15
$\begingroup$

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

  • 0
    if you're in a class, go to office hours. if you're on your own, go look at abelian groups first, since your ring is an abelian group under addition, and your quotient ring is a quotient group.2011-02-25

1 Answers 1

26

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*}

  • 0
    @Tyler: When you have two subsets of$a$ring/group, $S$, and $T$, then $S+T$ denotes the set $\{s+t\mid s\in S,t\in T\}$. So, yes. $I+I = \{a+b\mid a,b\in I\} = I$: Given any $a,b\in I$, $a+b\in I$, so $I+I\subseteq I$; and given any $x\in I$, since $0\in I$ you have $x=x+0\in I+I$; so $I\subseteq I+I$. So you have equality.2011-02-25