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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.

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Needs are relative to purposes. Need to do what? –  Qiaochu Yuan Sep 12 '11 at 19:58
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Without an identity element, you cannot define inverses. –  Rasmus Sep 12 '11 at 20:00
    
Do you want an additive identity, or a multiplicative identity? You original post had $0$ in it, now it says "identity element". –  Arturo Magidin Sep 12 '11 at 20:13
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@Stephan: in your last sentence, what is "it"? –  Qiaochu Yuan Sep 12 '11 at 21:10
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Is it so hard to imagine a "field" without a zero? Why is this such a silly question to you. I wanted to know if, like @Zhen wrote, there is a mathematical structure like a field but without a zero. So if I add x and -x it would disappear instead of becoming a 0. Is it possible to have such a structure and how does it look like and what cant you do with it. In my humble opinion some of you take this Q&A site too serious and should stick to questions of their level instead of making some novices feel dumb and get lost in personal issues... –  Stephan Schielke Sep 13 '11 at 6:03

6 Answers 6

up vote 4 down vote accepted

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

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..."


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...

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Do you get advertising revenue from wikipedia or something ??! $$$$;) –  The Chaz 2.0 Sep 14 '11 at 1:17
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Haha I wish :) I just usually err on the side of putting more links in my answers, than fewer (as long as they're all relevant). –  Zev Chonoles Sep 14 '11 at 1:26
    
And I appreciate it! I recently saw a diagram that related (if memory serves...) monoids, fields, groups, etc. Maybe I can find the link... –  The Chaz 2.0 Sep 14 '11 at 1:29

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.

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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.


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.

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Hmmm. If Axiom 3 is dropped and Axiom 4 is replaced with 4a: 'For every $a, b\in F$ there exists $c\in F$ s.t. $a+c=b$' then you still recover 3 by letting $b=a$ in 4a. But if we restrict 4a to $b\neq a$ then there aren't any immediately obvious 'hooks' to get back a zero... –  Steven Stadnicki Jul 1 '12 at 16:10

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.

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I am confused, isn't this answer essentially the same as those in math.stackexchange.com/questions/164811 which garnered over 30 upvotes combined, and not a single downvote? Why is this answer being downvoted? –  Asaf Karagila Jun 30 '12 at 23:11
    
@Ben: there will be a edit log! But okay... –  Asaf Karagila Jul 1 '12 at 15:20

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.

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The idea that axioms in a formal language "make reference" to semantic concepts like the existence of an additive identity is a somewhat naive philosophical stance that seems entirely antithetical to the relentless syntacticism you have been espousing on this site. If I wanted to play the usual Doug Spoonwood role I would say: "The existence of an additive identity comes as a certain WFF. You either have that WFF as an axiom or you don't. Other axioms might imply the existence of an additive identity, but that doesn't come as the same thing as having the additive identity axiom." –  Pete L. Clark Sep 13 '11 at 14:59
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I confess that I am not too surprised that I inadvertently clarified your position. But...you now believe that informal mathematical statements may be formalized as WFFs in multiple ways and that this is not problematic? That's good news. –  Pete L. Clark Sep 14 '11 at 0:17
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By the way, the standard symbols for universal and existential quantifiers are $\forall$ and $\exists$, not ! and @. The advantage of using $\forall$ and $\exists$ is that a worldwide mathematical/philosophical audience will understand you without your having to define your symbols in each comment...and incidentally, you will look less silly. –  Pete L. Clark Sep 14 '11 at 0:20
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I intentionally did not put all the parentheses in to see if you now understand that people refer to WFFs by standard shorthands rather than writing them out formally. (Why did you not complain about the $\ldots$? That's definitely forbidden in WFFs.) If you are still hung up on these syntactic points, then I suspect that what I am actually trying to say is falling on deaf ears. (By the way, your language is not very polite. I have a PhD in mathematics and have published a paper in mathematical logic. Are you sure you want to accuse me of not understanding WFFs?) –  Pete L. Clark Sep 14 '11 at 1:30
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$@$Doug: do you **really** not understand that I can unambiguously specify a WFF without writing it out in full, and that reasoning with and proving things about WFFs is not the same as writing them down? And why do you refuse to use $\forall$ and $\exists$? I am really having a hard time believing that you are making these comments in good faith. –  Pete L. Clark Sep 14 '11 at 5:18

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.

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Of course your axioms do not imply $x+ 0 = x$ for all $x$. Take for instance any set $X$ with more than one element: call one of the elements $0$. Define $-x = x$ for all $x \in X$ and $x+y = 0$ for all $x,y \in X$. Then $0$ is an absorbing element, not an identity....But this structure is nothing like a field -- indeed, you only have one binary operation! So what is your point here? –  Pete L. Clark Sep 14 '11 at 1:18
    
@Pete As I try to clear up in the edit, we now have a structure which satisfies the property of having an inverse (3. comes as equivalent to such a property, as I understand it), and satisfies addition and commutation. In other words, it satisfies all axioms of an additive Abelian group without the existence of an additive identity, and it comes as consistent. So, if you join in the other axioms for a field under a formulation like that of Asaf's, the set of axioms for a field without that of an additive inverse is NOT inconsistent. So, such structures DO come as possible. –  Doug Spoonwood Sep 14 '11 at 1:48
    
$@$Doug: if you are saying that your modified set of axioms is consistent, then this is obvious because any field satisfies these axioms. If you are saying this axiom system together with the additional axiom that $0$ is not an additive identity is consistent: yes, this is true. Take any field $F$ and redefine the addition operation to be identically $0$ as above, define $-x = x$ for all $x$, and keep the same multiplication operation. –  Pete L. Clark Sep 14 '11 at 5:11
    
Two questions: 1) Who said that such structures were not possible? Certainly not me: I was the one who gave you a model! (You said "I simply don't see how this implies...", which is not a proof by anyone's standard.) 2) In what sense is a structure in the above example "like a field" as the OP requests? What results can you prove about such structures? Being given by a similar list of axioms is a very superficial similarity between structures. –  Pete L. Clark Sep 14 '11 at 5:11
    
@Pete No you didn't say such weren't possible. Sorry if it seemed like I implied that you did. The OP as it stands currently says this... and yes I certainly did not give a proof. Yes, this answer doesn't give as much information as the OP requests. But, I'd think you'll probably have to use a formulation like this, because of what Arturo says (if he's correct) "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." –  Doug Spoonwood Sep 14 '11 at 12:32

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