Show that $\mathbb{Z}$ and $2\mathbb{Z}$ are not isomorphic as rings.

Question

My attempt: Suppose $\mathbb{Z}$ and $2\mathbb{Z}$ are isomorphic as rings, Let $\phi: \mathbb{Z} \rightarrow 2\mathbb{Z}$ be the isomorphism. Then we have $\phi(4) = \phi(2) + \phi(2) = 2n + 2n = 4n\,$ and $\phi(4) = \phi(2)\phi(2) = 4n^2$ and so $n = 0$ or $n = 1$. If $n = 0$, then $\phi$ is not surjective, which contradicts the fact that $\phi$ is an isomorphism. If $n = 1$, then $\phi(3) = 3 \notin 2\mathbb{Z}$, which again gives us a contradiction.

I think this is not a duplicate. The proposed duplicate proves the theorem, but this question asks for a critique of OP's own proof. Is that particular proof also valid? — David K, Apr 08 '15 at 13:12

score 2 · Accepted Answer · answered Apr 08 '15 at 12:20

Your reasoning is correct. However, you should say that you define $n=\phi(1)$. Also $n=1$ is already ruled out by $n=\phi(1)\in 2\mathbb Z$.

The more elegant approach to this problem would be to show that $2\mathbb Z$ has no multiplicative identity.

score 2 · Answer 2 · edited Apr 24 '19 at 17:55

2

An easy way for me:

Since $1$ is a generator of $\mathbb Z$ and $\phi$ is an isomorphism so $\phi(1)$ is a generator of $\mathbb 2Z$

Then $\phi(1)=2$ or $\phi(1)=-2$

Then $\phi(1)=\phi(1 \cdot 1)=\phi(1) \cdot \phi(1)=4$ in both cases

But same element can't be mapped to two different elements hence a contradiction

edited Apr 24 '19 at 17:55

Alex Johnson

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answered Apr 08 '15 at 13:50

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Show that $\mathbb{Z}$ and $2\mathbb{Z}$ are not isomorphic as rings.

2 Answers2