This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
In this post, we define the notion of a quotient ring. Recall from §Quotient Groups that for any subgroup \(H\) of a group \(G\), the quotient \(G/H\) is always defined as a set, yet it need not carry a group structure; for that, \(H\) must be a normal subgroup. Likewise, for a ring \(A\) and a subring \(S\), the quotient \(A/S\) need not inherit a ring structure.
Definition of Quotient Rings
First, if we ignore the multiplicative structures on \(A\) and \(S\), then \(S\) is a subgroup of \(A\). Since \(A\) is an abelian group, \(A/S\) inherits an abelian group structure. To define a ring structure on this quotient, a similar property must hold for multiplication. That is, for any two elements \(\alpha+S\) and \(\alpha'+S\) in \(A/S\), their product
\[(\alpha+S)(\alpha'+S)\overset{?}{=}\alpha\alpha'+S\]must be defined as above. On the other hand, for any \(xx'\in S\),
\[(\alpha+x)(\alpha'+x')=\alpha\alpha'+x\alpha'+\alpha x'+xx'\]and since \(xx'\in S\), the above equation can hold only if \(x\alpha',\alpha x'\in S\). That is, for any element \(x\in S\) and any \(\alpha\in A\), both \(\alpha x\in S\) and \(x\alpha\in S\) must hold; hence \(S\) must be a two-sided ideal of \(A\). From this discussion we obtain the following.
Definition 1 Let \(A\) be a ring and \(\mathfrak{a}\) a two-sided ideal thereof. The ring \(A/\mathfrak{a}\) defined as above is called the quotient ring of \(A\) by \(\mathfrak{a}\).
Then the following holds.
Proposition 2 Let \(A\) be a ring and \(\mathfrak{a}\) a two-sided ideal. Then the following hold:
- The map \(\pi:A\rightarrow A/\mathfrak{a}\) defined by \(\alpha\mapsto \alpha+\mathfrak{a}\) is a ring homomorphism.
- For a ring homomorphism \(\phi:A \rightarrow B\), if \(\phi(\mathfrak{a})=\{0\}\), then there exists a unique \(\bar{\phi}:A/\mathfrak{a}\rightarrow B\) such that \(\phi=\bar{\phi}\circ\pi\).
Proof
- That \(\pi\) is an abelian group homomorphism with respect to addition follows immediately from the results of §Quotient Groups. That \(\pi\) preserves multiplication is likewise clear from the above discussion, and one easily verifies that \(1+\mathfrak{a}\) is the multiplicative identity in \(A/\mathfrak{a}\).
-
First, regard \(\phi\) as an abelian group homomorphism. Then the subgroup \(\mathfrak{a}\) of \(A\) is contained in \(\ker \phi\) by the given condition, so there exists a unique group homomorphism \(\bar{\phi}:A/\mathfrak{a}\rightarrow B\) such that \(\phi=\bar{\phi}\circ\pi\). (§Isomorphisms, ⁋Proposition 3) Now take any two elements \(\alpha+\mathfrak{a}, \beta+\mathfrak{a}\) in \(A/\mathfrak{a}\). Then
\[(\alpha+\mathfrak{a})(\beta+\mathfrak{a})=\alpha\beta+\mathfrak{a}=\pi(\alpha\beta)\]so
\[\bar{\phi}((\alpha+\mathfrak{a})(\beta+\mathfrak{a}))=\bar{\phi}(\pi(\alpha)\pi(\beta))=\bar{\phi}(\pi(\alpha\beta))=\phi(\alpha\beta)=\phi(\alpha)\phi(\beta)=\bar{\phi}(\pi(\alpha))\bar{\phi}(\pi(\beta))=\bar{\phi}(\alpha+\mathfrak{a})\bar{\phi}(\beta+\mathfrak{a})\]and therefore \(\bar{\phi}\) preserves multiplication. Similarly, \(\bar{\phi}(1+\mathfrak{a})=\bar{\phi}(\pi(1))=\phi(1)=1\), so \(\bar{\phi}\) sends \(1\) to \(1\).
The following theorem may be regarded as the ring-homomorphism version of §Isomorphisms.
Theorem 3 Let \(\phi:A \rightarrow B\) be a ring homomorphism with kernel \(\ker \phi\) and image \(\im\phi\). Then the following hold:
- \(\ker \phi\) is a two-sided ideal of \(A\), and the map \(\alpha+\ker \phi \mapsto \phi(\alpha)\) is a well-defined isomorphism \(A/\ker \phi \rightarrow \im \phi\).
- 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 \(A\), and \(S\cap\ker \phi\) is a two-sided ideal of \(S\); moreover, there is an isomorphism \((S+\ker \phi)/\ker \phi\cong S/(S\cap \ker f)\).
- If \(\mathfrak{a}, \mathfrak{b}\) are two-sided ideals of \(A\) satisfying \(\mathfrak{b}\subseteq \mathfrak{a}\), then \(\mathfrak{a}/\mathfrak{b}\) is a two-sided ideal of \(A/\mathfrak{b}\) and \((A/\mathfrak{b})/(\mathfrak{a}/\mathfrak{b})\cong A/\mathfrak{a}\).
- 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 ideals of \(A\) containing \(\mathfrak{a}\).
As with Proposition 2, the proof proceeds almost exactly as in §Isomorphisms; the only additional step is to verify that the resulting group homomorphisms are indeed ring homomorphisms.
References
[Bou] Bourbaki, N. Algebra I. Elements of Mathematics. Springer. 1998.
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