3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Are you talking about this post? $\endgroup$ – rschwieb Oct 16 '13 at 13:18
  • $\begingroup$ @rschwieb no,that's coincidence. $\endgroup$ – jujuju Oct 16 '13 at 21:30
  • $\begingroup$ OK, just checking :) $\endgroup$ – rschwieb Oct 17 '13 at 12:42
2
$\begingroup$

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:

  1. $I$ is an additive subgroup of $R$; and
  2. $ir\in I$ 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

$\endgroup$
  • $\begingroup$ Footnote: I glossed over the details what operations we are talking about on the quotient so that the big picture isn't lost in details. I'm happy to supply details/pointers in the comments to elaborate, if necessary. $\endgroup$ – rschwieb Oct 15 '13 at 13:04
2
$\begingroup$

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 $1$.

$\endgroup$
  • $\begingroup$ could you explain what is Quotient ring? $\endgroup$ – jujuju Oct 15 '13 at 4:15
  • $\begingroup$ @jujuju A quotient ring is a ring $R$ divided out by an ideal $I$. The elements of this quotient ring, denoted by $R/I$, are equivalence classes of the form $[r] = r + I$ with $r \in R$. What we basically do is setting all elements of $I$ to zero, as $[r] = [r']$ if and only if $r$ and $r'$ differ by an element in $I$. $\endgroup$ – Arthur Oct 15 '13 at 4:28
  • $\begingroup$ @jujuju An example of a quotient ring is $\mathbb{Z}/n\mathbb{Z}$. Here we divide out by the ideal $n \mathbb{Z}$ of all multiples of $n$. So we basically identify all these multiples of $n$ with zero, and work modulo $n$. $\endgroup$ – Arthur Oct 15 '13 at 4:33
  • $\begingroup$ Sorry,there are no introduction about ideal in my book.I know ideal is a set by wiki,but r is a element.[r]=a set+a element?I still dont know what is quotient ring. $\endgroup$ – jujuju Oct 15 '13 at 4:43
  • $\begingroup$ The set $r + I$ is defined as $\{r+i : i \in I\}$. We set all the elements in $r + I$ equivalent to each other, and denote this equivalence class by $[r]$. The quotient ring is the ring $\{[r] : r \in R\}$, with natural addition and multiplication. $\endgroup$ – Arthur Oct 15 '13 at 4:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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