Let A be a ring, in class we defined the following two definitions of connected:
A is Zariski connected if Spec A is connected with the Zariski topology.
A is ring connected if
$$a+b=1, ab=0\Rightarrow a = 0, b = 1$$
(or $b=0, a=1$)
I want to know if these are equivalent with a proof or counter example. I have succeded in showing Zariski connected implies ring connected. But all atempts at the reverse are failing, and I don't know where to start searching for a counterexample.
Any help would be appreciated.
After a little thought, this does turn out to be equivalent to the definition of "connected" that I have seen, which is "no nontrivial idempotents."
It's known that $\mathrm{Spec}(R)$ is connected iff $R$ has no nontrivial idempotents (see here).
Certainly the condition you mention implies there are no nontrivial idempotents (just consider $e$ and $(1-e)$.)
Conversely: suppose that there are $a,b$ such that $a+b=1$ and $ab=0$ and there are only trivial idempotents. If either one is zero we are done, so without loss of generality suppose both are nonzero.
The condition that $a+b=1$ tells us $(a)$ and $(b)$ are coprime ideals. As such, the Chinese Remainder Theorem tells us that $(ab)=(a)(b)=(a)\cap (b)$ and that $R\cong R/(a)\times R/(b)$ as rings. If we assumed that there are only trivial idempotents, one of these halves would have to be zero, for example, $(b)=R$, but that can't be the case since $b$ is clearly not a unit (it is a zero divisor.). Similarly $(a)= R$ leads to a contradiction.