In terms of the usual definitions of a field, why is $\mathbb{R}/2\mathbb{Z}$ not a field? I'm trying to figure out which of the axioms (binary operation [so associativity and distributivity], additive/multiplicative inverses, commutativity, and the existence of an additive/multiplicative identity) fail to be satisfied for $\mathbb{R}/2\mathbb{Z}$ to not be a field. I realize that $2\mathbb{Z}$ is not an ideal of $\mathbb{R}$ and that furthermore my group is not even an integral domain (e.g., $\sqrt{2} \cdot \sqrt{2} = 0$), so it cannot possibly be a field. I just can't seem to determine what property of fields (in terms of the axioms) that it does not satisfy. To be clear, I am using the notation $\mathbb{R}/2\mathbb{Z}$ to refer to the set of cosets of reals such that two reals are in the same coset if their difference is an even integer. Where have I gone wrong with this?
 A: You already have a problem showing $\mathbb R/2\mathbb Z$ is even a ring.
The operation used on cosets for quotient rings ($(x+I)(y+I):=xy+I$) does not produce a well-defined multiplication on your cosets.
You need the thing in the bottom of the quotient to be an ideal of $\mathbb R$ for the operation to be well-defined (and there are only two ideals, $\{0\}$ and $\mathbb R$ itself.)
A: Multiplication of cosets via representatives is not well-defined.
For instance, note that $0+2\mathbb{Z}  =2+2\mathbb{Z}$. But if you try to define
$$(a+2\mathbb{Z})(b+2\mathbb{Z}) = ab+2\mathbb{Z}$$
then the answer depends on which representative you use:
$$\begin{align*}
\left(\frac{1}{2}+2\mathbb{Z}\right)(0+2\mathbb{Z}) &= 0+2\mathbb{Z}\\
\left(\frac{1}{2} + 2 \mathbb{Z}\right)(2+2\mathbb{Z}) &= 1+2\mathbb{Z}
\end{align*}$$
but $0+2\mathbb{Z} \neq1+2\mathbb{Z}$.
So the multiplication operation on cosets via representatives is not well defined; you don’t have a multiplication on cosets (at least, not an obvious one, and not one inherited from multiplication in $\mathbb{R}$.

Just as in the case of groups where we can define an operation on cosets via representatives if and only if  the subgroup is normal, 
in any ring $R$, if $T$ is a subring of $R$, then we can define multiplication of cosets in $R/T$ using representatives if and only if $T$ is an ideal.
Theorem. Let $R$ be a ring, and let $T$ be a subring. The operation on cosets $R/T$ given by
$$(r+T)(s+T) = rs+T$$
is well defined if and only if $T$ is a two-sided ideal of $R$.
Proof. The standard proof shows that if $T$ is a two-sided ideal, then the multiplication is well defined.
Conversely, assume the multiplication is well defined, and let $a\in T$, $r\in R$. Since $a+T=0+T$, we have that, because multiplication is well defined,
$$\begin{align*}
0+T = (r+T)(0+T) &= (r+T)(a+T) = ra+T\\
0+T = (0+T)(r+T) &= (a+T)(r+T) = ar+T\\
\end{align*}$$
This means $ra,ar\in T$. 
Thus, for $a\in T$ and $r\in R$, $ar,ra\in T$. This proves $T$ is a two-sided ideal. $\Box$
