# Field with one element and Chaitin's Constant

Some quotes from Wikipedia:

In mathematics, the field with one element is a suggestive name for an object that should behave similarly to a finite field with a single element, if such a field could exist. This object is denoted $$F_1$$, or, in a French–English pun, Fun. The name "field with one element" and the notation $$F_1$$ are only suggestive, as there is no field with one element in classical abstract algebra.

The possibility of studying the mathematics of $$F_1$$ was originally suggested in 1956 by Jacques Tits, published in Tits 1957, on the basis of an analogy between symmetries in projective geometry and the combinatorics of simplicial complexes. $$F_1$$ has been connected to noncommutative geometry and to a possible proof of the Riemann hypothesis. Many theories of $$F_1$$ have been proposed, but it is not clear which, if any, of them give $$F_1$$ all the desired properties.

https://en.wikipedia.org/wiki/Field_with_one_element

The axioms of a field are as follows:

Formally, a field is a set $$F$$ together with two operations called addition and multiplication. An operation is a mapping that associates an element of the set to every pair of its elements. The result of the addition of $$a$$ and $$b$$ is called the sum of $$a$$ and $$b$$ and denoted $$a + b$$. Similarly, the result of the multiplication of $$a$$ and $$b$$ is called the product of $$a$$ and $$b$$, and denoted $$ab$$ or $$a⋅b$$. These operations are required to satisfy the following properties, referred to as field axioms. In these axioms, $$a$$, $$b$$ and $$c$$ are arbitrary elements of the field $$F$$.

Associativity of addition and multiplication: $$a + (b + c) = (a + b) + > c$$ and $$a · (b · c) = (a · b) · c$$.

Commutativity of addition and multiplication: $$a + b = b + a$$ and $$a · > b = b · a$$.

Additive and multiplicative identity: there exist two different elements $$0$$ and $$1$$ in $$F$$ such that $$a + 0 = a$$ and $$a · 1 = a$$.

Additive inverses: for every $$a$$ in $$F$$, there exists an element in $$F$$, denoted $$−a$$, called the additive inverse of $$a$$, such that $$a + > (−a) = 0$$.

Multiplicative inverses: for every $$a ≠ 0$$ in $$F$$, there exists an element in $$F$$, denoted by $$a^{−1}$$, $$1/a$$, or $$\frac{1}{a}$$, called the multiplicative inverse of $$a$$, such that $$a · a^{−1} = 1$$.

Distributivity of multiplication over addition: $$a · (b + c) = (a · b) > + (a · c)$$.

This may be summarized by saying: a field has two operations, called addition and multiplication; it is an abelian group under the addition, with 0 as additive identity; the nonzero elements are an abelian group under the multiplication, with 1 as multiplicative identity; the multiplication is distributive over the addition.

https://en.wikipedia.org/wiki/Field_(mathematics)

Chaitin's constant is defined as follows:

Let $$P_F$$ be the domain of a prefix-free universal computable function $$F$$.

The constant $$\Omega_F$$ is then defined as $$\Omega _{F}=\sum _{p\in > P_{F}}2^{-|p|}$$,

where

$$\left|p\right|$$ denotes the length of a string $$p$$. This is an infinite sum which has one summand for every $$p$$ in the domain of $$F$$. The requirement that the domain be prefix-free, together with Kraft's inequality, ensures that this sum converges to a real number between $$0$$ and $$1$$. If $$F$$ is clear from context then $$\Omega_F$$ may be denoted simply $$\Omega$$, although different prefix-free universal computable functions lead to different values of $$\Omega$$.

https://en.wikipedia.org/wiki/Chaitin%27s_constant

As I understand it the digits of a Chaitin constant are undecidable, hence it is undecidable whether the set $$\{\Omega_F\}$$ together with the definitions $$\Omega_F + \Omega_F = \Omega_F$$ and $$\Omega_F \centerdot \Omega_F = \Omega_F$$ for addition and multiplication respectively, forms a field.

Thus we can assume that it does, taking this assumption as an axiom.

Does the assumed field $$\{\Omega_F\}$$ meet the requirements for a field with one element, does it have the necessary properties for why such an object was originally suggested?

• The last sentence of your first quoted paragraph says : "The name "field with one element" and the notation $F_1$ are only suggestive, as there is no field with one element in classical abstract algebra." This means that the "field with one element" is NOT defined as a one-element set endowed with some operations, as you suggest. And in any case, the only singleton where "usual definitions of real addition and multiplications" can define operations is $\{0\}$. – Arnaud D. Mar 5 at 11:19
• One consequence would be that the single element in the field would be both the zero - and the one - element. I do not think that this is allowed in a field. – Peter Mar 5 at 11:34
• You are mixing two relatively unrelated things. The field with one element does not exist - it is an analogy only. Every field has at least two elements. There is no real $x$ so that $\{x\}$ is a field. – Carl Mummert Mar 5 at 12:20
• @YeatsL: when they say "the digits are undecidable" they only mean there is no computer program to print them out. The digits for each Chaitin number are completely well defined and specifically specified, and there is one Chaitin number for each universal prefix free code. In the end the Chaitin numbers are just a particular family of uncomputable real numbers. There are many other uncomputable real numbers as well, and nothing about them is contradictory. They behave just like all other real numbers. Some people give too much hype to Chaitin numbers, which are not really very special. – Carl Mummert Mar 5 at 12:36
• @YeatsL Digits are, in particular, numbers; numbers follow specific laws, namely that a number is always equal to itself. It doesn't matter at all whether we know the number or not, whether we can compute it or not; if it is a number, it is equal to itself. – lisyarus Mar 5 at 12:53

You are mixing two relatively unrelated things. The field with one element does not exist - it is an analogy only. Every field has at least two elements. There is no real $$x$$ so that $$\{x\}$$ is a field.