The definition of a quotient ring is a straightforward generalization of the ring structure we defined for modular rings \(\Z/m\Z\) in Section 1.2. Since an ideal \(I\) of a ring \(R\) is in particular a subgroup of \((R,+)\text{,}\) and a normal one at that (since \((R,+)\) is abelian), we can form the coset space
As usual, once we know the two operations \(+\) and \(\cdot\) are well defined on \(R/I\text{,}\) the fact that they satisfy the ring axioms follows almost immediately from their being defined in terms of the operations of \(R\text{.}\)
The triple \((R/I,+,\cdot)\) is a ring, called the quotient ring of \(R\) by \(I\text{.}\) Its additive identity is \(0=0+I\text{,}\) and its multiplciative identity is \(1=1+I\text{.}\)
Let \(n\) be a positive integer. Show that the ring \(\Z/n\Z\) is equal to the quotient ring \(\Z/I\text{,}\) where \(I=\angvec{n}=\{kn\mid k\in \Z\}\text{.}\)
We have seen previously that as a group, \(\Z/n\Z\) is equal to the quotient group \(\Z/\angvec{n}\text{.}\) Moreover, it is clear that the coset multiplication operation is identical to the modular multiplication operation defined in Section 1.2. Thus the two rings are identical.
The connection between ring quotients and group quotients is strong. Indeed, by definition a ring quotient is equal to the group quotient \(R/I\text{,}\) thinking of \(I\) as a subgroup of \((R,+)\text{,}\) together with an extra layer of strucutre coming from the coset multiplication. It is not all that surprising, then, that ring quotients enjoy many of the same properties as group quotients, starting with a universal mapping property.
Let \(I\) be an ideal of the ring \(R\text{,}\) and let \(\pi\colon R\rightarrow R/I\) be the corresponding quotient map. Given any ring homomorphism \(\phi\colon R\rightarrow S\) satisfying \(I\subseteq \ker \phi\text{,}\) there exists a unique ring homomorphism \(\overline{\phi}\colon R/I\rightarrow S\) satisfying
We get most of this result for free from Theorem 1.13.1. The only thing we need to be careful about is that all maps in question satisfy the additional ring homomorphism axioms (ii) and (iii) from Definition 2.5.1, and this is straightforward.
The following corollary is simply a compact restatement of Theorem 2.6.2 using the language of \(\Hom(R,S)\) and can be handy for computing the set of homomorphisms between rings.
is an injection onto the set of all \(\phi\in \Hom(R,S)\) satisfying \(I\subseteq \ker \phi\text{.}\) In other words, we can identify \(\Hom(R/I, S)\) with the set
Of course, we also get an analogue of the isomorphism theorems for rings. We state them here without proof, and in somewhat terse fashion. Notice that some things are simplified by the fact that the normality issue does not arise (since \((R,+)\) is an abelian group).
Given a ring homomorphism \(\phi\colon R\rightarrow S\) with kernel \(I=\ker\phi\text{,}\) we have \(R/I\cong im\phi\) via the isomorphism \(\overline{\phi}(r+I)=\phi(r)\text{.}\)
Given a subring \(S\subseteq R\) and ideal \(I\subseteq R\text{,}\)\(S+I\) is a subring of \(R\text{,}\)\(S\cap I\) is an ideal of \(S\text{,}\) and \((S+I)/I\cong S/S\cap I\text{.}\)
Let \(\phi\colon R\rightarrow S\) be a surjective ring homomorphism. If \(J\) is an ideal of \(R\) containing \(\ker \phi\text{,}\) then \(\phi(J)\) is an ideal of \(S\text{,}\) and we have
In particular, considering the quotient map \(\pi\colon R\rightarrow R/I\text{,}\) we see that if \(I\subseteq J\) is a chain of ideals in \(R\text{,}\) then \(J/I\) is an ideal of \(R/I\text{,}\) and we have
We claim that \(\R[x]/(x^2+1)\cong \C\text{.}\) Define \(\phi\colon \R[x]\rightarrow \C\) as \(\phi(f)=f(i)\text{:}\) i.e., \(\phi\) is the evaluation at \(i\) map, thinking of \(\R\) as a subring of \(\C\text{.}\) This a ring homomorphism according to Theorem 2.5.7, and is easily seen to be surjective, since \(a+bi=\phi(a+bx)\) for any \(a,b\in \R\text{.}\) We claim that \(\ker\phi=(x^2+1)\text{.}\) Indeed, we see that
Thus \(x^2+1\in \ker\phi\text{.}\) It follows from Corollary 2.5.18 that \((x^2+1)\subseteq \ker\phi\text{.}\) Conversely, suppose \(f\in \ker\phi\text{,}\) and thus \(f(i)=0\text{.}\) Applying polynomial division (with remainder), we can write
But in \(\C\text{,}\) we have \(a+bi=0\text{,}\)\(a,b\in \R\text{,}\) if and only if \(a=b=0\text{.}\) Thus \(r(x)=0\text{,}\) and we see that \(f(x)=q(x)(x^2+1)\in (x^2+1)\text{.}\) Thus \(\ker\phi=(x^2+1)\text{,}\) and we conclude that
We have \(\Z/n\Z=\Z/I\text{,}\) where \(I=(n)\text{.}\) Let \(\pi\colon \Z\rightarrow \Z/I\) be the quotient map. We have \(\ker\pi=(n)\text{.}\) By the fourth isomorphism theorem, the ideals of \(\Z/n\Z\) are in bijective correspondence with the ideals of \(\Z\) that contain \((n)\text{.}\) A previous result tells us that these are precisely ideals of the form \((a)\text{,}\) where \(a\geq 0\) and \(a\mid n\text{.}\) Since the bijection is given by \(J\mapsto \pi(J)\text{,}\) we see that the ideals of \(\Z/n\Z\) are preciseley those of the form \((\overline{a})\text{,}\) where \(a\geq 0\) and \(a\mid n\text{.}\) In particular, all ideals of \(\Z/n\Z\) are principal, and there are finitely many of them.
Let \(n\) be a positive integer. Determine, up to isomorphism, all the quotients of \(\Z/n\Z\) by an ideal. Your answers should be expressed as familiar rings.
Let \(I=(n)\text{,}\) so that \(\Z/n\Z=\Z/I\text{.}\) From the previous example, the ideals of \(\Z/n\Z\) are precisely those of the form \((\overline{a})\text{,}\) where \(a\geq 0\) and \(a\mid n\text{.}\) Furthermore we have for any such \(a\text{,}\)
