Why must be the additive and multiplicative identities in a field be different?

I was recently reading about fields like $\mathbb {Z}_p$ and I'm wondering what's the reason they can't be the same element.

Is it about the additive identity being the only element with no multiplicative inverse? To be honest it's the only thing that comes to mind but it still doesn't tell me why it should be like that.

It is possible to define a field with just one element, which has to be the additive and multiplicative identity at the same time. Most definitions exclude this from being a field. If you have at least two elements in your field and try to make the identities the same you fail. Call the common identity $0$ and the other element $a$. But then $$0 \cdot a=a\\(0-0)\cdot a=a\\0\cdot a - 0 \cdot a=a\\a-a=a\\0=a$$

• Ohhh ok, thanks! So, for example, when it is stated that "$\mathbb {Z}_p$ is a field iff $p$ is prime, they mean "non trivial" field? – Francisco José Letterio Nov 16 '17 at 18:53
• $\Bbb Z_1$ satisfies all the field axioms unless you have one that says $0 \neq 1$. Most people seem to include that, some do not. Forbidding the one element field avoids having it be an exception to many of the theorems, which is why it is more convenient. If you forbid it, you don't need "non-trivial" in your statement, for example. – Ross Millikan Nov 16 '17 at 19:07
• @RossMillikan: It depends on how you formulate the field axioms. One possible formulation is: (1) $(K,+)$ is an abelian group; (2) $(K\setminus\{0\},\cdot)$ is an abelian group, (3) The distributive law holds. This formulation is not easily adapted to permit $0=1$. – celtschk Nov 16 '17 at 20:46
• It should be noted that what is usually meant when one says "field with one element" is not the structure you are describing. Instead, it is usually called the zero or trivial ring. – tomasz Nov 16 '17 at 22:04
• I've never seen this object referred to as literally being a field. It doesn't really behave anything like a field. – Morgan Rodgers Nov 17 '17 at 7:05

Let $x$ be any element of your field. Then $1=0$ implies $$x = 1x = 0x =0.$$ Hence your field has only one element. So, you can allow it, but then your field is necessarily trivial. So, if you want a field with more than one element, you have to have $1 \neq 0$.

• I think in the spirit of this question we have to show that $0x=0$. It is not an axiom. – Ross Millikan Nov 16 '17 at 17:51
• It is however about ten seconds and two further equalities, that should be reasonably obvious (or an excellent problem to work on, for those learning). @RossMillikan – Nij Nov 16 '17 at 18:55
• @RossMillikan it is a well known property of rings and fields. $0=(0x)-(0x) = ((0+0)x)-(0x) =(0x)+(0x)-(0x)=0x$ – Henry Nov 16 '17 at 19:00
• @Henry I think Ross Millikan knows this. I don't think he commented because he wondered whether it was actually true. but rather because this question is about the utmost basics and axioms, and in that case we really should rely on as few theorems as possible, if nothing else so to be certain that we're not circular. – Arthur Nov 18 '17 at 13:57

A field is the same thing as a commutative simple ring. But the trivial ring is too simple to be simple.

If $\Bbb F$ is any field for which

$\mathbf 1_{\Bbb F} = \mathbf 0_{\Bbb F}, \tag 1$

we have, for any $\theta \in \Bbb F$,

$\theta = \theta \mathbf 1_{\Bbb F} = \theta \mathbf 0_{\Bbb F} = \mathbf 0_{\Bbb F}, \tag 2$

so $\Bbb F$ has only one element, $\mathbf 0_{\Bbb F}$; if we want to avoid the trivial case

$\Bbb F = \{ \mathbf 0_{\Bbb F} \}, \tag 3$

we must assume that

$\mathbf 1_{\Bbb F} \ne \mathbf 0_{\Bbb F}. \tag 4$

Note Added in Edit, Thursday 16 November 2017 9:52 AM PST: In "the spirit" of Ross Millikan's comment, we have:

$\mathbf 0_{\Bbb F} = \mathbf 0_{\Bbb F} + \mathbf 0_{\Bbb F}, \tag 5$

since $\mathbf 0_{\Bbb F}$ is the additive identity. Then

$\theta \mathbf 0_{\Bbb F} = \theta(\mathbf 0_{\Bbb F} + \mathbf 0_{\Bbb F}) = \theta \mathbf 0_{\Bbb F} + \theta \mathbf 0_{\Bbb F}, \tag 6$

by the distributive law, whence

$\theta \mathbf 0_{\Bbb F} = \mathbf 0_{\Bbb F}. \tag 7$

End of Note.

• I think in the spirit of this question we have to show that $\theta 0_\Bbb F=0_\Bbb F$ It is not an axiom. – Ross Millikan Nov 16 '17 at 17:50
• @RossMillikan: Done! – Robert Lewis Nov 16 '17 at 18:00

If $0=1$ in a ring, as others have commented, then that ring contains only a single element. There are many properties that a field has that simply do not hold for the ring $\{0\}$, even though there are no zero divisors. (I am going to go ahead and say it is "not considered a field by anybody", in contrast to some of the other answers. I cannot find a single book or article where this is object is literally considered a field, though it is sometimes treated as a generalization of a field for certain combinatorial reasons.)

The fact that we do not have zero divisors in a field $\mathbb{F}$ means that $\mathbb{F} \setminus \{0\}$ gives us a (commutative) group. This will obviously not work if we only have one element in the field, because throwing away $0$ would leave us with the empty set.

We often want to work with polynomials with coefficients from a field. How does this work if $0$ is the only field element?

Looking at ideals of a ring $R$, we define a maximal ideal to be a proper ideal I (so $I \subset R$) which is not contained in any other ideal. Then a field can be characterized as a ring for which $\{0\}$ is the only maximal ideal. But if $0$ is the only element of your field, $\{0\}$ is not a proper ideal.

For these reasons, we require a field to have a multiplicative identity $1 \neq 0$. There are some cases where a "field of one element" gives some useful interpretation, there are some geometric concepts that lend themselves to thinking in this way, and it comes up in category theory to help unify some concepts, but when this comes up there are usually bright disclaimers saying this is not literally a field.