Is it possible to have a field without an additive identity?

Question

If I drop the axiom that Zero is the identity of an addition what consequences does this entail? What do I need to change to my axiomatization?

By definition it is not possible, but are there mathematical structures like a field but without an additive identity?

I don't have a concrete example/issue but I am interested in this question.

Answer 1

Yes. It is part of the definition of a field (and, more generally, a ring).

People do study various alterations to the concept of field, such as

• division rings
• near fields
• quasifields
• semifields

but, I think it is worth mentioning, all of these have an identity element for the $+$ operation. It is a pretty fundamental thing to want one's mathematical structures to have. I'm not saying that structures without an additive identity aren't worth studying, just that they would behave so differently from fields I don't see any reason to even say "It's like a field, but..."

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Edit: Looking at the page on semifields, it appears that an alternative definition of the term does in fact allow for not having an additive identity. So much for objects getting names that make sense...

Answer 2

Yes, by definition. The need for an identity element is deeply entrenched in group theory (otherwise there could be no distinction between groups and monoids), and the fact that the additive group of a field is, indeed, a group, is similarly entrenched in field theory.

Of course one could investigate the consequences of the field axioms without an additive identity -- but one would have to be careful about selecting which axioms that means; the usual textbook presentations rarely distinguish between variants of the axioms that can be shown to be equivalent given the existence of an additive identity. In any case, whatever results you reach then would certainly not be results about fields as everyone understand them. Unless, that is, you happened to pick a selection of axioms that imply that zero must exist anyway. For example, it seems to be hard to require any useful properties of subtraction without allowing a proof that $x-x=y-y$ for all $x$ and $y$, and that this common value is an additive identity.

Answer 3

The usual definition of a field is:

Definition. A set $F$ together with two binary operations $+$ and $\times$ is a field if and only if:

1. $+$ is associative;
2. $+$ is commutative;
3. $+$ has a neutral element $0$;
4. For every $a\in F$ there exists $b\in F$ such that $a+b=0$.
5. $\times$ is associative;
6. $\times$ is commutative;
7. $\times$ distributes over $+$;
8. There is an element $1\neq 0$ such that $a\times 1=a$ for all $a\in F$;
9. For each $a\in F$, if $a\neq 0$ then there exists $x\in F$ such that $a\times x = 1$.

Now, you can certainly ask whether the axioms are independent. For example, it is easy to see that if you drop the assumption of commutativity for $+$, then you can deduce it from the other eight axioms; on the other hand, the real quaternions show that you cannot drop the assumption that $\times$ is commutative and deduce it from the other axioms.

So you could be asking whether Axiom 3 is independent of the other axioms. The main difficulty with dropping Axiom 3 is that without it Axiom 4 becomes unintelligible, and Axiom 9 is also problematic. So before we drop Axiom 3, we need to replace Axioms 4 and 9 with something else that, together with Axiom 3 give a field, but which make sense in the absence of Axiom 3.

There are plenty of ways of defining "abelian group" without specifying the existence of a neutral element (e.g., "for every $a,b\in F$ there exists $x\in F$ such that $a+x=b$") replacing 4 with something like this will automatically imply the existence of a $0$. Do you have something specific in mind?

(For a similar question, see for example this sci.math post by Dave Rusin where he discusses the independence of the axioms of a vector space, where he faces a similar problem with dropping "existence of $0$")

Added. While I was writing this, the title of the post was changed to "without an identity" instead of "without a $0$". You run into similar problems: if you drop Axiom 8, you need to replace Axiom 9 with something that still makes sense; depending on what you replace it with, it may or may not imply the existence of a multiplicative identity in the presence of the other axioms. Again, the question is whether you have something in mind or not.

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If the question is just a poorly phrased way of asking if the one element ring is a field, the answer is that it is not considered to be a field. There are good reasons for this, even though the one element ring satisfies all the axioms except for the $1\neq 0$ clause of Axiom 8.

Answer 4

This depends on the axioms, of course. There are two common ways to axiomatize fields. I was taught using the second approach. In both a field is composed of $\langle F,0,1,+,\cdot\rangle$

The First Way: Groups.

1. $\langle F,0,+\rangle$ is an abelian group;
2. $\langle F\setminus\{0\},1,\cdot\rangle$ is an abelian group;
3. Distributivity, that is $a\cdot(b+c)=a\cdot b+a\cdot c$

In this axiomatization, since a group cannot be empty, we have to have $0\neq 1$, otherwise the second condition will not hold.

The Second Way: Specifying the axioms.

1. $\forall a\forall b(a+b=b+a)$ (Commutativity of addition)
2. $\forall a(a+0=a)$ (Zero is the identity of addition)
3. $\forall a\exists b(a+b=0)$ (Addition is invertible)
4. $\forall a\forall b\forall c(a+(b+c)=(a+b)+c)$ (Addition is associative)
5. $\forall a\forall b(a\cdot b=b\cdot a)$ (Multiplication is commutative)
6. $\forall a(a\cdot 1=a)$ (One is the identity of multiplication)
7. $\forall a\exists b(a\neq 0\rightarrow a\cdot b=1)$ (Non-zero elements are multiplication-invertible)
8. $\forall a\forall b\forall c(a\cdot(b\cdot c)=(a\cdot b)\cdot c))$ (Multiplication is associative)
9. $\forall a\forall b\forall c(a\cdot(b+c)=(a\cdot b)+(a\cdot c))$ (Distributivity of multiplication over addition)

Note that these axioms are satisfied by $\{e\}$, by interpreting $e=0=1$. This sort of axiomatization allows the field with one element to exist.

If however, to the second approach we add $0\neq 1$ then we have equivalence between the two ways.

In both these cases we have $1$ defined in the language, and so we have to interpret it somehow, even if we do allow $0=1$, we still have $1\in F$. Simply because it is embedded into the definition of a field and cannot be avoided.

Answer 5

If I understand the definition of a field via abelian groups and distributivity as in Asaf's answer correctly, the perhaps surprising answer is no, you don't need to have an identity element in your axiom set for a field. The trick lies in that you don't need an identity element in the axioms for a group. Group theory, in fact, can get founded upon a single axiom, or several axioms which make no reference to a neutral element, if you really want that. I've only glanced at things here, so see 1 2 and 3 for particular axiom sets. Once you pick an appropriate axiom set for groups, you use the first part of Asaf's procedure for defining a field, which implies no mention of an identity element in the axiom set for fields. The existence of inverses also doesn't mean one needs to talk about the identity element in the axiom set. One could rewrite such existence statements via unary functions as one might do in the context of universal algebra, and this does capture the concept here, since inverses can get proved unique.

Answer 6

Suppose that we talk about a unary function "-" defined on F with -(0)=0, and -(-(x))=x. Note that instead of talking about a "inverses" for a field or group, we could just talk about those structures with such a unary function. Now let ⟨F,+, -⟩ satisfy the following properties:

1. For all x, y, (x+y)=(y+x)

2. For all x, y, z ((x+y)+z)=(x+(y+z))

3. For all x, (x+-(x))=0

I simply don't see how this implies that "for all x, (x+0)=x", so I don't see why we can't consider a field-like structure which drops only the axiom of an additive identity. How does this work? We basically have a commutative, associative structure which has an inverse also, since 3. comes as basically equivalent to the notion of having an additive inverse. So, now look at Asaf's way of defining a field. Instead of having ⟨F,0,+⟩ as an Abelian group, we have ⟨F,0,+⟩ as a structure where + satisfies association, commutation, and property 3. above. We leave the rest of Asaf's way of defining a field intact, so it comes as possible to only drop the axiom of an additive inverse for a field.