# 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?

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: $$m = \dim \mathbb Q^m = \dim\ker\psi + \dim\operatorname{im}\psi = \dim 0 + \dim \mathbb Q^n = 0 + n = n$$

• Maybe you want to say explicitly that in this case there is no need to assume that the rings are isomorphic in order to find $m=n$, that is, the group homomorphism is also a homomorphism of $\mathbb Z$-modules. – user26857 Sep 28 '15 at 16:54
• two questions:1.how is this extension done?2.I dont get how it follows from rank-nullity.Is it possible for you to give some more details. – Learnmore Sep 28 '15 at 17:26

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 $\mathbb{Z}^m$?

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 $m=n$.