Quotient Rings and Ring Homomorphisms

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

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

Definition of Quotient Rings

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

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

should be defined as above. On the other hand, for any \(x,x'\in S\) we have

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

so for the above formula to hold we must always have \(x\alpha'+\alpha x'+xx'\in S\). In particular, setting \(x'=0\) forces \(x\alpha'\in S\) for any \(\alpha'\in A\), and setting \(x=0\) forces \(\alpha x'\in S\) for any \(\alpha\in A\). Thus \(S\) must be a two-sided ideal of \(A\). Conversely, if \(S\) is a two-sided ideal, then all three terms \(x\alpha'\), \(\alpha x'\), \(xx'\) lie in \(S\), so the above multiplication is well defined independent 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 of \(A\) by \(\mathfrak{a}\).

Then the following holds.

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

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

Proof

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

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

   and at this point one can check that \(1+\mathfrak{a}\) becomes the multiplicative identity of \(A/\mathfrak{a}\).

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

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

   so the identity

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

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

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

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

Proof

For 1 and 3 one proceeds almost exactly as in §Group Homomorphisms, and checks that the group homomorphisms obtained there are actually ring homomorphisms in the same way as in part 2 of Proposition 2.

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

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

where \(\alpha\alpha'\in S\), and since \(\ker \phi\) is a two-sided ideal by 1, the three terms in parentheses all lie in \(\ker \phi\). Adding \(1\in S\) 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 \(y\in S\cap\ker \phi\) both \(\alpha y\) and \(y\alpha\) belong to \(S\) and \(\ker \phi\), so this is a two-sided ideal of \(S\). Now consider the composition

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

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

That the two correspondences \(\bar{\mathfrak{b}}\mapsto\pi^{-1}(\bar{\mathfrak{b}})\) and \(\mathfrak{b}\mapsto\pi(\mathfrak{b})\) are inverses of each other and preserve inclusions follows from §Group Homomorphisms, ⁋Theorem 7 (The fourth isomorphism theorem). It remains to check that these correspondences send 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 \(\bar{\mathfrak{b}}\) 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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