# Is {0} a field?

1. Aug 18, 2010

### murmillo

1. The problem statement, all variables and given/known data
Is {0} a field?

2. Relevant equations
field is ring where every non-zero a has a multiplicative inverse

3. The attempt at a solution
I thought {0} was a field, since it seems trivially true that every non-zero element has a multiplicative inverse. Since there are no non-zero elements in {0}. But today I came across the proposition that if R is a field, it has exactly two ideals. Which suggests to me that {0} is not a field, because then R would have only one ideal.

2. Aug 18, 2010

Every field has at least two distinct elements (i.e. the additive and the multiplicative identity), so {0} can't be a field.

3. Aug 18, 2010

### Hurkyl

Staff Emeritus
It depends on how you define field. It's rather typical to include $0 \neq 1$ as a field axiom.

Last edited: Aug 18, 2010
4. Aug 18, 2010

### Fredrik

Staff Emeritus
When 0≠1 is not included in the axioms, the characteristic of the field is defined as the smallest positive integer p such that $\sum_{i=1}^p 1=0$, or as 0 if no such integer exists. The field of real numbers has characteristic 0. The only field with characteristic 1 is the trivial field {0}. (Because 1=0 implies x=1x=0x=0). The characteristic is always 0,1 or a prime number.

5. Aug 18, 2010

### gomunkul51

6. Aug 18, 2010

### murmillo

radou: I think we must show that {0} is not a field first, then use that to prove that every field must have at least two elements.

I guess the reason why {0} is not a field is because (according to Artin) part of the definition of a field is that if F is a field then F-{0} is a group. But if {0} were a group I would have an empty group, which does not contain an identity element.

Yes, I think some books state 1 is not 0 in their definition of field, but in my notes it just said a ring where "every non-zero element has an inverse." I don't think that definition is correct. It should have said a non-zero ring where every non-zero element has an inverse, or something like that.

Thanks, guys.

7. Aug 18, 2010

### Fredrik

Staff Emeritus
The first of those two references includes the requirement 1≠0. The other one doesn't. As I (and Hurkyl) said above, both definitions make sense, but are not equivalent. {0} is a field according to one of the definitions, but not the other. I don't see a problem with letting {0} be considered a field. It's already considered a group, a vector space, etc.

8. Aug 18, 2010

### gomunkul51

an Empty Set has no elements that could satisfy any of the additive and multiplicative rules !
thus it is not a field :)

*** BUT If by {0} you meant not an empty set but a set with one element 0 (zero), it could be a field :)
I think it obeys all the rules..

P.S. the additive and multiplicative identities don't have to be 1 and 0 (could be any other element), they just have to exist in the set (field).

Last edited: Aug 18, 2010
9. Aug 19, 2010

### Fredrik

Staff Emeritus
Yes, {0} is a set with one element. The empty set is written either as $$\emptyset$$ or {}.

10. Aug 20, 2010

A field is a commutative ring with unity such that all nonzero elements are units.

{} is a commutative ring with no nonzero elements, so it does not violate this definition.

Why doesn't this suffice? Or does the empty set, by the same reasoning as above, not have a 1?

11. Aug 20, 2010

### Hurkyl

Staff Emeritus
{} has neither a 0 nor a 1, so it's not a ring. :tongue:

(The zero ring does have a 1. It's just equal to 0)

12. Aug 21, 2010

### HallsofIvy

Staff Emeritus
What "empty" group are you talking about? The set of objects, {0}, with addition is a group as is the set of invertible members with multiplication. (Note that 0*0= 0 and if 0 is the only member then 0 is also the multiplicative identity so 0 has a multiplicative inverse.)

13. Aug 21, 2010

### gomunkul51

I misunderstood {0} as being an empty set {}.

Last edited: Aug 21, 2010
14. Aug 21, 2010

### Office_Shredder

Staff Emeritus
The empty group isn't a group! It needs an identity element