8.3: Ideals and Quotient Rings
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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.
Let \(R\) be a ring and let \(I\) be a subset of \(R\).
- \(I\) is a left ideal (respectively, right ideal) of \(R\) if \(I\) is a subring and \(rI\subseteq I\) (respectively, \(Ir\subseteq I\)) for all \(r\in R\).
- \(I\) 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.
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 \(R\).
If \(R\) a commutative ring and \(I\) is an ideal of \(R\), then \(R/I\) is a commutative ring.
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.
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.
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 \(1\neq 0\).
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.
The following facts are easily verified.
- \((A)\) is the intersection of all ideals containing \(A\).
- If \(R\) is commutative, then \((a)=aR:=\{ar\mid r\in R\}\).
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 \(n\).
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 \((2,x)\).
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.
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).
If \(R\) is a field, then every nonzero ring homomorphism from \(R\) into another ring is an injection.