Field axioms: Why do we have $ 1 \neq 0$? In the definitions of a field, we have $ 1 \neq 0$.
I know that in regular multiplication $0 \times 1=0$ but for reciprocal we don't have inverse of $0$.
But all the spaces and different definitions of multiplications that are satisfying field axioms, why do we need $ 1 \neq 0$?
Please do not use too technical of terminology. I am reading Baby Rudin right now.
 A: As pointed out already in another answer, a ring in which $1=0$ just consists of a single element as for example 
$$a=1a=0a=(0+0)a = (1+1)a= a+a$$ 
shows $0=a$ for each $a$.
And, one does not want this structure, just $\{0\}$, to be a field, since this would be inconvenient, since then on would one would write all the time let $K$ be a field other than the trivial field, instead of just writing let $K$ be a field. 
For example, there is no reasonable notion of a (non-trivial) vectorspace over that "field" so what is $K^2,K^3$ and so on in that case? Again the trivial vectorspace, but then the dimension of $K^n$ is not $n$ anymore for this "field". 
Polynomials over that ring make not much sense either and one can continue in this way.
A: If $1 = 0$, the field is only $\Bbb K = \{0\}$. $$x = x \cdot 1 = x \cdot 0 = 0, \quad \forall\, x \in \Bbb K$$
A: If 1=0, by definition of these two elements, one would have, for every element $a$ of this "field,"  $a=1a=0a=0$ as @quid has shown, in other words the ground set for this field would simply be a singleton. So there would be not much excitement or usefulness in this field. It is like lack of excitement in considering log base 1 -- it is no fun, or for that matter considering 1 to be a prime number. 
