Is there an analogue of finite fields for products of two prime powers? - MathOverflow

Is there an analogue of finite fields for products of two prime powers?

2012-01-12T10:53:36Z

The collection of prime powers can be characterized in the following way:

There is a field with $q$ elements if and only if $q$ is a prime power. Furthermore if it exists then this field is unique up to isomorphism.

My question is,

Is there an analogous characterization for positive integers which are products of two distinct prime powers?

To avoid triviality let us say that the characterization should be presented independently of the prime factorization of the number involved. So we are looking for a statement like,

"There exists a set from a certain class which has property $P(n)$ if and only if $n$ is a product of two distinct prime powers,"

and both the set from the relevant class and the property $P(n)$ should be defined in a way that does not inherently depend on the prime factorization of $n$.

Answer by Goldstern for Is there an analogue of finite fields for products of two prime powers?

2012-01-12T14:53:06Z

For any (first order, but other variants are also reasonable) formula $\varphi$ without free variables, finite model theory defines $S(\varphi)$, the "spectrum" of $\varphi$, as the set of all positive natural numbers $n$ such that there exists a structure of size $n$ satisfying $\varphi$.

(As far as I remember, there is no nice characterization of those subsets of $\mathbb N$ which are spectra. In particular: it is still open whether the set of all spectra of first order formulas is closed under complements.)

It would be reasonable to define $S^*(\varphi)$ as the set of all $n$ such that there is a unique (up to isomorphism) structure of size $n$ satisfying $\varphi$.

As Emil Jeřábek has implicitly pointed out in his first comment, there is a first order formula $\varphi_{\text{product of fields}} $ such that $S^*(\varphi_{\text{product of fields}} )=S(\varphi_{\text{product of fields}} ) = $ the set of all products of two prime powers. (The formula is really quite explicit; I do not give it here as it would not add any relevant information to my answer, I think.)

I know that this answer is in a sense trivial. But I don't see a formal criterion that will distinguish the trivial from the nontrivial answers.

Answer by Kevin O'Bryant for Is there an analogue of finite fields for products of two prime powers?

2012-01-12T22:01:35Z

There are at most 2 groups (up to isomorphism) of order $n$, and there is not a field of order $n$, if and only if $n$ is the product of two distinct primes.

This doesn't cover "prime powers", but at least it's nontrivial.

Answer by Noam D. Elkies for Is there an analogue of finite fields for products of two prime powers?

2012-01-13T02:07:46Z

If nothing else then $n$ is a product of powers of two distinct primes iff there is a unique pair of relatively prime natural numbers, other than the trivial $\lbrace 1,n \rbrace$, whose product is $n$.

Answer by felix for Is there an analogue of finite fields for products of two prime powers?

2012-01-14T15:50:06Z

A natural number $n$ is the product of precisely two prime powers if and only if there exists an abelian group of order $n$ having precisely two maximal subgroups. (And that group is unique up to isomorphism.)