Nillradical is prime ideal, then the ring is not a product ring.

Question

Let $A$ be a commutative ring and $nill（A）$　is not a prime ideal. This is just a characterization of $SpecA$ to be irreducible.

Then, according to the argument of general topology, irreducible deduces connected. So,to translate this into ring theory, We should like to say 'if $nill（A）$　is not a prime ideal, then $A$ cannot be a product of non-zero ring.'

But I cannot show the last proposition with ring theory. Thank you for your help, my teachers.

Answer 1

I suppose the rings have unit. Suppose that $A=U\times V$. $(1,0)(0,1)=(0,0)$ implies $(1,0)\in Nil(A)$ or $(0,1)\in Nil(A)$ contradiction since $(1,0)$ and $(0,1)$ are not nilpotent.