Quotient Rings, the Ring Isomorphism Theorems

Section 9.2 Quotient Rings, the Ring Isomorphism Theorems

“Fools ignore complexity. Pragmatists suffer it. Some can avoid it. Geniuses remove it.”
―Alan Perlis

Subsection Quotient Rings

Lemma 9.16. Additive Cosets form Abelian Group.

For a two-sided ideal \(I\) of \(R\text{,}\) the set of additive cosets modulo \(I\) is \(R/I=\{r+I : r\in R\}.\) This is an abelian group with respect to addition given by \((r + I) + (s + I) = (r +s ) + I\text{.}\)

Definition 9.17. Quotient Ring.

For a two-sided ideal \(I\) of \(R\) The quotient ring of \(R\) modulo \(I\) is the set \(R/I\) with addition defined as above and multiplication given by \((r + I) \cdot (s + I) = (rs) + I\text{.}\)

Exercise 9.18. Quotient Rings are Rings.

Given a two-sided ideal \(I\) of \(R\text{,}\) prove that the quotient ring \(R/I\) is indeed a ring.

Just like we had with groups, it turns out our dear friend \(\Z/n\) has been a quotient all along.

Example 9.19. Quotients of \(\Z\).

If \(I=\igen n\) is an ideal in the ring \(\Z\text{,}\) then the quotient ring \(\Z/n\) is the familiar ring \(\Z/n\text{.}\)

And, once again, we have a quotient map that turns out to be a homomorphism.

Exercise 9.20. Quotient Map is Surjective Ring Map.

Prove that the canonical quotient map \(q:R\to R/I\) is a surjective ring homomorphism.

Last but certainly not least, we have an analogue of Theorem 3.39 for ideals.

Once again, I would like to suggest that “ideal subgroup” is a better name for a normal subgroup.

Theorem 9.21. Ideal iff Kernel of Ring Map.

Let \(R\) be a ring. A subset \(I\) of \(R\) is an ideal of \(R\) if and only if there exists a ring homomorphism \(\varphi\) such that \(\ker(\varphi)=I\text{.}\)

Subsection The Ring Isomorphism Theorems

We arrive at the isomorphism theorems for rings. We’ve seen most of this stuff before in different packaging, so lets jump right in.

Theorem 9.22. UMP for Quotient Rings.

If \(f: R \to S\) is a ring homomorphism and \(I \subseteq R\) is an ideal such that \(I \subseteq \ker(f)\text{,}\) there exists a well defined ring homomorphism \(\overline{f}: R/I \to S\) such that \(\overline{f} (r+I) = f(r)\text{.}\) Furthermore, if \(f\) is surjective then \(\overline{f}\) is surjective and if \(I=\ker(f)\) then \(\overline{f}\) is injective.

Proof.

Ignoring \(\cdot\) for a minute, we know that there is a unique homomorphism \(\overline{f}\) of abelian groups from \((R/I, +)\) to \((S, +)\) such that \(\overline{f} (r+I) = f(r)\text{.}\) It remains only to check that \(\overline{f}\) preserves multiplication: Given elements \(r + I, s + I \in R/I\text{,}\) their product is \(rs + I\text{,}\) and we have

\begin{equation*} \overline{f}(rs + I) = f(rs) = f(r)f(s) = f(r + I) f(s +I), \end{equation*}

since \(f\) preserves multiplication.

Theorem 9.23. First Isomorphism Theorem for Rings.

If \(f: R \to S\) is a ring homomorphism, then \(R/\ker(f)\cong \im(f)\) via the map \(\overline{f}\) given by \(\ov{f}(r + \ker(f)) = f(r)\text{.}\)

Proof.

The map \(\overline{f}\) is a well-defined ring homomorphism by Theorem 9.22. By the First Isomorphism Theorem for Groups for groups, the map \(\ov{f}\) is bijective, finishing the proof.

Theorem 9.24. Second Isomorphism Theorem for Rings.

Let \(S\) be a subring and let \(I\) be an ideal of \(R\text{.}\) Then \(S + I = \{s + i \mid s \in S, i \in I\}\) is a subring of \(R\text{,}\) \(S \cap I\) is an ideal of \(S\text{,}\) and

\begin{equation*} \frac{S+I}{I}\cong \frac{S}{S\cap I}. \end{equation*}

Theorem 9.25. Third Isomorphism Theorem for Rings.

If \(R\) is a ring and \(I \subseteq J\) are two ideals of \(R\text{,}\) then \(J/I\) is an ideal of \(R/I\) and

\begin{equation*} \frac{R/I}{J/I} \cong R/J \text{ via } (r + I) + J/I \mapsto r + J. \end{equation*}

Theorem 9.26. Lattice Isomorphism Theorem for Rings.

Suppose \(R\) is a ring and \(I\) is a two-sided ideal of \(R,\) and write \(\pi: R \to R/I\) for the quotient ring homomorphism. There is a bijection

\begin{equation*} \Psi:\{\text{subrings of }R\text{ containing }I\}\to \{\text{subrings of }R/I\}, \Psi(S)=\pi(S)=S/I \end{equation*}

with inverse

\begin{equation*} \Phi:\{\text{subrings of }R/I\}\to\{\text{subrings of }R\text{ containing }I\}, \Phi(S)=\pi^{-1}(S). \end{equation*}

Moreover this bijection induces a bijection between

\begin{equation*} \{\text{ideals of}\;R\text{ containing }I\}\leftrightarrow \{\text{ideals of }R/I\} \end{equation*}

since \(J\) is an ideal of \(R\) containing \(I\) if and only if \(\Psi(J)\) is an ideal of \(R/I\text{.}\)

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