In this article, we define the notion of a quotient ring. Recall from §Quotient Groups that for any subgroup \(H\) of a group \(G\), the set \(G/H\) is always defined as a set, but it does not always carry a group structure; for this, we needed the condition that \(H\) be a normal subgroup. Similarly, for a ring \(A\) and a subring \(S\), the set \(A/S\) does not always carry a ring structure.
Definition of Quotient Rings
First, if we forget the multiplicative structure of \(A\) and \(S\), then \(S\) is a subgroup of \(A\). Since \(A\) is an abelian group, \(A/S\) has an abelian group structure. For a ring structure to be defined on this, a similar property must hold for the multiplication. That is, for any two elements \(\alpha+S\) and \(\alpha'+S\) of \(A/S\), their product
\[(\alpha+S)(\alpha'+S)\overset{?}{=}\alpha\alpha'+S\]should be defined as above. Now, for any \(xx'\in S\),
\[(\alpha+x)(\alpha'+x')=\alpha\alpha'+x\alpha'+\alpha x'+xx'\]and since \(xx'\in S\), for the above equation to hold, we must have \(x\alpha',\alpha x'\in S\) always. That is, for an element \(x\) of \(S\), when we take any \(\alpha\in A\), both \(\alpha x\in S\) and \(x\alpha\in S\) must hold, so \(S\) must be a two-sided ideal of \(A\). 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 holds:
- The function \(\pi:A\rightarrow A/\mathfrak{a}\) 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 \(\bar{\phi}\) from \(A/\mathfrak{a}\) to \(B\) such that \(\phi=\bar{\phi}\circ\pi\) holds.
Proof
- That \(\pi\) defines an abelian group homomorphism with respect to addition follows immediately from the results of §Quotient Groups. That \(\pi\) preserves multiplication is also clear from the above discussion, and we can verify that \(1+\mathfrak{a}\) is the multiplicative identity of \(A/\mathfrak{a}\).
-
First, consider \(\phi\) as an abelian group homomorphism. By the given condition, 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 \(\phi=\bar{\phi}\circ\pi\) holds. (§Isomorphisms, ⁋Proposition 3) Now choose any two 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 by the equation
\[\bar{\phi}((\alpha+\mathfrak{a})(\beta+\mathfrak{a}))=\bar{\phi}(\pi(\alpha)\pi(\beta))=\bar{\phi}(\pi(\alpha\beta))=\phi(\alpha\beta)=\phi(\alpha)\phi(\beta)=\bar{\phi}(\pi(\alpha))\bar{\phi}(\pi(\beta))=\bar{\phi}(\alpha+\mathfrak{a})\bar{\phi}(\beta+\mathfrak{a})\]we see 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 considered as the ring homomorphism version of §Isomorphisms.
Theorem 3 For a ring homomorphism \(\phi:A \rightarrow B\) with kernel \(\ker \phi\) and image \(\im\phi\), the following holds:
- \(\ker \phi\) is a two-sided ideal of \(A\), and \(\alpha+\ker \phi \mapsto \phi(\alpha)\) defines a well-defined isomorphism \(A/\ker \phi \rightarrow \im \phi\).
- 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 \(S\), with an isomorphism \((S+\ker \phi)/\ker \phi\cong S/(S\cap \ker f)\).
- 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 \((A/\mathfrak{b})/(\mathfrak{a}/\mathfrak{b})\cong A/\mathfrak{a}\) holds.
- 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 \(\mathfrak{a}\).
The proof follows almost identically to what was done in §Isomorphisms, as in Proposition 2, with the only additional step being 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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