8.3: Ideals and Quotient Rings

Recall that in the case of a homomorphism $\phi$ of groups, the cosets of $\ker(\phi)$ have the structure of a group (that happens to be isomorphic to the image of $\phi$ by the First Isomorphism Theorem). In this case, $\ker(\phi)$ is the identity of the associated quotient group. Moreover, recall that every kernel is a normal subgroup of the domain and every normal subgroup can be realized as the kernel of some group homomorphism. Can we do the same sort of thing for rings?

Let $\phi:R\to S$ be a ring homomorphism with $\ker(\phi)=I$. Note that $\phi$ is also a group homomorphism of abelian groups and the cosets of $\ker(\phi)$ are of the form $r+I$. More specifically, if $\phi(r)=a$, then $\phi^{-1}(a)=r+I$.

These cosets naturally have the structure of a ring isomorphic to the image of $\phi$: $$\begin{aligned} (r+I)+(s+I) & = (r+s)+I\\ (r+I)(s+I) & = (rs)+I\end{aligned}$$ The reason for this is that if $\phi^{-1}(a)=X$ and $\phi^{-1}(b)=Y$, then the inverse image of $a+b$ and $ab$ are $X+Y$ and $XY$, respectively.

The corresponding ring of cosets is called the quotient ring of $R$ by $I=\ker(\phi)$ and is denoted by $R/I$. The additive structure of the quotient ring $R/I$ is exactly the additive quotient group of the additive abelian group $R$ by the normal subgroup $I$ (all subgroups are normal in abelian groups). When $I$ is the kernel of some ring homomorphism $\phi$, the additive abelian quotient group $R/I$ also has a multiplicative structure defined in (2) above, making $R/I$ into a ring.

Can we make $R/I$ into a ring for any subring $I$?

The answer is “no" in general, just like in the situation with groups. But perhaps this isn’t obvious because if $I$ is an arbitrary subring of $R$, then $I$ is necessarily an additive subgroup of the abelian group $R$, which implies that $I$ is an additive normal subgroup of the group $R$. It turns out that the multiplicative structure of $R/I$ may not be well-defined if $I$ is an arbitrary subring.

Let $I$ be an arbitrary subgroup of the additive group $R$. Let $r+I$ and $s+I$ be two arbitrary cosets. In order for multiplication of the cosets to be well-defined, the product of the two cosets must be independent of choice of representatives. Let $r+\alpha$ and $s+\beta$ be arbitrary representatives of $r+I$ and $s+I$, respectively ( $\alpha,\beta\in I$), so that $r+I=(r+\alpha)+I$ and $s+I=(s+\beta)+I$. We must have $$\begin{aligned} (r+\alpha)(s+\beta)+I & =rs+I.\end{aligned}$$ This needs to be true for all possible choices of $r,s\in R$ and $\alpha, \beta\in I$. In particular, it must be true when $r=s=0$. In this case, we must have $$\begin{aligned} \alpha\beta+I & =I.\end{aligned}$$ But this only happens when $\alpha\beta\in I$. That is, one requirement for multiplication of cosets to be well-defined is that $I$ must be closed under multiplication, making $I$ a subring.

Next, if we let $s=0$ and let $r$ be arbitrary, we see that we must have $r\beta\in I$ for every $r\in R$ and every $\beta\in I$. That is, it must be the case that $I$ is closed under multiplication on the left by elements from $R$. Similarly, letting $r=0$, we can conclude that we must have $I$ closed under multiplication on the right by elements from $R$.

On the other hand, if $I$ is closed under multiplication on the left and on the right by elements from $R$, then it is clear that relation (4) above is satisfied.

It is easy to verify that if the multiplication of cosets defined in (2) above is well-defined, then this multiplication makes the additive quotient group $R/I$ into a ring (just check the axioms for being a ring).

We have shown that the quotient $R/I$ of the ring $R$ by a subgroup $I$ has a natural ring structure if and only if $I$ is closed under multiplication on the left and right by elements of $R$ (which also forces $I$ to be a subring). Such subrings are called ideals.

