If rings $\mathbb Z^m$ and $\mathbb Z^n$ are isomorphic then $m=n$ It is given that the rings $\mathbb Z^m$ and $\mathbb Z^n$ are isomorphic.
Show that $m=n$.
My try:
I want to show that $m\leq n$ and $n\leq m$ .
Suppose $m\leq n$.Let $\phi:\mathbb Z^m\to \mathbb Z^n$ be the given isomorphism.Then since $(\underbrace{1,1,\ldots, 1}_{m \text{ times}})$ is the identity of $\mathbb Z^m$ so $\phi(\underbrace{1,1,\ldots, 1}_{m \text{ times}})=(\underbrace{1,1,\ldots, 1}_{n \text{ times}})$.
Also $\mathbb Z^m$ has $m$ generators, $\{(1,0,\ldots, 0),(0,1,\ldots,0),\dots,(0,0,\ldots,1)\}$. Since $\phi $ is an isomorphism it must map a generator to a generator.
Now obviously $\mathbb Z^n$ has a set of $n$ generators given by $(1,0,0\ldots,0),(0,1,0\ldots0),(0,0,1,\ldots0),\dots,(0,0,\ldots, 1)$.Let $A$ be the set of generators of $\mathbb Z^n$. Then if it can be shown that card $A$ is $n$ ,then we cannot  map  a set of $m$ elements injectively to a set of $n$ elements and hence a contradiction is reached and then we are done.
How to proceed from here?
 A: Every group homomorphism $\phi:\mathbb Z^m\to \mathbb Z^n$ extends uniquely to a homomorphism $\psi:\mathbb Q^m\to \mathbb Q^n$, which is automatically a linear transformation of vector spaces over $\mathbb Q$. The result then follows from the rank-nullity theorem.
Here are some details:
Every element of $\mathbb Q^m$ can be written in the form $x=(\dfrac{a_1}{b},\dots,\dfrac{a_m}{b})$ with $a_1, \dots, a_m, b \in \mathbb Z$. Then $bx \in \mathbb Z^m$ and so $b\psi(x)=\psi(bx)=\phi(bx)$, which gives $\psi(x)=\dfrac{1}{b}\phi(bx)$. This proves uniqueness and also gives a formula for existence of $\psi$.
The rank-nullity theorem implies that  $m=n$ if $\psi:\mathbb Q^m\to \mathbb Q^n$ is bijective:
$$
m = \dim \mathbb Q^m
= \dim\ker\psi + \dim\operatorname{im}\psi
= \dim 0 + \dim \mathbb Q^n
= 0 + n = n
$$
A: Unfortunately, your strategy won't work: the ring $\mathbb{Z}^n$ can be generated by fewer than $n$ elements.  For instance, you can check that $\mathbb{Z}^4$ is generated as a ring by the two elements $(1,0,1,0)$ and $(1,1,0,0)$.  In fact, it turns out that in general the minimum number of elements needed to generate $\mathbb{Z}^n$ as a ring is $\lceil\log_2 n\rceil$, though this takes some work to prove.
As mentioned in the other answers, you can actually prove that $\mathbb{Z}^m$ and $\mathbb{Z}^n$ are not even isomorphic as groups, so they can't be isomorphic as rings.  But if you just want to show they aren't isomorphic as rings, there is a simpler argument.  Here's a hint: can you characterize the units of $\mathbb{Z}^n$?  How many units are there in $\mathbb{Z}^n$, and how many units are there in $\mathbb{Z}^m$?
A: If $\mathbb Z^m$ is isomorphic to $\mathbb Z^n$ as rings, then they are isomorphic as abelian groups as well, but since $\mathbb Z^m$ is free abelian group of rank $m$, and $\mathbb Z^n$ is free abelian group of rank $n$, it follows that $m=n$.
