Is there any elegent way to formally prove that the ring $\mathbb Z/(8)$ cannot be decomposed to a product of rings? So far, my idea is that:
As a additive group, $\mathbb Z/(8)$ is the cyclic group $\mathbb Z_8$. So to find two rings that give a product ring that is isomorphic to $\mathbb Z_8$. The order of the product ring must be 8.
We know that $|H*K| = {|H| \; |K|}$. As the factors of 8 are 1,2,4,8, and we cannot involve the trivial ring, we need to try groups of order 2 and order 4.
There is only one structure of group of order 2:$\mathbb Z_2$.
But there are two structures of group of order 4: $\mathbb Z_4$ and the Klein four group. Neither of these $2$ groups gives us a cyclic group of order $8$.
Hence the $\mathbb Z_8$ cannot be decomposed to the product of 2 rings.
But I find this method tedious...
Could someone give a simpler and nicer way to prove it? Thanks so much!
 A: My favorite way to do this is that ring decompositions of a ring correspond to central idempotents. Nontrivial decompositions correspond to nontrivial idempotents.
In this case, six trivial checks suffice to prove it can't be decomposed:
$$
2^2\equiv 4\\
3^2\equiv 1\\
4^2\equiv 0\\
5^2\equiv 1\\
6^2\equiv 4\\
7^2\equiv 1\\
$$
Or, if that's too ugly, think about what it would mean to be idempotent:
If $8$ divides $x-x^2 = x(1-x)$ would imply that $2$ divides one of $x$ or $1-x$. Of course it can't divide both, so once you decide which one is divisible by $2$, it is also divisible by $8$ (and hence is zero in $\mathbb Z/(8)$.) At that point you know one of $x$ and $1-x$ is zero and the other is $1$.
Another reason that this ring can't be decomposed is that it's a local ring, that is, it has a unique maximal ideal. Any ring with a nilpotent maximal ideal is local, and that's the case here.
However the check above works even for rings that aren't local, like integral domains. (And it is not always necessary to manually check every element as I did above.)
A: For this particular case, the order of an element of $\mathbb Z_2\oplus H$ it at most 4: $$4(0,x)=(0,4x)=(0,0)$$ and $$4(1,x)=(0,4x)=(0,0).$$ 
In $\mathbb Z_8$, the element 1 has order 8. So $\mathbb Z_8$ and $\mathbb Z_2\oplus H$ cannot be isomorphic. 
