Why is $I$ often an ideal in quotient ring $A/I$?

Question

When talking about quotient ring $A/I$, where $A$ is a ring, $I$ is often assumed to be an ideal. Why is this so? What makes ideals very important when discussing quotient ring?

Answer 1

First, to construct the quotient space, you want the relation $\sim$ defined by $x\sim y\iff x-y\in I$ to be an equivalence relation. Then you want to be able to define operations for the cosets so that $[x]+[y]=[x+y]$ and $[xy]=[x][y]$ are well defined and give $A/I$ the structure of a ring. From these requirements it follows that $I$ is an ideal. That is, you cannot form the quotient ring if $I$ is not an ideal.

Answer 2

Here is a fairly intuitive way to see why $I$ must be an ideal for the quotient to be a ring.

When we take a quotient $R/I$, what we would like to do is identify all elements of $I$ with a single element, and then identify other elements "as needed".

Recalling that $R$ is in particular an additive group, to get a nice quotient we want $I$ to be a subgroup (and fortunately, normality is then automatic as the group is abelian). This means that the additive structure on $R/I$ should be the one we are used to by taking a quotient of groups, and in particular, we need to identify all elements of $I$ with $0$.

Now (writing $\overline{y} = y + I$ for the class containing $y$ in $R/I$), we get that for any $r\in R$ and $x\in I$ we have $$\overline{rx} = \overline{r}\cdot\overline{x} = \overline{r}\cdot\overline{0} = \overline{r\cdot 0} = \overline{0}$$

So for all $r\in R$ and $x\in I$ we see that $\overline{rx} = \overline{0}$ in $R/I$. But the way we did the construction, we not only wanted to identify everything in $I$ with $0$, we did not want to identify anything else with $0$ (since we wanted to take the usual quotient of groups). So the above precisely says that $rx\in I$.

In some sense this means that the fact that being an ideal is sufficient to define a ring structure on the quotient is actually a bit of a surprise (we might well have needed something stronger, as we have above just used that multiplying with $0$ gives $0$). But it is of course an easy exercise that being an ideal is sufficient.