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.

You can't "set" $n$ to be something; it is given to you (also, I am sure that it is to be implicitly assumed that $n$ is an integer).
Remember that
$$n\mathbb{Z}=\{\ldots,-2n,-n,0,n,2n,\ldots\}$$
So, do you think you can find a unit for $n\mathbb{Z}$ when $n>1$?
If $n\mathbb{Z}$ were a field, every non-zero element of $n\mathbb{Z}$ would have a multiplicative inverse. Which element of $n\mathbb{Z}$ do you think is the multiplicative inverse of $n$, for example?
