# What is a quotient ring and cosets?

So I am trying to understand what a coset is and what a quotient ring is.

So I am going to tell you guys what I know. And please let me know if my thinking is right or wrong, and if I am missing something. For the rest of this post, assume R is a ring and I is an ideal of that ring.

So $I = (m)$ is a principal ideal generated by $m$ where $m \in R$.

Now the congruence class of $a$ and $I$ is denoted by $[a]_m$ but this congruence class can also be written as $a + (m)$ or simply $a + I$. now this congruence class is obviously a set. So is this the coset of it? So for any $a$ that you choose in $R$ and you "add" the ideal to it (generated by m which is also in R) you get a coset.

The quotient ring R/I just means ALL the cosets of I in R. So does this say that if hypothetically speaking m = 3, then 1 + (3) is one coset, 2 + (3) is another coset and hence the quotient ring is ALL the cosets for every possible a?

I hope I make sense. If someone could send me a link to an easy (introduction to algebra) article or a "tutorial", that would be appreciated. I am using Hungerford Algebra.

If anyone can explain this to me in easy english, that is appreciated.

Thanks

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"So $I=(m)$ is a principal ideal..." The word "So" is not correct here, since not every ideal in every ring is principal. Did you mean " Say $I=(m)$ is a principal ideal..." ? – Arturo Magidin Feb 25 '11 at 4:25
what is the difference between a principal and a "regular" ideal? I guess "say" would work but I wanna know the difference. – Tyler Hilton Feb 25 '11 at 4:34
@Tyler: every principal ideal is an ideal, but not every ideal is principal (there are some rings where they are, but those are very special rings). For example, if $R=\mathbb{Z}[x]$, the ring of all polynomials with integer coefficients, the set $I$ of all polynomials that have even constant coefficient is an ideal (check that this is the case), but it cannot be written as $(p(x))$ for some polynomial $p(x)\in\mathbb{Z}[x]$: such a polynomial would have to divide $2$, and also divide $x$, and you'll see that the only polynomials that do that are $\pm 1$, which are not in the ideal. – Arturo Magidin Feb 25 '11 at 4:40
@Tyler: Just to be clear: an ideal is any nonempty subset $I$ such that for all $a,b$, if $a,b\in I$ then $a-b\in I$, and if $a\in I$ and $x\in R$, then $xa,ax\in I$. Taking "multiples of $m\in R$" gives one example of ideals, but they aren't the only ones in general, as the example above shows. – Arturo Magidin Feb 25 '11 at 4:49
@Tyler: P.S. You have to be very careful with adjectives in ring theory. As it happens "regular ideal" is a term of art, it describes a specific kind of ideal... – Arturo Magidin Feb 25 '11 at 5:03

Remember that every equivalence relation induces a partition on the set on which you have defined the relation. An ideal $I$ defines an equivalence relation on the set $R$ by saying that $a\sim b$ if and only if $a-b\in I$; we express this by saying that $a$ is congruent to $b$ modulo $I$.

That means that $R$ is partitioned into equivalence class under this "congruent to modulo $I$" relation. The equivalence classes are called "cosets" (I claimed some time ago this is short for "congruence set", but have been unable to substantiate this; but you can surely imagine that it is). The cosets are the equivalence classes.

Now, what is the equivalence class of an $a\in R$? it is the set of all things that are congruent to $a$ modulo $I$; this consists exactly of all elements of the form $a+x$ with $x\in I$, so we write it as $$a + I = \{ a+x \mid x\in I\}.$$ This is its description as a set. If we want to think of it in terms of the equivalence relation and remember that it is the equivalence class of $a$, then we use the standard notation for equivalence classes and write $[a]$ (or $[a]_I$, or $[a]_m$, to remind us also of which ideal $I$ we are dealing with).

When you ask if $a+I$ is "the coset of it", it is unclear to me who "it" is. But, $a+I$ is the equivalence class of $a$, so it is the coset of $a$ (since "the coset of $x$" just means "the equivalence class of $x$ under the equivalence relation 'congruent modulo $I$'").

