1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.