Steps to prove or disprove if two rings are isomorphic So i'm struggling on how to prove if two rings are not isomorphic to one another. My professor told me that if a ring is not isomorphic to another, the best way to prove that this is true is to find a preserved property of isomorphisms that is not held.
So i considered the following:
1.Q and R (quotients and rationals)
2.$~~\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}$ and $\mathbb{Z}/16\mathbb{Z}$ (Z mod 4 cross Z mod 4 and Z mod 16)
I cannot seem to think of any of the properties:
communicative, identity, integral domain, and field property that do not hold for rings. My professor told me this it isnt enough to give an example mapping like F: Q -> R and show that it isnt isomorphic.
Hence, how can i show that these two problems above arent isomorphic?
Can anyone give me some finite steps to prove is something is isomorphic to something or not?
 A: HINT: $Z_{16}$ has an element of additive order $16$. Does $\Bbb Z_4\times\Bbb Z_4$?
Added: You don’t want to confine your attention to ‘big’ properties like commutativity or being an integral domain; often the differences are only to be found at a more detailed level.
A: 1.
The rationals $\mathbb Q$ and the reals $\mathbb R$ are not isomorphic because they don't have the same cardinality.
If you want an algebraic reason, there is an element $u \in \mathbb R$ such that $u^2=2$ but there is not such element in $\mathbb Q$.
In fact, every ring homomorphism $\mathbb Q \to \mathbb R$ must be the inclusion, and so cannot be surjective.
2.
Consider the idempotents in each ring, that is the solutions of $x^2=x$. 
The idempotents of $\mathbb Z_{16}$ are $0$ and $1$ but there are at least $4$ idempotents in $\mathbb Z_{4}\times \mathbb Z_{4}$, namely $(0,0), (1,0), (0,1), (1,1)$.
A: As to your final question, no: there is no general routine that would always work in finitely many steps to decide whether two rings are isomorphic.  At your stage you will probably benefit more from Brian M. Scott's tips on how to approach the specific problem you raised.  But your closing question is meaningful, and its answer is no.