The reason that when you "add" the ideal you get the coset is just because of what the definition of the equivalence relation is: every element of the form $a+i$ is congruent to $a$ modulo $I$, because $a-(a+i) = -i\in I$; and if $b$ is congruent to $a$ modulo $I$, then $a-b=i$ for some $i\in I$, so we get that $b=a-i$. That is, every element of the form $a+x$ with $x\in I$ is in the coset, and everything in the coset is of the form $a+x$ with $x\in I$. So the notation $a+I$ is both suggestive and useful.

Now, the set $R/I$ is just the set of equivalence classes; as a set, the elements are the cosets. Each coset has many different names, since $[a]_I = [b]_I$ whenever $a\sim b$.

As a ring, $R/I$ is the ring whose elements are the cosets/equivalence classes, and whose operations $\oplus$ and $\odot$ are defined by \begin{align*} [x]_I \oplus [y]_I &= [x+y]_I,\\\ [x]_I\odot[y]_I &= [x\cdot y]_I \end{align*} where $+$ and $\cdot$ are the operations in $R$. (We later drop the distinction between $+$ and $\oplus$, but the point here is that they are operations defined on different sets, so they are really different functions, though very closely related).

In your example, yes: $R/I$ is the collection of all cosets, but remember that the same coset may have many different names. So, for instance, if $R=\mathbb{Z}$ and $I=(3)$, then $R/I$ consists of all cosets, which are sets of the form $$a + (3) = \{ \ldots, a+3(-2), a+3(-1) , a+3(0), a+3(1), a+3(2), a+3(4),\ldots\}$$ for all $a$. But as it happens, every single coset is equal to either $0+(3)$, $1+(3)$, or $2+(3)$, so in fact $R/I$ has only three elements, even though each of those elements has infinitely many names: \begin{align*} 0+(3) &= 3+(3) = 6+(3) = 9+(3) =\cdots = 3k+(3),\\ 1+(3) &= 4+(3) = 7+(3) = 10+(3) = \cdots = 3k+1 + (3)\\ 2+(3) &= 5+(3) = 8+(3) = 11+(3) = \cdots = 3k+2 + (3) \end{align*}

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In your last example, why can $a$ only be $0, 1, 2$? – Tyler Hilton Feb 25 '11 at 5:07
@Tyler Hilton: $a$ can be anything, but no matter what $a$ is, $a+(3)$ will be equal to one and only one of $0+(3)$, $1+(3)$, or $2+(3)$. For example, $-5+(3)$ is perfectly fine, it just happens to be the same set as $1+(3)$. As to how I can tell that those three are enough, it's because every number in $\mathbb{Z}$ can be written uniquely as $a=3k+r$, with $0\leq r\lt 3$ (division algorithm), and it is easy to check that the set $a+(3)$ will be the same as the set $r+(3)$ (just check equality as sets). – Arturo Magidin Feb 25 '11 at 5:10
but how can -5 be the same as 1? I know that we learn congruent classes last samester so I mean, i know WHY -5 is also 1. but how can you see it here? I think I am getting confused about (3). How do you add a single element to a set? srry, just read your edit now. Still kinda confused :( – Tyler Hilton Feb 25 '11 at 5:13
@Tyler: $-5$ is not the same as $1$! I'm saying the equivalence class of $-5$, under the relation "is congruent to modulo $I$" is the same as the equivalence class of $1$ under that equivalence relation. As to how you "add elements to sets", it's explained in the post. "$a+(3)$" is just shorthand for the set $\{a+x\mid x\in (3)\}$, that is, the set of all sums of $a$ with elements of the set $(3)$. – Arturo Magidin Feb 25 '11 at 5:16
@Tyler: As to why the equivalence classes are the same, because $-5\equiv 1$ modulo $(3)$. Or, we want to show that$$\{-5+x\mid x\in(3)\} = \{1+y\mid y\in (3)\}$$ as sets. So, pick $-5+x$ in the LHS; then $x=3k$ for some $k$, and $-5+x = 3(-2)+1+3k = 1+3(k-2)$, which is in the RHS. So $\subseteq$ holds. And if $1+y$ is in the RHS, then $y=3\ell$ for some $\ell$ (since $y\in(3)$), and $1+y = 1+3\ell = -5 + 3(\ell+2)$, which is in the LHS. So $\supseteq$ also holds, and we have equality of sets. – Arturo Magidin Feb 25 '11 at 5:19