Quotient Rings and Ring Isomorphisms

In this post we define the notion of a quotient ring. Recalling when we defined §Quotient Groups, for any group \(G\) and any subgroup \(H\), \(G/H\) is always defined as a set, but it did not always have a group structure, and for this, the condition that \(H\) is a normal subgroup was required. Likewise, for a ring \(A\), the manner in which a quotient can be defined is restricted.

Definition of Quotient Rings

First, if \(A\) is an abelian group and \(S\) is its subgroup, an abelian group structure exists on \(A/S\). For a ring structure to be defined on top of this, a similar property must hold for the multiplicative structure as well. That is, for any two elements \(\alpha+S\), \(\alpha'+S\) of \(A/S\), their product

\[(\alpha+S)(\alpha'+S)\overset{?}{=}\alpha\alpha'+S\]

must be defined as above. On the other hand, since for any \(x,x'\in S\)

\[(\alpha+x)(\alpha'+x')=\alpha\alpha'+x\alpha'+\alpha x'+xx'\]

in order for the above formula to hold, \(x\alpha'+\alpha x'+xx'\in S\) must always hold. In particular, setting \(x'=0\), for any \(\alpha'\in A\) we must have \(x\alpha'\in S\), and setting \(x=0\), for any \(\alpha\in A\) we must have \(\alpha x'\in S\). That is, \(S\) must be a two-sided ideal of \(A\). Conversely, if \(S\) is a two-sided ideal, then the three terms \(x\alpha'\), \(\alpha x'\), \(xx'\) all belong to \(S\), so the multiplication above is well-defined independently of the choice of representatives. From this discussion we obtain the following.

Definition 1 Let a ring \(A\) and a two-sided ideal \(\mathfrak{a}\) be given. The ring \(A/\mathfrak{a}\) defined as above is called the quotient ring by \(\mathfrak{a}\) of \(A\).

Then the following holds.

Proposition 2 For a ring \(A\) and a two-sided ideal \(\mathfrak{a}\), the following hold.

1. Defined by \(\alpha\mapsto \alpha+\mathfrak{a}\), the function \(\pi:A\rightarrow A/\mathfrak{a}\) is a ring homomorphism.
2. For a ring homomorphism \(\phi:A \rightarrow B\), if \(\phi(\mathfrak{a})=\{0\}\), then from \(A/\mathfrak{a}\) to \(B\) there exists a unique ring homomorphism \(\bar{\phi}\) such that \(\phi=\bar{\phi}\circ\pi\) holds.

Proof

1. That \(\pi\) defines an abelian group homomorphism with respect to addition is a result of §Quotient Groups. That \(\pi\) preserves multiplication is obtained from the following calculation

   \[\pi(\alpha)\pi(\alpha')=(\alpha+\mathfrak{a})(\alpha'+\mathfrak{a})=\alpha\alpha'+\mathfrak{a}=\pi(\alpha\alpha')\]

   and here one can verify that \(1+\mathfrak{a}\) becomes the \(1\) of \(A/\mathfrak{a}\).

2. First regard \(\phi\) as an abelian group homomorphism. Then by the given condition, the subgroup of \(A\), \(\mathfrak{a}\), is contained in \(\ker \phi\), so from \(A/\mathfrak{a}\) to \(B\) there exists a unique group homomorphism \(\bar{\phi}:A/\mathfrak{a}\rightarrow B\) such that \(\phi=\bar{\phi}\circ\pi\) holds. (§Group Isomorphisms, ⁋Proposition 3)
   Now arbitrarily choose two elements of \(A/\mathfrak{a}\), \(\alpha+\mathfrak{a}, \beta+\mathfrak{a}\). Then since

   \[(\alpha+\mathfrak{a})(\beta+\mathfrak{a})=\alpha\beta+\mathfrak{a}=\pi(\alpha\beta)\]

   by the following equation,

   \[\bar{\phi}((\alpha+\mathfrak{a})(\beta+\mathfrak{a}))=\bar{\phi}(\pi(\alpha)\pi(\beta))=\bar{\phi}(\pi(\alpha\beta))=\phi(\alpha\beta)=\phi(\alpha)\phi(\beta)=\bar{\phi}(\pi(\alpha))\bar{\phi}(\pi(\beta))=\bar{\phi}(\alpha+\mathfrak{a})\bar{\phi}(\beta+\mathfrak{a})\]

   \(\bar{\phi}\) preserves multiplication. Similarly, from \(\bar{\phi}(1+\mathfrak{a})=\bar{\phi}(\pi(1))=\phi(1)=1\), \(\bar{\phi}\) sends \(1\) to \(1\).

