Prove that in a ring with at least two elements $0\neq 1$.

Question

Let R be a non-trivial ring then prove $0\neq 1$.

Answer 1

Assume, to the contrary, that 1 = 0. Since $R$ is a non-trivial ring, there is an element $a\in ~R$ such that $a\neq0$. However, then $a=a·1=a·0=0,$ which is a contradiction.

Hence $1\neq0.$

Answer 2

Hint: Show by considering $x(0+0)$ that $x\cdot 0=0$.

Answer 3

An element $y$ which has both $0$- and $1$-element properties, leaves no room for other elements in the ring $R$. For an arbitrary $x \in R$ one has $$ x =^1 x y =^0 x (y+y)=^D xy + xy =^1 x + x \Rightarrow x = 0 $$ The uniqueness of the $0$ element would give that $R = \{ 0 \}$.