# A “generalized field” with $q$ elements, when $q$ is any number?

It is well-known that if a finite field has $q \in \mathbb{N}$ elements, then $q$ is prime power and $q > 1$.

However, various modification of the concept of a "field" have been made in order to make sense of $\mathbb{F}_1$, the field with $1$ "element". See for instance Mapping $\mathbb{F}_1$-land for an overview. In combinatorics, the notion of a q-analog has a structural interpretation only when $q$ is a prime power (now including $q=1$), which is kind of awkward.

Question. Can you think of a reasonable notion of a "generalized field" such that, for every natural number $q \geq 1$ there is a "generalized field" $\mathbb{F}_q$ with $q$ "elements"?

Here are some minimal requirements: The category of fields should have a fully faithful functor to the category of generalized fields. If $q$ is a prime power, then this embedding should map $\mathbb{F}_q$ to $\mathbb{F}_q$. Every $\mathbb{F}_q$-"module" should be free. There should be a morphism $\mathbb{F}_q \to \mathbb{F}_{q'}$ if and only if $v_p(q)|v_p(q')$ for each prime $p$. The number of monic irreducible polynomials of degree $n$ over $\mathbb{F}_q$ (suitably defined) should be $\sum_{d|n} \mu(d) \cdot q^{n/d}$.

• Hmm, an immediate thought is to do this via some sort of quantum group, starting with $q$ as a formal parameter and then specializing. But that is only a rough idea, and I have no idea how one might do it precisely. Ohh, and that title is great (Mapping Fun-land). – Tobias Kildetoft Jul 9 '15 at 9:59
• What is $v_p(\cdot)$ here? – Travis Willse Jul 9 '15 at 10:34
• $v_p$ denotes the multiplicity of $p$. – Martin Brandenburg Jul 9 '15 at 10:36
• +1 for a question that had to be asked. Food for thought: Proto-abelian categories (a generalization of abelian categories) and a category of vector spaces over $\mathbb{F}_1$ are defined in Dyckerhoff's lecture notes math.uni-bonn.de/people/dyckerho/notes.pdf , and they satisfy at least one of the many wishes one might have (namely, exhibiting various combinatorial structures as $\mathbb{F}_1$-analogues of linear-algebraic structures). There is also Tesler's thesis on semi-primary lattices math.ucsd.edu/~gptesler/Nthesis.pdf , which I wish I had the time to read :/ – darij grinberg Sep 27 '15 at 22:35

Here is a very simple answer, which works at least for $q>1$. If $q = q_1 \dotsc q_l$ with coprime prime powers, then let $\mathbb{F}_q = \mathbb{F}_{q_1} \times \dotsc \times \mathbb{F}_{q_l}$ as a commutative ring. These are precisely those finite reduced commutative rings $R$ such $R/pR$ is a field for all prime numbers $p$. These rings have all the desired properties, except perhaps for the number of irreducible polynomials (this notion is not well-behaved over rings with zero divisors).
• Isn't this just $\mathbb Z/q\mathbb Z$... ? – Dustan Levenstein Nov 17 '15 at 4:18
• No. (A counterexample is $q=4$.) – Martin Brandenburg Nov 17 '15 at 9:55