Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$n \mathbb{Z}$ 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.

share|improve this question
    
Did you see my explanation of why < and > are incorrect notation on your earlier question? –  Zev Chonoles Mar 30 '13 at 5:32
1  
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

3 Answers 3

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.

share|improve this answer

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?

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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