Is $2\mathbb{Z}$ ring isomorphic to $4\mathbb{Z}$? A ring isomorphism $f: R \rightarrow S$ satisfies these properties:
a) $f(a + b)$ = $f(a) + f(b)$, for all $a, b \in R$.
b) $f(ab) = f(a)f(b)$, for all $a, b \in R$.
I'm inclined to believe that the rings $2\mathbb{Z}$ and $4\mathbb{Z}$ are isomorphic because the groups $2\mathbb{Z}$ and $4\mathbb{Z}$ are isomomorphic, but I'm having some trouble applying ring isomorphism axioms to integer rings in general. Any kind of walkthrough would be tremendously helpful in understanding ring homomorphisms. Thank you.
EDIT: Ah sorry, I forgot to include that $f$ is a bijection between $R$ and $S$.
 A: Let $\phi\colon 2\mathbb{Z}\rightarrow 4\mathbb{Z}$  be an arbitrary ring homomorphism. Then there are some $j,k\in \mathbb{Z}$ such that $\phi(2)=4j$ and $\phi(4)=4k$. Then
$$
4k=\phi(4)=2\cdot \phi(2)=2\cdot 4j=8j, \quad 4k=\phi(4)=(4j)^2=16j^2.
$$
Hence we obtain $8j=16j^2$, so that $j=0$, and the only morphism is the trivial one, and the rings are not isomorphic.
A:  First, note that $2\mathbb{Z}$ and $4\mathbb{Z}$ are not rings - at least, not in the sense of the definition of ring isomorphism you've given. That definition assumes that rings have a multiplicative unity, and neither $2\mathbb{Z}$ or $4\mathbb{Z}$ do. EDIT: The OP has corrected their definition of isomorphism, so this paragraph is irrelevant now.
That said, we can talk about rings without identity - and the definition of an isomorphism between such is the same as what you've given, minus the third clause. So: are they?
In order to get some intuition for what the answer should be, think about some possible maps between them. For example, a reasonable first guess would be the map $f:2\mathbb{Z}\rightarrow 4\mathbb{Z}: x\mapsto 2x$. This is a natural bijection between the two sets, and it is an isomorphism of the additive groups. So: is it a ring isomorphism? HINT: think about $f(2\times 2)$ versus $f(2)\times f(2)$ . . .
Now intuitively, you should have some sense about whether these two rings-without-identity are isomorphic. Do you see how to prove that? HINT: think about the possible places that any putative isomorphism can send $2$ . . .
A: Let $f\colon 2\Bbb Z\to 4\Bbb Z$ be additive and multiplicative map. Then $$2f(2) = f(2\cdot 2) = f(2)f(2) \implies (f(2)-2)f(2) = 0.$$ Since $f(2)\in 4\Bbb Z$, then $f(2)\neq 2$. Hence, $f(2) = 0$. That is, the only rng homomorphism is zero morphism, which is obviously not bijective.
A: Since both rings don't have a 1 we are interested in isomorphisms of rngs. So skip the third axiom. As groups there are only two isomomorphsims: They are induced by $2 \mapsto 4$ and $2\mapsto -4$ (for a group homomorphis it is enough to tell what happens to the generators). So we have to check if one of theses satisfies the second condition. Unfortunately $f(2)^2=16$ but $f(2^2)=8$. So none of these is a rng homo. Hence the two rngs are not isomorphic.
A: I'm not familiar with the third one being included in the definition, and here you have a problem because neither $2Z$ nor $4Z$ has a $1$. 
If there exists an isomorphism of rings between them, say $f$, then $f(4) = f(2 \times 2) = 16 k^2$ where $f(2) = 4k$. But $f(4) = f(2) + f(2) = 8k$. So $8k = 16k^2$, i.e. $k=0$, but then $f(2) = f(4) = 0$, so $f$ is not bijective.
A: One direct way to see that two rings are non-isomorphic is to write down an equation that has a different number of solutions in the two rings.
In this case, $2\mathbb{Z}$ has two solutions to the equation $x\cdot x = x+x$, while $4\mathbb{Z}$ has only one.
