Isomorphism from Ring $\mathbb Z$ to $\mathbb Z \times \mathbb Z$ I'm to show there does not exist any ring isomorphism from $\mathbb Z$ to $\mathbb Z \times \mathbb  Z$.
Note: I know how possible ring homomorphims would look like but need to find suitable contadiction! Please help out!
 A: What is Z*Z? If this is $\mathbb{Z}\times\mathbb{Z}$ then they are not isomorphic because $\mathbb{Z}$ has not divisors of zero, but  $\mathbb{Z}\times\mathbb{Z}$ has them ($(1,0)(0,1)=(0,0)$).
A: Although the question has been answered, let me sketch some general procedure how to show that two rings or in fact any two mathematical objects are not isomorphic: One has to find invariants. The invariant should take isomorphic objects to equal invariants, or rather isomorphic invariants if these invariants are not just numbers. If we even take into account arbitrary homomorphisms, these invariants are called functors. One can argue that functors are one of the most important inventions of $20$th century pure mathematics.
For example, we have the functor which maps a ring $R$ to its group of units $R^*$. Isomorphic rings have isomorphic group of units. Now $\mathbb{Z}^* = \{\pm 1\}$ has $2$ elements, but $(\mathbb{Z} \times \mathbb{Z})^* = \{(\pm 1, \pm 1), (\pm 1,\mp 1)\}$ has four elements. Hence these groups are not isomorphic. Hence the rings $\mathbb{Z}$ and $\mathbb{Z} \times \mathbb{Z}$ are not isomorphic.
Here is another functor: It maps a ring $R$ to the $R/2R$. Notice again that $\mathbb{Z}/(2)$ has two elements, but $(\mathbb{Z} \times \mathbb{Z})/(2) = \mathbb{Z}/(2) \times \mathbb{Z}/(2)$ has four elements. Actually this functor also goes from abelian groups to abelian groups. It can be used to show that $\mathbb{Z}^n \cong \mathbb{Z}^m$ as abelian groups if and only if $n=m$.
