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One can find various proofs on this site showing that $2\mathbb{Z}$ and $3\mathbb{Z}$ are not isomorphic by supposing there is an isomorphism and computing what happens to certain elements and deriving a contradiction.

However, I am wondering if my proof is also valid? I have not found a posted question that uses this proof:

Note that $2\mathbb{Z}$ and $3\mathbb{Z}$ are both ideals of $\mathbb{Z}$ because they are subrings that are closed under multiplication.

So, then $\mathbb{Z}$/$2\mathbb{Z}$ and $\mathbb{Z}$/$3\mathbb{Z}$ are both quotient rings. However, $\mathbb{Z}$/$2\mathbb{Z} = \{\bar{0}, \bar{1}\}$ while $\mathbb{Z}$/$3\mathbb{Z} = \{\bar{0}, \bar{1}, \bar{2}\}$, so the quotient rings are not isomorphic because they have a different cardinality. Can I conclude that $2 \mathbb{Z}$ and $3 \mathbb{Z}$ are not isomorphic, because otherwise their quotient rings would be isomorphic?

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    $\begingroup$ No, you cannot conclude that. $\endgroup$ – Mariano Suárez-Álvarez Jul 2 '17 at 19:29
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    $\begingroup$ Why not, though? $\endgroup$ – user389056 Jul 2 '17 at 19:32
  • $\begingroup$ There are rings $R$ which contain ideals $I$ and $J$ which are isomorphic as rings but such that $R/I$ and $R/J$ are not isomorphic. That the quotients $R/I$ and $R/J$ be isomorphic or not tells you nothing about possible isomorphisms between $I$ and $J$. $\endgroup$ – Mariano Suárez-Álvarez Jul 2 '17 at 20:08
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    $\begingroup$ (In any case, it is not I who should explain why not but you who should explain why yes! If you want to use an argument to conclude something, it is you who have to come up with a justification. That' s how math works, really.) $\endgroup$ – Mariano Suárez-Álvarez Jul 2 '17 at 20:11
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You are implicitly claiming the following: if two ideals $I_1,I_2$ in a ring $R$ are isomorphic as rings, then the quotient rings $R/I_1$ and $R/I_2$ are isomorphic.

To see that this is false, consider the following example, let $\mathbb Z ^0$ be the ring whose underlying additive group is $(\mathbb Z, +)$, but with trivial multiplication, i.e. $\forall a,b \in \mathbb Z^0: ab = 0$. Then every subgroup of $(\mathbb Z, +)$ is an ideal of $\mathbb Z^0$. The subgroups $2\mathbb Z^0$ and $3 \mathbb Z^0$ are both isomorphic to $\mathbb Z^0$ itself, but the quotients $\mathbb Z^0 / 2 \mathbb Z^0$ and $\mathbb Z^0 / 3\mathbb Z^0$ are not.

If you want a unital example, consider $R = \displaystyle \prod_{n=1}^{\infty}\mathbb Z$, then $ I =\{0\} \times \displaystyle \prod_{n=2}^{\infty}\mathbb Z$, is an ideal which is isomorphic to $R$ itself as a ring (the isomorphism is a simple right shift). But we have $R/R = 0$, whereas $R/I \cong \mathbb Z$

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    $\begingroup$ Nice. And if you want a unital example with proper ideals, just do another shift. $\endgroup$ – Harald Hanche-Olsen Jul 2 '17 at 20:15
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    $\begingroup$ If you want an unital example, just add units to the first one you gave. $\endgroup$ – Mariano Suárez-Álvarez Jul 2 '17 at 20:23
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These are rings without unity. In $2\Bbb Z$, $x=2$ solves $x^2=x+x$. What about in $3\Bbb Z$?

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  • $\begingroup$ I like that you wrote $x+x$ so that its meaning is clear when you consider $3\mathbb{Z}$ also. $\endgroup$ – Malkoun Jul 2 '17 at 19:40
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    $\begingroup$ It's a fine answer, but it does not answer the question, if you read it more carefully. (Yes, I have done this myself, jumping into answer mode too quickly. It is easily done.) $\endgroup$ – Harald Hanche-Olsen Jul 2 '17 at 19:47
  • $\begingroup$ @HaraldHanche-Olsen It does answer the question in the title. Frequently the question in the title is quite different from the question in the text $\smile$. $\endgroup$ – Lord Shark the Unknown Jul 2 '17 at 19:48
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    $\begingroup$ This is why I included the proof-verification tag. $\endgroup$ – user389056 Jul 2 '17 at 19:49
  • $\begingroup$ @user389056, while the tag is useful, you will certainly agree that your title is not. $\endgroup$ – Mariano Suárez-Álvarez Jul 2 '17 at 20:15

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