# 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.

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Did you see my explanation of why < and > are incorrect notation on your earlier question? –  Zev Chonoles Mar 30 '13 at 5:32
Presumably, you mean to say "My professor's solution says that $R$ is not unitary unless $n=1$." –  Zev Chonoles Mar 30 '13 at 5:32
Math Damon: We encourage those who ask questions and receive answers to 1) upvote answers that are helpful, and 2) to select a helpful answer and accept it. To accept an answer, simply click on the $\checkmark$ to the left of the answer you'd like to accept. (You can accept only one answer per question, but upvote as many as you'd like). Bonus: you get 2 reputation points for each answer you accept! –  amWhy Apr 1 '13 at 20:04

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.

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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?

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While it’s perfectly true that for any $x\in\Bbb R$ you can define $x\Bbb Z=\{xn:n\in\Bbb Z\}$, this set is closed under multiplication if and only if $x\in\Bbb Z$. Thus, even if the context and the use of the letter $n$ didn’t give it away, you could deduce that $n$ must be an integer. And you should certainly understand that $n$ is a single fixed entity that is given to you, not an arbitrary quantity or one taking multiple values.

By definition $n\Bbb Z$ is unitary if and only if it contains $1$, the multiplicative identity, which is the case if and only if there is a $k\in\Bbb Z$ such that $nk=1$, i.e., if and only if $n=\frac1k$ for some non-zero integer $k$. Clearly the only values of $n$ making $n\Bbb Z$ unitary are $1$ and $-1$: no other reciprocal of a non-zero integer is an integer.

And under $n\Bbb Z$ is never a field: either it’s not unitary, or it’s simply $\Bbb Z$, which lacks multiplicative inverses.

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