Is the ring $3\mathbb Z$ a ring homomorphic image of the ring $2\mathbb Z$.

Question

Is the ring $2\mathbb Z$ isomorphic to the ring $3\mathbb Z$ ?

Solution: Let if possible,$\phi:\mathbb {2Z\to 3Z}$ be a ring isomorphism.

Then $\phi$ is a group isomorphism between the additive groups $2\mathbb Z$ and $3\mathbb Z$.

Now,$\phi(2)$ completely determines $\phi$.Let $\phi(2)=3x$ where $x\in \mathbb Z$.

Then $\phi(2y)=3xy $ for all $y\in \mathbb Z$.

Now,$\phi$ is surjective,so $3\in 3\mathbb Z$ has a preimage.So,$\phi(2y)=3$ for some $y\in \mathbb Z$.

Which implies,$3xy=3\implies xy=1$.So,$x,y\in \{-1,1\}$.But then $\phi(2)=\pm3$ and hence $\phi(4)=\phi(2.2)=\phi(2).\phi(2)=9$.

Again,$\phi(4)=\phi(2+2)=\phi(2)+\phi(2)=\pm 6$,but that is not possible.

So,$2\mathbb Z$ and $3\mathbb Z$ are not ring isomorphic.

Note: $3\mathbb Z$ is not a ring homomorphic image of $2\mathbb Z$,as we did not make use of injectivity of $\phi$ in the above proof.

Is the above conclusion correct?

Your title and body disagree. Ring homomorphic or ring isomorphic? — Clement Yung, May 27 '20 at 05:46
@ClementYung Actually I wanted to say that the proof of isomorphism can be carried in the same manner even for epimorphism question as we do not use injectivity. — Kishalay Sarkar, May 27 '20 at 05:57

score 2 · Accepted Answer · answered May 27 '20 at 12:56

Your argument is correct.

But if you strip it of all inessential parts, you get a shorter, in my opinion also clearer, argument that actually proves something more.

The key point is, as you did, to look at $\phi(4)$, where $\phi$ is an arbitrary rng-homomorphism from $2\mathbb{Z}$ to some sub-rng of $\mathbb{Z}$. Since $4 = 2\cdot 2 = 2 + 2$ it follows that $$0 = \phi(4) - \phi(4) = \phi(2\cdot 2) - \phi(2+2) = \phi(2)^2 - \bigl(\phi(2) + \phi(2)\bigr) = \phi(2)\cdot \bigl(\phi(2) - 2\bigr)\,.$$ Thus either $\phi(2) = 0$ or $\phi(2) = 2$. Hence either $\phi = 0$ or $\phi$ is the inclusion of $2\mathbb{Z}$ in $\mathbb{Z}$, or $\phi$ is the identity map on $2\mathbb{Z}$.

In particular, the only rng-homomorphism $2\mathbb{Z} \to 3\mathbb{Z}$ is the zero map.

score -2 · Answer 2 · answered May 27 '20 at 05:53

-2

Your argument seems correct. But there's an easier way: if $2\Bbb Z$ and $3\Bbb Z$ were isomorphic, then the quotient rings $\Bbb Z/3\Bbb Z$ and $\Bbb Z/2\Bbb Z$ would be isomorphic too. But this can't be possible since $\Bbb Z/3\Bbb Z$ has precisely 3 elements and $\Bbb Z/2\Bbb Z$ has precisely 2 elements.

answered May 27 '20 at 05:53

General Grievous

2,121

Does it work?Because as groups $2\mathbb Z$ and $3\mathbb Z$ are isomorphic although the groups $Z/2Z$ and $Z/3Z$ are not isomorphic. – Kishalay Sarkar May 27 '20 at 05:58
It works because these rings are not isomorphic as rings. – General Grievous May 27 '20 at 06:01
No,I am not saying that, I mean to say that if $R$ has two ideals $I\simeq J$ then does imply $R/I\simeq R/J$. – Kishalay Sarkar May 27 '20 at 06:02
Yes that is correct. Therefore the contrapositive of this statement is: if $R/I $ is not isomorphic to $R/J$ then $I$ cannot be isomorphic to $J$ – General Grievous May 27 '20 at 06:05
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@PankajTiwary For groups this is not true.If $G$ is a group and $H,K$ are normal subgroups of $G$ such that $H\simeq K$,still $G/H\simeq G/K$ is not true.For example $G=\mathbb Z$ and $2Z\simeq 3Z$ as groups,but $Z/2Z \not \simeq Z/3Z$ as groups – Kishalay Sarkar May 27 '20 at 06:05
Yes that is also correct. But we do not require that fact here. – General Grievous May 27 '20 at 06:07
Your argument does not work. – Kishalay Sarkar May 27 '20 at 06:21
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The question, Pankaj, is whether you can prove (or give a reference proving) that if ideals $I$ and $J$ are isomorphic as rings then $R/I$ and $R/J$ are isomorphic as rings? – Gerry Myerson May 27 '20 at 06:22
@PankajTiwari Can you show that $R/I\simeq R/J$ if $I\simeq J$. – Kishalay Sarkar May 27 '20 at 06:29
@PankajTiwari Also let me make clear two facts. – Kishalay Sarkar May 27 '20 at 06:31
@PankajTiwari In my convention,a ring may not contain idenity and need not be commutative.And a ring homomorphism may not map $1$ to the $1$ of the codomain ring.It just preserves the addition and multiplication. – Kishalay Sarkar May 27 '20 at 06:32
I apologize if I was not clear. Whatever I said applies for the ring $\Bbb Z$. Further, a ring isomorphism requires the identity to be mapped to the identity. This is the most popular definition – General Grievous May 27 '20 at 06:34
@PankajTiwari I am not following that convention,even authentic books like Dummit Foote does not require a ring to have idenity.And a homomorphism need not preserve identity. – Kishalay Sarkar May 27 '20 at 06:35
@PankajTiwari Please answer as per my convention. – Kishalay Sarkar May 27 '20 at 06:35
It is upto you to decied what definition what you want to use. Atiyah-Macdonald(among others) uses the definition I mentioned. – General Grievous May 27 '20 at 06:36
@PankajTiwari But can you prove the fact you are claiming using your convention? – Kishalay Sarkar May 27 '20 at 06:37
look at this https://math.stackexchange.com/questions/2528762/are-10-mathbb-z-and-20-mathbb-z-isomorphic-to-each-other-or-not-as-rings?rq=1 – General Grievous May 27 '20 at 06:38
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You use the rings $2Z$ and $3Z$ in your answer, Pankaj, but these rings have no multiplicative identity. – Gerry Myerson May 27 '20 at 06:40
Yes I realize my mistake now. It seems I've been using the definition @KishalaySarkar said without realizing. My bad, I hope things are clear now with the link I've given. I apologize for the confusion. – General Grievous May 27 '20 at 06:41
@PankajTiwari Your answer also needs an edit,as it is not correct...you can edit it. – Kishalay Sarkar May 27 '20 at 06:51

Is the ring $3\mathbb Z$ a ring homomorphic image of the ring $2\mathbb Z$.

2 Answers2