Definition: Ideal

Let $R$ be a ring and let $I$ be a subset of $R$.

1. $I$ is a left ideal (respectively, right ideal) of $R$ if $I$ is a subring and $rI\subseteq I$ (respectively, $Ir\subseteq I$) for all $r\in R$.
2. $I$ is an ideal (or two-sided ideal) if $I$ is both a left and a right ideal.

Here’s a summary of everything that just happened.

Theorem $\PageIndex{1}$

Let $R$ be a ring and let $I$ be an ideal of $R$. Then the additive quotient group $R/I$ is a ring under the binary operations: $$\begin{aligned} (r+I)+(s+I) & = (r+s)+I\\ (r+I)(s+I) & = (rs)+I\end{aligned}$$ for all $r,s\in R$. Conversely, if $I$ is any subgroup such that the above operations are well-defined, then $I$ is an ideal of $R$.

Theorem $\PageIndex{2}$

If $R$ a commutative ring and $I$ is an ideal of $R$, then $R/I$ is a commutative ring.

Theorem $\PageIndex{3}$

Suppose $I$ and $J$ are ideals of the ring $R$. Then $I\cap J$ is an ideal of $R$.

As you might expect, we have some isomorphism theorems.

Theorem $\PageIndex{4}$: First Isomorphism Theorem for Rings

If $\phi:R\to S$ is a ring homomorphism, then $\ker(\phi)$ is an ideal of $R$ and $R/\ker(\phi)\cong \phi(R)$.

We also have the expected Second, Third, and Fourth Isomorphism Theorems for rings. The next theorem tells us that a subring is an ideal if and only if it is a kernel of a ring homomorphism.

Theorem $\PageIndex{5}$

If $I$ is any ideal of $R$, then the natural projection $\pi:R\to R/I$ defined via $\pi(r)=r+I$ is a surjective ring homomorphism with $\ker(\pi)=I$.

For the remainder of this section, assume that $R$ is a ring with identity $1\neq 0$.

Definition: Ideal Generated by $A$ and Principal Ideal

Let $A$ be any subset of $R$. Let $(A)$ denote the smallest ideal of $R$ containing $A$, called the ideal generated by $A$. If $A$ consists of a single element, say $A=\{a\}$, then $(a):=(\{a\})$ is called a principal ideal.

Remark $\PageIndex{1}$

The following facts are easily verified.

1. $(A)$ is the intersection of all ideals containing $A$.
2. If $R$ is commutative, then $(a)=aR:=\{ar\mid r\in R\}$.

Example $\PageIndex{1}$: Principal Ideals

In $\mathbb{Z}$, $n\mathbb{Z}=(n)=(-n)$. In fact, these are the only ideals in $\mathbb{Z}$ (since these are the only subgroups). So, all the ideals in $\mathbb{Z}$ are principal. If $m$ and $n$ are positive integers, then $n\mathbb{Z}\subseteq m\mathbb{Z}$ if and only if $m$ divides $n$. Moreover, we have $(m,n)=(d)$, where $d$ is the greatest common divisor of $m$ and $n$.

Exercise $\PageIndex{1}$

Consider the ideal $(2,x)$ in $\mathbb{Z}[x]$. Note that $(2,x)=\{2p(x)+xq(x)\mid p(x),q(x)\in\mathbb{Z}[x]\}$. Argue that $(2,x)$ is not a principal ideal, i.e., there is no single polynomial in $\mathbb{Z}[x]$ that we can use to generate $(2,x)$.

Theorem $\PageIndex{6}$

Assume $R$ is a commutative ring with $1\neq 0$. Let $I$ be an ideal of $R$. Then $I=R$ if and only if $I$ contains a unit.

Theorem $\PageIndex{7}$

Assume $R$ is a commutative ring with $1\neq 0$. Then $R$ is a field if and only if its only ideals are $(0)$ and $R$.

Loosely speaking, the previous results say that fields are “like simple groups" (i.e, groups with no non-trivial normal subgroups).

Corollary $\PageIndex{1}$: Homomorphism From Field

If $R$ is a field, then every nonzero ring homomorphism from $R$ into another ring is an injection.