If I'm not mistaken, in a ring with identity, the additive identity cannot have a multiplicative inverse. I'm trying to prove this.
Here's my attempt so far:
Suppose $0\cdot a=1$
$$0\cdot a=1$$
$$0\cdot (a + n) = 0\cdot a + 0\cdot n$$
It's easy to see here that if $n$ were a positive integer (same works for negative):
$$n=1+1+1+\cdots$$
$$0\cdot a + 0\cdot (1 + 1 + 1+\cdots )$$
$$0 \cdot a + 0 + 0 + 0 + \dots = 0\cdot a$$
$$\therefore 0\cdot n = 0$$
If $a$ is some whole number of multiplicative inverses summed together, then this is a contradiction.
This hinges on the assumption that $n$ is some integer multiple of the multiplicative unit, which it may not be. This also got me thinking about whether the distributivity of multiplication over addition is discretized in whole terms (you can only use operators with whole numbers of terms) or whether the concept of distrubitivity actually goes deeper than this.
Mainly I'd like a proof that the additive identity can not have a multiplicative inverse and why $0\neq 1$. But some deep insight into why the properties of a ring lead to this would be a bonus.
Best Answer
$$0\cdot a= (0+0)\cdot a=0\cdot a+0\cdot a$$ Now substract $(0\cdot a)$ on both sides (we can do this because $\forall r\ \exists(-r)\mid r-r=0)$:$$0=0\cdot a$$ This means that $0$ can only have a multiplicative inverse if $0\cdot \tilde a=0=1$ for some $\tilde a$. This then implies that we have $x=x\cdot1=x\cdot0=0\ \ \forall x$, hence we live in the zero-ring $\{0\}$.
So for any ring $R$, we have that $0$ has a multiplicative inverse $\iff$$R=\{0\}$