Classifying Algebraic Structures as Fields I just have a quick question!
We got a question asking us which of the following algebraic structures were fields, I got every one of them except the last one.  I don't even know how to begin how to attack the question.
Is real numbers modulo 2$\pi$ a field?
 A: If you want the quotient to be a ring, then you must take the quotient by an ideal. The ideal generated by $2\pi$ is all of $\Bbb R$, so this quotient is the zero ring, which is not considered to be a field.
This is nothing special with $2\pi$: every nonzero number will make the quotient zero. If you try to mod out by zero, you will of course get just $\Bbb R$ back, and so that would be a field!
If you have convinced yourself the thing you are modding out is not all of $\Bbb R$, then unfortunately you are thinking of something which isn't an ideal, and so the quotient by this object will not even be a ring (and of course could not be a field, then.)
A: It's not a field. In fact, if $\mathbb{R}$ mod $2\pi$ is referring to the group of radian angle measures under addition, then it doesn't even have a good multiplication operation. It has well-defined addition, but multiplication is a problem because the remainder you get when subtracting the closest integer multiple of $2\pi$ depends on what representative you pick mod $2\pi$.  For example, 
$$1\times 1 = 1$$
However, 
$$(1+2\pi)\times(1+2\pi)=1+4\pi + 4\pi^2$$
which does not differ by an integer multiple of $2\pi$ from $1$. In the example of angle measures, $1$ and $1+2\pi$ "should" represent the same angle, but there is no good way to "multiply angles" so that $1$ and $1+2\pi$ will be guaranteed to get the same answer.
EDIT: a propos of your comment to rschweib, "ring" is the same definition as "field" except elements aren't required to have multiplicative inverses so there's not necessarily a good division operation. For example, the integers form a ring because you can add, subtract and multiply, but not necessarily divide. To be a field, you also need to be able to divide. When we say, "it's not even a ring!" we mean not only do we not have division, we don't even have multiplication.
A: For $K$ to be a field, you need to first state: "What is the set, what is the addition operator, and what is the multiplication operator?"  Once you have these understood, you need to pass the following two tests:


*

*Is $K$ an abelian group under +? 

*Is $K - \{0\}$ an abelian group under $\times$?


So in your specific example, the set is infinite: it's $\mathbb{R} \bmod 2\pi$, which is essentially the interval $[0, 2\pi)$.  The addition and multiplication operators are simply the same ones from $\mathbb{R}$ but modulo $2\pi$.
So now apply the two tests above.  Is $[0,2\pi)$ an abelian group under addition mod $2\pi$?  You have to go through the requirements for abelian groups (well-defined operation, closure, associativity, existence of an identity, and inverses for all elements).  You should be able to convince yourself that the answer is "yes."  
Now do this again for the second requirement: is $(0,2\pi)$ an abelian group under multiplication mod $2\pi$?  (Notice I changed the left bracket to a paren to indicate the exclusion of 0 here.)  As others have already pointed out, the operation is not well-defined: by saying that our set is "equivalent" to the interval $(0,2\pi)$ we are grouping all multiples of $2\pi$ together into one equivalence class called $0$ (because after all, $0=2\pi =4\pi =6\pi= \cdots = -2\pi = -4\pi=\cdots$ when considered mod $2\pi$).  In order for multiplication to be "well defined" it must be true that you can choose any representation of a set member and multiply it by any other set member and always get the same result.  But, as Ben Blum-Smith already pointed out, this fails in our setting.  So multiplication is not well-defined and we don't have an abelian group.  (Once you find the operation is not well-defined, you can't even sensibly ask about the other group properties, so you stop here.)

Of course this is all very long-winded.  If you are taking a test, you would immediately suspect this thing isn't a field.  You don't have to find all the reasons why, just find one reason.  And with some experience, you would quickly "sense" that something isn't going to work with multiplication and you would zero in on this "not well-defined" problem.
A: The circle (or torus) quotient group $\rm\ R/G = \Bbb R/2\pi\Bbb Z\ $ does not form a ring, because, generally
Theorem $\ $ Coset multiplication $\rm\:(r\! +\! G)(s\! +\! G) = rs\! +\! G\:$ is well-defined $\rm\!\iff\! G\,$ is an ideal of $\rm\,R$  
Thus the additive quotient group $\rm\:R/G\:$ inherits the multiplication of $\rm\:R\:$ iff $\rm\:G,\:$ in addition to being an additive subgroup of $\rm\,R,\:$ is, $ $ further, $ $ closed under multiplication by all elements of $\rm\,R.$  
In your example, it is easy to construct explicit examples illustrating that such coset multiplication is not well-defined.
