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.

Solution 1:

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 $$0 \cdot a=a\\(0-0)\cdot a=a\\0\cdot a - 0 \cdot a=a\\a-a=a\\0=a$$

Solution 2:

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 $1 \neq 0$.

Solution 3:

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.

Solution 4:

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