I have a doubt in the definition of Quotient Rings Take $R$ to be a ring and $I$ to be an ideal in it.Let us label all element of $I$ be $a_{i}$

The Quotient Rings are defined as $\frac{Q}{I} = {x+I} $ where ${x+I}$ are the set of all cosets. Now the multiplication and addition are defined usually $(x+I)+(y+I)=(x+y)+I$ and $(x+I)*(y+I)=(x*y)+I$
Now i have observed a special property in addition of cosets $(x+I)+(y+I)=(x+y)+I$ here what i mean it is just not defined by addition but what actually holds $(x+I)+(y+I)=x+a_{1}+y+a_{2}=x+y+a_{1}+a_{2} \subset (x+y)+I$ This will hold since the group defined is abelian and for the other direction $\forall a_{i} (x+y)+a_{i}=x+0+y+a_{i}\subset (x+I)+(y+I)$ . I have proved this for ideals but it can be proved in the same idea for Quotient groups. However this just doesn't hold for multiplication for Quotient rings i mean I am not able to prove the other direction $ \forall i \ x*y +a_{i} = (x+a_{j})*(x+a_{k}) $ . I am fairly confident that this itself won't hold, can anyone give me an example for this.

What i think this means that addition of cosets is obvious in the sense you add all pair of elements in the coset and you get a new coset, which will be defined as the addition of cosets, however not for multiplication of cosets. My question is what is the reasoning behind this definition? Why not choose multiplication of cosets to be multiplication of each pair of elemnts and forming a new coset. Granted it will not be an existing coset but i think we can choose something stronger than an ideal for this new definition


It would really be better to stop thinking of $x+I$ as a set and starting thinking of it as a point, for that it what it actually is in the quotient group. After the cosets have been identified as equivalence classes, it becomes unnecessary to discuss them as sets from that moment on. You need only refer to the equivalence relation. Continuing to think of them as sets can be complicated and hard to work with, as you have discovered.

The definition of $(x+I) + (y+I)=x+y+I$ is a well-defined operation since if $x+I=x'+I$ and $y+I=y'+I$, then $x-x', y-y'\in I$, so their sum $x+y-x'-y'\in I$, hence $x+y+I=x'+y'+I$, and thus the operation is well-defined. (Normality of $I$ appeared indirectly in the step where I commuted the terms.)

As for multiplication, begin the same way with $x,x', y,y'$. Because $I$ is an ideal, $(x-x')y\in I$, so $xy + I=x'y+I$. Since $x'(y-y')$ is also in $I$, $x'y+I=x'y'+I$. Combining these two equalities, we have $xy+I=x'y'+I$. Thus multiplication is well-defined.

My next question is can we choose a stronger condition than ideal so that multiplication of cosets(points) is equal to multiplication of individual elements of the set

Maybe, but consider that this does not lead to good results in all cases. Suppose that $xI=yI=I^2=\{0\}$. For example, you could take $R=\Bbb Z/8\Bbb Z$, the ideal $I=4\Bbb Z/8\Bbb Z$, and $x=y=2+8\Bbb Z$. Then $(x+i)(y+i')=xy$ for all choices of $i,i'$, so in fact $(x+I)(y+I)=\{xy\}$ as sets. But this does not change the fact that $(x+I)(y+I)=xy+I$ as cosets. Setwise multiplication is not really the goal of multiplying cosets in quotient rings.

The example I picked is pretty dramatic, but I imagine it happens much more frequently than you think that $(x+I)(y+I)\subsetneq xy+I$ as sets.

Here's another example that comes to mind. Consider $R=F[x,y,z]/(xz,yz)$ and $I=(z)$. You're always going to have $(x+(z))(y+(z))\subseteq xy+(z^2)\subsetneq xy+(z)$.

Finding conditions on $R$ to produce all of $xy+I$ would require, for all choices of $x,y\in R$ and $k\in I$, to find an $i$ and $j$ such that $xy+xj+iy+ij=k$. That sounds rather hard.

  • $\begingroup$ Ok i get what you mean, My next question is can we choose a stronger condition than ideal so that multiplication of cosets(points) is equal to multiplication of individual elements of the set $\endgroup$ – Varun Sep 25 '15 at 14:23

Multiplication of cosets works as such; for $i,j \in I$ and $a,b \in R$ we have that $a+i \in a + I$ and $b+j \in b+J$, so choosing $a+i$ and $b+j$ as our coset representatives, consider the multiplication $(a+i)(b+j)$.

This is of course $ab+ib+aj+ij$, but since $I$ is an ideal, $ib, aj, ij$ are all in $I$. So then the coset $ab+I$ contains $ab+ib+aj+ij$, but we may choose any $I$ element as our representative.

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    $\begingroup$ True, but what of the other direction . $\endgroup$ – Varun Sep 25 '15 at 13:50

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