Can 3-fields be consistently defined and constructed?

[Damidami]

I was thinking about some similarities in the definitions of group and field, and if it would be possible to generalize in some sense, like follows.

A field is basically a set F, such that (F,+) is a commutative group with identity 0, and (F-{0}, .) is a commutative group with identity 1, and . distributes over +, that is a.(b+c) = a.b + a.c

If I call n the number of operations in the algebraic structure, then n=2 for fields, if I set n=1 there is left only the first operation + (and no distributive law) so I get the definition of a commutative group.

Is it possible to generalize and get some consistent definition of an "n-field"? That is, for example, a 3-field has 3 operations (+, . ,*) with 3 distinct identities (0,1,e) such that (F-{0,1},*) is a commutative group with identity e and
1) a.(b+c) = a.b + a.c (. distributes over +)
2) a*(b.c) = a*b . a*c (* distributes over .)

I can't find any inconsistency, but neither can I construct an example of such a 3-field.

If it could be possible, one in principle could prove things about n-fields, and then letting n=2 one would get a proof about fields, and letting n=1 one would get a proof about commutative groups (that's the goal I had in mind).

Can anyone construct an example of a 3-field, or show is has no sense at all?
Thanks.

 

[homeomorphic]

I don't know, but n-field might bad name because that sounds like a name that would be reserved for categorifications of fields. Actually, for some reason that I don't quite understand (more out of negligence than the difficulty of understanding it, I suspect), rings are not easily categorified. So, rather than categorifying the integers, you can categorify the non-negative numbers. There's some issue with additive inverses. The non-negative numbers form what's call a rig, which is basically a ring without additive inverses. A categorification of that would be called a 2-rig. Just an aside.

It's easy to imagine a field with another binary operation defined on it, but the problem is compatibility with the other two operations. And for that matter, how do you decide how the third operation should be related to the other two?

 

[Damidami]

Hi homeomorphic,

I don't know much about cateogory theory. As to the name, we can change it, instead of an n-field, it could be named an n-groupfield (I like the "n" attached to the name, so one can fastly know how many operations are defined on the structure).

Anyway, before worring about the name I was worring about if it even has any sense to define such structure. I still couldn't construct an example.

As to how to decide how the third operation should be related to the other two, that would be the distributive laws:

2) a*(b.c) = a*b . a*c (* distributes over .)
maybe I should add also this one
3) a*(b+c) = a*b + a*c (* distributes over +)
(I'm not sure if some of them could be deduced as a theorem).

When constructing a group, the one with smallest number of elements is the trivial group $G = (0,+)$ that only has one element (the identity). When constructing a field, the smallest one is $F=(\{0,1\},+,.) = F_2$

I was thinking that maybe the smallest "3-groupfield" would have 3 elements (the identities), so in principle it only lasts to construct the 3 "multiplication tables" for +, . and *. (Each of the 3 multiplication tables is a 3x3 array with the outcome of each opearation)

Does it have sense? What do you think?

Thanks,
Damián.

 

[Hurkyl]

I'm going to use # for the third operation of your 3-fields, so there isn't any confusion with multiplication.

Consider a field F. To define a third operation as you suggest, we can completely ignore +. Let's ignore 0 for the moment too.

The multiplicative structure of F-{0} is an abelian group. To turn F into a 3-field, we must first find an ordinary field whose additive group is isomorphic to (F-{0}, *).

The first point to note is that if F contains a finite field of order q, then F-{0} has a cyclic subgroup of order q-1. This implies:

• q-1 is either prime or 1.
• If q is not 2, then F-{0} must consist entirely of roots of unity

If F does not contain a finite field -- i.e. if F contains the rational numbers -- then the same argument applies: (F-{0}) contains a cyclic subgroup of order 2, and a cyclic subgroup of infinite order, and therefore cannot be isomorphic to the additive group of a field.


I believe you should be able to define a 3-field structure on GF(4) -- the finite field of 4 elements. I haven't worked out what to do with 0, but I imagine you can just say 0#x=1 for all x. Same with GF(8), GF(32), and so forth.


The remaining case are transcendental extensions of GF(2). e.g. let F be the field of rational functions over GF(2). This won't work, because (F-{0}, *) is a direct sum of copies of Z. But if we take the algebraic extension of F by adding all n-th roots to everything... or maybe just take the algebraic closure... then (F-{0}, *) is a direct sum of copies of infinitely many copies of Q. Now that can be isomorphic to the additive group of a field.

 

[Damidami]

Hi Hurkyl

I'm going to use # for the third operation of your 3-fields, so there isn't any confusion with multiplication.

Consider a field F. To define a third operation as you suggest, we can completely ignore +. Let's ignore 0 for the moment too.

The multiplicative structure of F-{0} is an abelian group. To turn F into a 3-field, we must first find an ordinary field whose additive group is isomorphic to (F-{0}, *).

Completly agree up to here.
The rest is probably right, I have to think more about it.

I wanted to say I think I found an example of a 3-field.
I'll fix $+$ for the first operation, $\cdot$ for the second operation, and $*$ for the thirth operation.