The following theorem can be regarded as the ring homomorphism version of §Group Isomorphisms.

Theorem 3 For a ring homomorphism \(\phi:A \rightarrow B\), its kernel \(\ker \phi\), and its image \(\im\phi\), the following hold.

1. \(\ker \phi\) is a two-sided ideal of \(A\), and \(\alpha+\ker \phi \mapsto \phi(\alpha)\) defines a well-defined isomorphism \(A/\ker \phi \rightarrow \im \phi\).
2. For a subring of \(A\), \(S\), the set \(S+\ker \phi=\{\alpha+x\mid\alpha\in S, x\in\ker \phi\}\) is a subring of \(A\), \(S\cap\ker \phi\) becomes a two-sided ideal of \(S\), and there exists an isomorphism \((S+\ker \phi)/\ker \phi\cong S/(S\cap \ker \phi)\).
3. If two two-sided ideals of \(A\), \(\mathfrak{a}, \mathfrak{b}\), satisfy \(\mathfrak{b}\subseteq \mathfrak{a}\), then \(\mathfrak{a}/\mathfrak{b}\) is a two-sided ideal of \(A/\mathfrak{b}\) and \((A/\mathfrak{b})/(\mathfrak{a}/\mathfrak{b})\cong A/\mathfrak{a}\) holds.
4. For a two-sided ideal of \(A\), \(\mathfrak{a}\), there exists an inclusion-preserving bijection between the set of two-sided ideals of \(A/\mathfrak{a}\) and the set of two-sided ideals containing \(\mathfrak{a}\) in \(A\).

Proof

Parts 1 and 3 proceed almost identically to what was treated in §Group Isomorphisms; one only needs to verify that the group homomorphism obtained there is in fact also a ring homomorphism in the same manner as part 2 of Proposition 2.

In the case of 2, that \(S+\ker \phi\) is a subgroup under addition is the same as in the case of groups. For any \(\alpha,\alpha'\in S\) and \(x,x'\in\ker \phi\), in

\[(\alpha+x)(\alpha'+x')=\alpha\alpha'+(x\alpha'+\alpha x'+xx')\]

we have \(\alpha\alpha'\in S\), and since \(\ker \phi\) is a two-sided ideal by 1, the three terms in parentheses all belong to \(\ker \phi\). Adding \(1\in S\) to this, we obtain that \(S+\ker \phi\) is a subring of \(A\). Also, \(S\cap\ker \phi\) is a subgroup of \(S\) under addition, and for any \(\alpha\in S\) and \(y\in S\cap\ker \phi\), since both \(\alpha y\) and \(y\alpha\) belong to both \(S\) and \(\ker \phi\), this is a two-sided ideal of \(S\). Now considering the composition

\[S\hookrightarrow S+\ker \phi\longrightarrow (S+\ker \phi)/\ker \phi\]

this is surjective and its kernel is \(S\cap\ker \phi\), so applying 1 yields the desired isomorphism.

That the two correspondences \(\bar{\mathfrak{b}}\mapsto\pi^{-1}(\bar{\mathfrak{b}})\) and \(\mathfrak{b}\mapsto\pi(\mathfrak{b})\) in 4 are inverses of each other and preserve inclusion follows from §Group Isomorphisms, ⁋Theorem 7 (The fourth isomorphism theorem). What remains is that this correspondence sends two-sided ideals to two-sided ideals. First, for any \(\alpha\in A\) and \(x\in\pi^{-1}(\bar{\mathfrak{b}})\),

\[\pi(\alpha x)=\pi(\alpha)\pi(x)\in\bar{\mathfrak{b}}\]

so \(\alpha x\in\pi^{-1}(\bar{\mathfrak{b}})\), and thus \(\pi^{-1}(\bar{\mathfrak{b}})\) is closed under left multiplication. Similarly, since \(\pi\) is surjective, any element of \(A/\mathfrak{a}\) is of the form \(\pi(\alpha)\), and therefore

\[\pi(\alpha)\pi(x)=\pi(\alpha x)\in\pi(\mathfrak{b})\]

from which we can verify that \(\pi(\mathfrak{b})\) is closed under left multiplication. The case of right multiplication can be shown in the same way, and hence these are two-sided ideals.

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References

[Bou] Bourbaki, N. Algebra I. Elements of Mathematics. Springer. 1998.

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