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

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

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.

• Also answered here. Apr 8, 2015 at 12:22
• 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? Apr 8, 2015 at 13:12

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.

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