Let $R$ be a ring such that $a+b = a \cdot b, \; \forall \, a, b \in R$. Show that $R= \{0\}.$

Question

Let $R$ be a ring such that $a+b = a \cdot b, \; \forall \, a, b \in R$. Show that $R= \{0\}$.

I'm really lost. Does anyone have any idea?

Answer 1

Let $b=0$. Then we get $a+0=a\cdot0\implies a=0$. Since $a$ was arbitrary, $\mathcal R=\{0\}$

Answer 2

I was wondering if you could do this:

Assume it's not identically $0$ then there must exists an element such that it is not the zero element, say $a$ then observe $0\times a=0+a=a$...