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I am thinking about construction of a structure that "obeys" all of the axioms for a field except an axiom of distributivity of multiplication over addition, and I am not sure is that possible at all?

I mean, it ought to be possible, because, it seems that an axiom of distributivity is not necessarily redundant in a sense that it is implied by other axioms of a field.

But I am really really not sure.

Do you have somewhere already an example of such a structure, it could even be some familiar field with addition and multiplication specially defined to suit the purpose.

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    $\begingroup$ As I emphasized here in 2010 when discussing the the law of signs: the distributive law is the keystone of the ring structure, since it is the only law that connects the additive and multiplicative structures. Removing it from the ring axioms leaves a much poorer structure with two completely unrelated additive and multiplicative structures. $\endgroup$ Aug 4, 2023 at 16:40
  • $\begingroup$ This may be relevant: arxiv.org/pdf/0707.4024.pdf $\endgroup$
    – Anixx
    Sep 12, 2023 at 9:57

4 Answers 4

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Without the distributive axiom, there is no connection between the addition and multiplication operations at all (other than that they happen to be defined on the same set). So, you could take any set $S$ and consider any abelian group structure on $S$ at all as addition and any abelian group structure on $S\setminus\{0\}$ at all as multiplication (and $0$ times anything is $0$). Or, the multiplicative group structure could include $0$, with $0$ having an inverse, as long as $0$ is not the multiplicative identity.

For an explicit example, for instance, let $S=\{0,1\}$, with addition being the usual addition mod $2$, and multiplication being addition mod $2$ with the roles of $0$ and $1$ swapped (so $0\cdot 0=1$, $0\cdot 1=0$, $1\cdot 1=1$). This then satisfies all the field axioms except distributivity.

For another example, you could let $S$ be a set with $6$ elements, with addition having the structure of $\mathbb{Z}/6$ and multiplication of nonzero elements having the structure of $\mathbb{Z}/5$ (and $0$ times anything is $0$). This cannot be a field since there is no field with $6$ elements, but it satisfies all the axioms except distributivity.

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  • $\begingroup$ Hmm, $\overline{\mathbb R}$ is not distributive but algebraically quite useful. I think, at least sdistributivity regarding scalar multiplication is essential. $\endgroup$
    – Anixx
    Sep 9, 2023 at 3:50
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There are several families of algebraic structures that are "almost" fields.

  1. Skew fields (aka division rings) satisfy all of the conditions of a field, except that the multiplication is not required to be commutative. If we have a finite structure, there are no skew fields that are not in fact fields.
  2. Near fields satisfy all of the conditions of a skew field, except that we require only one of the distributive laws.
  3. Semifields satisfy all of the conditions of a skew field, except for associativity of multiplication.
  4. Quasifields (as mentioned by Berci) are essentially the weakest of the batch: they are not required to have associativity of multiplication, and are only required to have one of the distributive laws.

These are all interesting to consider as finite structures, because we do have nontrivial examples of finite near/semi/quasifields. There are even finite semifields with commutative multiplication that are not examples of finite fields.

The smallest example of a near field has order 9, and is due to Hall. We take an irreducible polynomial of $\mathbb{F}_{3}$, say $f(x) = x^{2}+x+2$. You use this to define a new multiplication on $\mathbb{F}_{3}^{2}$ by $$(x_{1},y_{1})\cdot(x_{2},0) = (x_{1}x_{2}, y_{1}x_{2}),$$ and $$(x_{1},y_{1})\cdot(x_{2},y_{2}) = (x_{1}x_{2}-y_{1}y_{2}^{-1}f(x_{2}), x_{1}y_{2}-y_{1}x_{2}+2y_{1}).$$ (The 2 as coefficient in the last term comes from the polynomial $f(x)$, if $f(x) = x^{2}+ax+b$, that coefficient is $-a$.)

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    $\begingroup$ Note that the multiplicative group of the given nearfield of 9 elements is the quaternion group, and we can represent it by additive basis $1,i$ over $\Bbb F_3$ and setting $j=i-1$ and $k=i+1$. $\endgroup$
    – Berci
    Jan 11, 2020 at 14:01
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Quasifields are not commutative and they obey distributivity only on one side. They arise in the study of (finite) projective planes.

In fact, there are many irregular projective planes that induce irregular field-like structure.
Briefly, remove a line from the given projective plane and regard it the 'line at infinity', then fix a line, on which easy geometric constructions - using parallels - lead to an 'addition' and a 'multiplication', which in general may satisfy only some relaxed versions of a few field axioms.

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Anti-dual numbers. Based on similar principles you can build an algebraic system taking any symmetric (against both x and y axes) and crossing x axis at 1 and -1 curve as a unit circle.

Using lemniscate you get anti-hyperbolic numbers. Using diagonal square you get numbers with taxicab/manhattan metric.

Let's take a Cartesian plane equipped also with polar coordinates.

Consider a contour around the origin, defined in polar coordinates as $r=r(\phi)$. The contour is symmetric against the $x$ and $y$ axis-es and crosses points $(1,0)$ and $(-1,0)$. The contour defines the set of points to which we assign magnitude $1$.

Now, an arbitrary point, corresponding to a 2-dimensional number on the plane $z=(a,b)$ is characterized by angle $\alpha(z)=\arctan(b/a)$, magnitude $M(z)=\frac{\sqrt{a^2+b^2}}{r(\alpha(z))}$ and argument $\operatorname{arg}(z)=\int_0^{\alpha(z)} r(\phi)^2 d\phi$.

The addition of numbers is defined element-wise as $(a_1,b_1)+(a_2,b_2)=(a_1+a_2,b_1+b_2)$.

The multiplication is defined in such a way that the arguments are added and magnitudes are multiplied: $\operatorname{arg}(uv)=\operatorname{arg}(u)+\operatorname{arg}(v)$ and $M(uv)=M(u)M(v)$. This uniquely gives us a point which corresponds to the product. Multiplication defined this way is commutative.

Given these definitions, the usual complex numbers correspond to $r(\phi)=1$, split-complex numbers correspond to $r(\phi)=\frac{1}{\sqrt{\cos ^2(\phi)-\sin ^2(\phi)}}$ and dual numbers correspond to $r(\phi)=\frac1{|\cos \phi|}$.

Now, the most promising shapes are the reciprocals of the shapes of split-complex and dual numbers, which are lemniscates:

$r(\phi)=\sqrt{\cos ^2(\phi)-\sin ^2(\phi)}$

enter image description here

In this system $M(z)=\frac{a^2+b^2}{\sqrt{2 a^2-\left(a^2+b^2\right)}}$ and $\operatorname{arg} z=\frac{a b}{a^2 + b^2}$ (this formula is valid only for the first quarter, otherwise a piecewise definition is needed).

and

$r(\phi)=|\cos(\phi)|$

enter image description here

In this system the magnitude is $M(z)=\frac{a^2 + b^2}a$ and argument is $\operatorname{arg} z=\frac{1}{4}\left(\frac{a b}{a^2+b^2}+\arctan\left(b/a\right)\right)$ (this formula is valid only for the first quarter, otherwise a piecewise definition is needed).

As well as hexagonally symmetrical "hyperbola"

$r(\phi)=\frac{1}{\sqrt{\cos ^2\left(\frac{3 \phi }{2}\right)-\sin ^2\left(\frac{3 \phi }{2}\right)}}$

enter image description here

It seems that the first two have infinitely large elements, for instance, in the second one it is $\omega=(0,1)$. The third one has zero divisors and (seemingly) idempotents.

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  • $\begingroup$ Why are the downvotes? $\endgroup$
    – Anixx
    Aug 5, 2023 at 1:04

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