If $R = \langle n \mathbb{Z}, +, \cdot\rangle$, is it a ring? Is it commutative, does it have a unity, is it a field? $n \mathbb{Z}$ with usual addition and multiplication.
Here is what I have: Let $R = \langle n \mathbb{Z}, +, \cdot\rangle$. $R$ is closed under $+$ and under $\cdot$. $R$ is a ring since $\langle n \mathbb{Z}, +\rangle$ is an abelian group, multiplication is associative, and left/right distributive laws hold.
Now, $R$ is a commutative ring since $nx \cdot ny = ny \cdot nx$. I also said $R$ is unitary by setting $n = 1, y = 1$. I also said it's a field since $n$ is an arbitrary value (the question doesn't specify that this be an integer).
My professor's solution says that $R$ is unitary unless $n=1$. It also says that $R$ is not a field. What am I doing wrong?
Thanks so much for helping.

By definition, $n\Bbb Z$ is the cyclic subgroup of $\Bbb Z$ generated by $n$, so $n$ must be an integer. I assume for the following that we are taking $n>0$. You only have a multiplicative identity if $n=1$--your professor's solution was probably meant to say that $R$ isn't unitary unless $n=1$. Since $n$ must be an integer, then $n\Bbb Z$ fails to be a field.
