# Are all the prime ideals isomorphic to $\mathbb{Z}$?

I know that is possible to define a bijective group homomorphism between $(2)$ and $\mathbb{Z}$ through the function $f(2x) = x$ and similarly for $(3)$ via $f(3x) = x$. But I thought that for two groups to be isomorphic they had to have the same order. If the order is infinity how could you tell the two groups say $(2)$ and $\mathbb{Z}$ are isomorphic to each other?

Could subgroups be isomorphic to the original groups?

-
There are prime ideals in the title, but not in the question. Why is that? –  user23211 Mar 30 '12 at 19:38
First isomorphism theorem? $f: p\mathbb{Z} \to \mathbb{Z},$ $\dfrac{p\mathbb{Z}}{\operatorname{ker}(f)} \cong \mathbb{Z}.$ –  user2468 Mar 30 '12 at 20:09

Every ideal of $\mathbb{Z}$ is also a subgroup. Every nontrivial subgroup of $\mathbb{Z}$ is infinite cyclic, hence isomorphic to $\mathbb{Z}$.

So for any $n\neq 0$, the map $f(x)=nx$ gives you a group homomorphism $\mathbb{Z}\to n\mathbb{Z}$.

The order is no good; while having the same cardinality is necessary for an isomorphism to exist, it is not generally sufficient.

-
For instance, $\mathbb{Z}$ is not isomorphic to $\mathbb{Q}$ as additive groups even though they have the same cardinality. Nor are $\mathbb{Z}$ and $\mathbb{Z}\times\mathbb{Z}$. –  lhf Mar 30 '12 at 19:51

You already showed $(2)$ and $\mathbb{Z}$ are isomorphic to each other, by explicitly writing down an isomorphism.

As an aside, because you have already shown them isomorphic, that implies they have the same cardinality.

The ring $\mathbb{Z}$ does have the property that every non-zero ideal of $\mathbb{Z}$ is isomorphic to $\mathbb{Z}$ as an abelian group.

More generally, there exist rings $R$ with the property that every ideal of $R$ is isomorphic to $R$ as an $R$-module (which, among other things, implies isomorphic as an abelian group). Principal ideal domains (PID), for example.

(aside: I'm not sure if there are any examples that are not PIDs)

-
A comment on your aside: this condition is equivalent to $R$ being a PID (of course, you want to require that any nonzero ideal is isomorphic to $R$). If $I \subset R$ is an ideal such that $R \cong I$ as $R$-modules, $I$ is generated by the image of $1$. To see that $R$ is a domain: if $x \in R$ is nonzero the ideal $xR$ is isomorphic to $R$, hence has no torsion, so $x$ is not a zero-divisor. –  Justin Campbell Mar 30 '12 at 21:49