A proof and question about Quotient ring I don't understand quotient rings very well,
and I am confused about the proof of "The quotient ring $\Bbb Z/(m)$ is a field if and only if $m$ is a prime."
I know what mod means. Help me understand the concept and proof, please.
 A: If $m = ab$ is composite, then in $\mathbb{Z}/m\mathbb{Z}$ we get $ab = 0$, and so $\mathbb{Z}/m\mathbb{Z}$ is not a field because there are zero divisors.
Now assume $p$ is prime. We want to find a multiplicative inverse of an element $0 \neq a \in \mathbb{Z}/p\mathbb{Z}$. Consider the numbers $a,2a,\ldots,(p-1)a$. We cannot have $sa = ta$ for $s \neq t$ as that would imply $(s-t)a \equiv 0$ mod $p$, and clearly $(s-t)a$ is not divisible by $p$ in $\mathbb{Z}$. So the $p-1$ elements $a,2a,\ldots,(p-1)a$ are nonzero and distinct in $\mathbb{Z}/p\mathbb{Z}$, and hence one of them must be equal to $1$.
A: It is a good idea to learn what an ideal is and what a quotient ring is simultaneously. You say you "know mod," but if you don't know what an ideal is, I think your knowledge of "mod" is not very thorough. But at least you have a start :)
The beginning of quotients
Given a group $G$ and a subgroup $H$, you can always split $G$ into pieces with a relation $a\sim b$ iff $ab^{-1}\in H$. What you get is several nonoverlapping subsets of $G$, and within each subset, all elements are mutually related. One central idea of taking a quotient is to think of each such subset as a single "point." The set of these new "big points" is written as "$G/H$".
Now $G/H$ is just a set usually, but it would be nice if it were a group. For this to happen, $H$ has to have special properties: $H$ needs to be a normal subgroup of $G$. Then $G/H$ has is a group in a nice way.
Moving to rings
Now if you have a ring $R$ with an additive subgroup $I$, you can form the set $R/I$ as above. What's nice is that since the additive group $R$ is abelian, all subgroups of $R$ are normal, so $R/I$ is guaranteed to be an additive group. 
But now we have more demands: we wish $R/I$ also had multiplication, so that it can become a ring! Just as in the group case, $I$ needs to have special properties: this time, $I$ has to be an ideal for $R/I$ to turn into a ring.
Fortunately, the definition of an ideal is not really that tough to understand. A nonempty subset $I$ of $R$ is an ideal if: 


*

*$I$ is an additive subgroup of $R$; and 

*$ir\in I$ and $ri\in I$ for all $i\in I$ and $r\in R$.


It's just an additive subgroup of $R$ that "absorbs" multiplication by elements of $R$.
More on understanding quotients
One way of understanding a quotient ring $R/I$ is by thinking of it as "like the old ring, except now things inside $I$ are like zero." In $\Bbb Z/m\Bbb Z$, all multiples of $m$ are now like zero. That's why modulo $8$, $15\equiv 7+8\equiv 7+0\equiv 7$. The same sort of idea applies to all quotient rings. 
Incidentally, there are lots of other questions asking for intuition about quotient rings (frequently polynomial rings) here on m.SE, so you might get a better idea if you try to read those too. Here are a few:
What is a quotient ring and cosets?  this one has a nice solution by Arturo Magadin.
Understanding the quotient ring $\mathbb{R}[x]/(x^3)$.
Understanding quotients of $\mathbb{Q}[x]$
Error in understanding the theorem about the invertibility of an element(coset) of a quotient ring
How to deal with polynomial quotient rings
