Show that in a field always $0\ne1$ Suppose that $F$ is a field and prove that $0\ne1$
According to the definition of a field I know that the zero element is different from the one element, but is there a scientific proof for that?
 A: In a field, $0\neq 1$ by definition. This is to exclude the zero ring from being a field.
This leads to a natural question which is an easy but illuminating exercise: If we are in a ring with $0=1$, show that every element equals $0$.
A: Some field axioms do not include $\,0 \ne 1\,$ but, rather, deduce it from other axioms. For example, the axioms may state that $\, F\setminus0\,$ is a monoid, hence the neutral element $1$ of the monoid is distinct from the neutral element $\,0\,$ of the additive group of $\,F.\,$ Therefore, in this axiomatization of a field, $\,0\ne 1\,$ is not an axiom but, rather, a derived theorem (albeit a trivial one). Thus it may well be that the exercise in your question was posed in the context of this or a similar axiom system.
Remark $\ $ This thread discusses the rationale for excluding the one-element ring from the class of fields, so it may be of interest. Beware that there are occasional (inconsequential) errors when using axioms involving monoids and unit groups,  as the linked comment shows.
A: For every $x \in F $ we have $$0 \cdot x = 0 $$ because $$0\cdot x = (0 + 0 ) \cdot x = 0\cdot x + 0 \cdot x$$ by the axioms of field.
But also $$1 \cdot x = x \ \ \forall x \in F$$
and so if $0 = 1 $ we have $$x = 1\cdot x = 0 \cdot x = 0 \ \ \forall x \in F$$
and then $F = 0 $
A: There is a principle in proof theory that if you can find a set of variables such that axioms hold but a conclusion does not, then it is impossible to prove the conclusion from the axioms.
For example, suppose you know the following axioms:
$$x > y \tag{A1}$$
$$x > z \tag{A2}$$
Call $P$ a proposition to be proven as $y > z$.  Well, if we let $x = 3, y = 1, z = 2$.  Note that under this assignment, A1 and A2 are true, but $P$ is false.  Therefore you cannot prove $P$ from A1 and A2.
There are 10 or so axioms of field theory.  The one element field satisfies the axioms of field theory.  The one element field also satisfies $0_F = 1_F$, therefore it is impossible to prove $0_F \ne 1_F$ from just field axioms.
