# I Isomorphism between 2Z and 3Z

1. May 2, 2017

### Mr Davis 97

My book is trying to show that the rngs $2 \mathbb{Z}$ and $3 \mathbb{Z}$ are not isomorphic. It starts by saying that if there were an isomorphism $\mu : 2 \mathbb{Z} \to 3 \mathbb{Z}$ then by group theory we would know that $\mu (2) = \pm 3$. It then goes on to show that this leads to a contradiction. My question has to do with why it must be true, if we assume $\mu$ is an isomorphism, that $\mu (2) = \pm 3$

2. May 2, 2017

### andrewkirk

$2\mathbb Z$ contains exactly two elements that can generate the ring on their own. Those elements are 2 and -2.
$3\mathbb Z$ contains exactly two elements that can generate the ring on their own. Those elements are 3 and -3.
Since the property of being able to generate the ring on its own is a ring property, it must be preserved by any ring isomorphism. Hence any ring isomorphism must map 2 to either 3 or -3 and must map -2 to the other one.

3. May 2, 2017

### Mr Davis 97

Okay, that makes sense. You need to preserve the structural properties.

To make the problem more general, how would you count how many homomorphisms there are, and not necessarily isomorphisms? I'm assuming that it might still involve the fact that $2 \mathbb{Z}$ is cyclic.

4. May 3, 2017

### andrewkirk

There are no ring homomorphisms between the two. The proof that there are no ring isomorphisms does not use bijectivity and hence applies equally well to ring homomorphisms.

There will be group homomorphisms between the two sets as abelian groups, but no ring homomorphisms.

In general, if you want to count homomorphisms, look at how many generators there are, and how many different ways they can be mapped while preserving the algebraic structure.

5. May 3, 2017

### Mr Davis 97

What if we make the distinction between rings with unity and rings without unity? Would homomorphisms exist for the latter but not the former?

6. May 3, 2017

### andrewkirk

I don't understand the scope of your question. Some pairs of rings can be connected by homorphisms and others cannot.

If we restrict ourselves to isomorphisms, IIRC all cyclic, unital rings of a given order are isomorphic. That is not the case for non-unital rings, as this example demonstrates.

7. May 3, 2017

### Mr Davis 97

Actually, scratch that. I was confused.

Could you walk me through application of the statement: "In general, if you want to count homomorphisms, look at how many generators there are, and how many different ways they can be mapped while preserving the algebraic structure"?

In this case, we know that $2 \mathbb{Z}$ has two generators, $-2$ and $2$. How do we determine how a homomorphism $\mu$ could map these while maintaining the algebraic structure? I don't see how in this case we use the facts $\mu (a+b) = \mu (a) + \mu (b)$ and $\mu (ab) = \mu(a) \mu(b)$

8. May 3, 2017

### andrewkirk

Say we want to find a homomorphism $f$ from $2\mathbb Z$ to some cyclic ring $R$ which is generated by $r$. Since 2 generates $2\mathbb Z$, $f(2)$ must generate $Im\ R$. To find $f(k)$ for $k\in 2\mathbb Z$ we use the linearity rules to observe that $f(k)=f\left(\sum_{j=1}^{k/2}2\right)=\sum_{j=1}^{k/2} f(2)$, which is the image of 2 added to itself $k/2$ times. Thus, knowing the action of $f$ on $2$ tells us the action of $f$ on any other element of the domain.

More generally, if the domain is any cyclic ring, we need only determine the action of the candidate homomorphism on a generator of the domain, in order to determine the entire map.

If $R$ contains an isomorphic copy of the domain, we can find a non-trivial isomorphism, eg if $R$ is the integers or Gaussian Integers. But otherwise it may not be possible, so that the only homomorphism is the trivial one that maps everything to zero. The example given is of that kind. It is the distributive law that creates the constraints.