standing at the origin in \mathbb{R}^2 and there is an infinitely thin tree placed at every integer lattice point, then \frac{6}{\pi^2} is the proportion of the lattice points that you can see. In this post I’d like to explain why this “should” be true. This will give me a chance to blog about some material from another math.SE answer of mine which I’ve been meaning to get to, and along the way we’ll reach several other interesting destinations.

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Another small example I noticed awhile ago and forgot to write up.

Prime numbers, as one of the most fundamental concepts in mathematics, have a way of turning up in unexpected places. For example, the life cycles of some cicadas are either 13 or 17 years. It’s thought that this is a response to predation by predators with shorter life cycles; if your life cycle is prime, a predator with any shorter life cycle can’t reliably predate upon you.

A month or so ago I noticed a similar effect happening in the card game BS. In BS, some number of players (usually about four) are dealt the same number of cards from a standard deck without jokers. Beginning with one fixed card, such as the two of clubs, players take turns placing some number of cards face-down in the center. The catch is that the players must claim that they are placing down some number of a specific card; Player 1 must claim that they are placing down twos, Player 2 must claim that they are placing down threes, and so forth until we get to kings and start over. Any time cards are played, another player can accuse the current player of lying. If the accusation is right, the lying player must pick up the pile in the center. If it is wrong, the accusing player must pick up the pile in the center. The goal is to get rid of all of one’s cards.

I’ve been playing this game for years, but I didn’t notice until quite recently that the reason the game terminates in practice is that 13, the number of types of cards in a standard deck, is prime. If, for example, we stopped playing with aces and only used 12 types of cards, then a game with 4 | 12 people need not terminate. Consider a game in which Player 1 has only cards 2, 6, 10, Player 2 has only cards 3, 7, J, Player 3 has only cards 4, 8, Q, and Player 4 has only cards 5, 9, K, and suppose that Player 1 has to play threes at some point in the game. Then no player can get rid of their cards without lying; since the number of players divides the number of card types, every player will always be asked to play a card they don’t have. Once every player is aware of this, every player can call out every other player’s lies, and it will become impossible to end the game reasonably.

More generally, such situations can occur if 13 is replaced by a composite number n such that the number of players is at least the smallest prime factor of n. This is because people who get rid of their cards will leave the game until the number of players is equal to the smallest prime factor of n, at which point the game may stall. But because 13 is prime, any game played with less than 13 people has the property that each player will eventually be asked to play a card that they have.

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In the first few lectures of Graph Theory, the lecturer (Paul Russell) presented a cute application of Ramsey theory to Fermat’s Last Theorem. It makes a great introduction to the general process of casting a problem in one branch of mathematics as a problem in another and is the perfect size for a blog post, so I thought I’d talk about it.

The setup is as follows. One naive way to go about proving the nonexistence of nontrivial integer solutions to x^n + y^n = z^n, n > 2 (that is, solutions such that x, y, z are not equal to zero) is using modular arithmetic; that is, one might hope that for every n > 2 it might be possible to find a modulus m such that the equation has no nontrivial solution \bmod m. To simplify matters, we’ll assume that x, y, z are relatively prime to m, or else there is some subtlety in the definition of “nontrivial” (e.g. we might have x, y, z not divisible by m but x^n \equiv 0 \bmod m.) Note that it might be the case that m is not relatively prime to a particular nontrivial solution in the integers, but if we can prove non-existence of nontrivial solutions for infinitely many m (in particular, such that any integer is relatively prime to at least one such m) then we can conclude that no nontrivial integer solutions exist.

By the Chinese Remainder Theorem, this is possible if and only if it is possible with m a prime power, say m = p^k. If p is relatively prime to n, this is possible if and only if it is possible with m = p. This is because given a nontrivial solution \bmod p we can use Hensel’s lemma to lift it to a nontrivial solution \bmod p^k for any k (and even to \mathbb{Z}_p), and the converse is obvious. (Again to simplify matters, we’ll ignore the finitely many primes that divide n.) So we are led to the following question.

For a fixed positive integer n > 2 do there exist infinitely many primes p relatively prime to n such that x^n + y^n \equiv z^n \bmod p has no nontrivial solutions?

As it turns out, the answer is no. In 1916 Schur found a clever way to prove this by proving the following theorem.

Theorem: For every positive integer k there exists a positive integer m such that if \{ 1, 2, ... m \} is partitioned into k disjoint subsets A_1, ... A_k, then there exists i such that there exist a, b, c \in A_i with a + b = c. In other words, the Schur number S(k) = m exists. (Note that I am using a convention which is off by 1.)

If we let p be a prime greater than S(n) and let the A_i be the cosets of the subgroup of n^{th} powers in (\mathbb{Z}/p\mathbb{Z})^{*}, which has index at most n, we obtain the following as a corollary.

Corollary: Fix a positive integer n > 2. For any sufficiently large prime p, there exists a nontrivial solution to x^n + y^n \equiv z^n \bmod p.

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