$\begin{array}{l | lll} + & 0 & 1 & e \\ \hline \\ 0 & 0 & 1 & e \\ 1 & 1 & e & 0 \\ e & e & 0 & 1 \end{array} \ \ \ \ \ \ \ \ \ \ \begin{array}{l | lll} \cdot & 0 & 1 & e \\ \hline \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & e \\ e & 0 & e & 1 \end{array} \ \ \ \ \ \ \ \ \ \ \begin{array}{l | lll} * & 0 & 1 & e \\ \hline \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 \\ e & 0 & 1 & e \end{array}$

Notice how $(F,+,\cdot)$ is the field $F_3$, and how $(\{1,e\}, \cdot, *)$ is the field $F_2$. That fixes the "last four places" in the table for $*$. The first row and column in that table I set to $0$ and it seems to work this way, but I'm not sure if it wouldn't also work with other values. That is, I'm not sure if this "3-Field" is unique up to "3-field isomorfism".

Am I making sense?
Thanks,
Damián.

 

[Damidami]

Just for clarifying, I want to add what's the current "definition" I'm making for an "n-field".

An n-field is a set $F$ together with $n \in \mathbb{N}$ operations, let's call them $+_1, +_2, \ldots, +_n$ each satisfying:

Let $a,b,c \in F$, and $1 \leq i \leq n$

1) Associativity.
$a +_i (b +_i c) = (a +_i b) +_i c$

2) Commutativity
$a +_i b = b +_i a$

3) Existence of identity
There exists $0_i \in F$ such that $a +_i 0_i = a$

We also require $0_i \neq 0_j$ for $1 \leq j < i$

4) Existence of inverses
For every $a \in F$ with $a \neq 0_j$ with $1 \leq j < i$ there exists $-a \in F$ such that $a +_i (-a) = 0_i$

5) Distributive over the last operation
For all $a,b,c \in F$, and $2 \leq i \leq n$ the following equality holds:
$a +_i (b +_{i-1} c) = (a +_i b) +_{i-1} (a +_i c)$

-----​

For $n=1$ we use the notation $+_1 = +$ and the definition of the 1-field coincides with the definition of the commutative group $F = (G,+)$

For $n=2$ we use the notation $+_1 = +$ and $+_2 = \cdot$, and we get the usual definition of the field $(F, +, \cdot)$

For $n=3$ I'm using the notation $+_1 = +$, $+_2 = \cdot$ and $+_3 = *$, and is the first new case.

 

[Hurkyl]

Notice how $(F,+,\cdot)$ is the field $F_3$, and how $(\{1,e\}, \cdot, *)$ is the field $F_2$.

Ah right; I forgot to check the opposite parity. This should be the only case where (F,+,.) is finite with odd cardinality.

That is, I'm not sure if this "3-Field" is unique up to "3-field isomorfism".

Every 3-field isomorphism is a field isomorphism of the ordinary fields you get by forgetting the third operation.

If (F,+,.) is the field of 3 elements, then the only field automorphism of (F,+,.) is the identity -- so every 3-field automorphism of (F,+,.,*) is also the identity.

There is only one way to extend the group (F - {0},.) to a field (F - {0},.,*).

Therefore, the number of non-isomorphic 3-fields with 3 elements comes down to how many different ways you can assign values to 0*x.

For some reason I had convinced myself that defining 0*x=0 wouldn't work, but I guess you checked it out and it does. Have you checked 0*x=1? I don't know if there are any other options.

 

[DonAntonio]

I was thinking about some similarities in the definitions of group and field, and if it would be possible to generalize in some sense, like follows.

A field is basically a set F, such that (F,+) is a commutative group with identity 0, and (F-{0}, .) is a commutative group with identity 1, and . distributes over +, that is a.(b+c) = a.b + a.c

If I call n the number of operations in the algebraic structure, then n=2 for fields, if I set n=1 there is left only the first operation + (and no distributive law) so I get the definition of a commutative group.

Is it possible to generalize and get some consistent definition of an "n-field"? That is, for example, a 3-field has 3 operations (+, . ,*) with 3 distinct identities (0,1,e) such that (F-{0,1},*) is a commutative group with identity e and
1) a.(b+c) = a.b + a.c (. distributes over +)
2) a*(b.c) = a*b . a*c (* distributes over .)

I can't find any inconsistency, but neither can I construct an example of such a 3-field.

If it could be possible, one in principle could prove things about n-fields, and then letting n=2 one would get a proof about fields, and letting n=1 one would get a proof about commutative groups (that's the goal I had in mind).

Can anyone construct an example of a 3-field, or show is has no sense at all?
Thanks.

I'm not sure whether the following is what you're looking for but you can take any field K and look at the structure of vector space over any subfield F of K (for definitiness, take K = ℂ , F = ℝ).

Of course, the operation "multiplication by scalar" is not between elements of ℂ but between elements of ℝ and ℂ .

In the above case we have an algebra K over a field F.


Tonio

 

[Damidami]

Hi DonAntonio,
Thanks for your answer.
I'm aware of those things, but it's different from what we are discussing here. In what we call a 3-field, the three binary operations have to be defined on the same set F, we use no external 'scalar' set, so that's a different issue.