## 6.2 Quotient Rings and Homomorphisms We now show that the set of congruence classes modulo an ideal is itself a ring. As you might expect, this is a straightforward generalization of what we did with congruence classes in $\mathbb{Z}$ and $F[x]$. However, you may not have expected these rings of congruence classes to have close connections with some topics studied in Chapter 3, isomorphisms and homomorphisms. These connections are explored in detail and provide new insight into the structure of rings. Let $I$ be an ideal in a ring $R$. The elements of the set $R/I$ are the cosets of $I$ (congruence classes modulo $I$), that is, all sets of the form $a + I = \{a + i \ | \ i \in I\}$. In order to define addition and multiplication of cosets as we did with congruence classes in $\mathbb{Z}$ and $F[x]$, we need ### Theorem 6.8 Let $I$ be an ideal in a ring $R$. If $a + I = b + I$ and $c + I = d + I$ in $R/I$, then $$(a + c) + I = (b + d) + I$$ and $$ac + I = bd + I.$$ #### Proof This is a generalization of Theorem 2.6, in slightly different notation. Replace "[a]" by "a + I" and copy the proof of Theorem 2.6, using Theorems 6.5 and 6.6 in place of Theorems 2.2 and 2.3. We can now define addition and multiplication in $R/I$ just as we did in $\mathbb{Z}_n$ and $F[x]/(p(x))$: The sum of the coset $a + I$ (congruence class of $a$) and the coset $c + I$ (congruence class of $c$) is the coset $(a + c) + I$ (congruence class of $a + c$). In symbols, $$(a + I) + (c + I) = (a + c) + I.$$ This statement may be a bit confusing because the plus sign is used with three entirely different meanings: - as a formal symbol to denote a coset: $a + I$; - as an operation on elements of $R$: $a + c$; - as the addition operation on cosets that is being defined.* The important thing is that, because of Theorem 6.8, coset addition is independent of the choice of representative elements in each coset. Even if we replace $a + I$ by an equal coset $b + I$ and replace $c + I$ by an equal coset $d + I$, the resulting coset sum, namely $(b + d) + I$, is the same as $(a + c) + I$. Multiplication of cosets is defined similarly and is independent of the choice of representatives by Theorem 6.8: $$(a + I)(c + I) = ac + I.$$ ### Example 1 If $I$ is the principal ideal $(3)$ in $\mathbb{Z}$, then addition and multiplication of cosets is the same as addition and multiplication of congruence classes in Section 2.2. Thus $\mathbb{Z}/I$ is just the ring $\mathbb{Z}_3$. ### Example 2† If $F$ is a field, $p(x)$ is a polynomial in $F[x]$, and $I$ is the principal ideal $(p(x))$, then cosets of $I$ are precisely congruence classes modulo $p(x)$, so that addition and multiplication of cosets are done exactly as they were in Section 5.2. Thus $F[x]/I$ is the congruence-class ring $F[x]/(p(x))$. ### Example 3 Let $I$ be the ideal of polynomials with even constant terms in $\mathbb{Z}[x]$. As we saw in Example 12 of Section 6.1, $\mathbb{Z}[x]/I$ consists of just two distinct cosets, $0 + I$ and $1 + I$. We have $(1 + I) + (1 + I) = (1 + 1) + I = 2 + I = 0 + I$, so that $2 \equiv 0 \pmod{I}$ and, hence, $2 + I = 0 + I$. Similar calculations produce the following tables for $\mathbb{Z}[x]/I$. It is easy to see that $\mathbb{Z}[x]/I$ is a ring (in fact, a field) isomorphic to $\mathbb{Z}_2$: Addition Table: $$ \begin{array}{c|ccc} + & 0 + I & 1 + I \\ \hline 0 + I & 0 + I & 1 + I \\ 1 + I & 1 + I & 0 + I \\ \end{array} $$ Multiplication Table: $$ \begin{array}{c|ccc} \cdot & 0 + I & 1 + I \\ \hline 0 + I & 0 + I & 0 + I \\ 1 + I & 0 + I & 1 + I \\ \end{array} $$ --- *This ambiguity can be avoided by using a different notation for cosets, such as $[a]$, and a different symbol for coset addition, such as $\oplus$. The notation above is customary, however, and once you're used to it, there should be no confusion. †Skip this example if you have not read Chapter 5. These examples illustrate the following theorem, which should not be very surprising in view of your previous experience with $\mathbb{Z}$ and $F[x]$. ### Theorem 6.9 Let $I$ be an ideal in a ring $R$. Then 1. $R/I$ is a ring, with addition and multiplication of cosets as defined previously. 2. If $R$ is commutative, then $R/I$ is a commutative ring. 3. If $R$ has an identity, then so does the ring $R/I$. #### Proof 1. With the usual change of notation ("$a + I$" instead of "$[a]$"), the proof of Theorem 2.7 carries over to the present situation since that proof depends only on the fact that $\mathbb{Z}$ is a ring. Don't take our word for it, though; write out the proof in detail for yourself. 2. If $R$ is commutative and $a, c \in R$, then $ac = ca$. Consequently, in $R/I$ we have $(a + I)(c + I) = ac + I = ca + I = (c + I)(a + I)$. Hence, $R/I$ is commutative. 3. The identity in $R/I$ is the coset $1_R + I$ because $(a + I)(1_R + I) = a1_R + I = a + I$ and similarly $(1_R + I)(a + I) = a + I$. The ring $R/I$ is called the **quotient ring** (or **factor ring**) of $R$ by $I$. One sometimes speaks of factoring out the ideal $I$ to obtain the quotient ring $R/I$. ### Homomorphisms Quotient rings are the natural generalization of congruence-class arithmetic in $\mathbb{Z}$ and $F[x]$. As is often the case in mathematics, however, a concept developed with one idea in mind may have unexpected linkages with other important mathematical concepts. That is precisely the situation here. We shall now see that the concept of homomorphism that arose in our study of isomorphism of rings in Chapter 3 is closely related to ideals and quotient rings. #### Definition Let $f: R \rightarrow S$ be a homomorphism of rings. Then the **kernel** of $f$ is the set $$K = \{ r \in R \ | \ f(r) = 0_S \}.$$ Thus, the kernel of $f$ is the subset of $R$ consisting of those elements of $R$ that $f$ maps to $0_S$ in $S$. Note that $0_R$ is in the kernel since $f(0_R) = 0_S$ by Theorem 3.10. However, the kernel may also contain nonzero elements. ### Example 4 In Example 6 of Section 3.3, we saw that the function $f: \mathbb{Z} \rightarrow \mathbb{Z}_6$ defined by $f(r) = [r] \in \mathbb{Z}_6$ is a homomorphism of rings. Its kernel $K$ contains many nonzero integers. For instance, $12 \in K$ because $f(12) = [12] = [0]$ in $\mathbb{Z}_6$. In fact, every multiple of 6 is in the kernel because $$ K = \{ r \in \mathbb{Z} \ | \ f(r) = [0] \} = \{ r \in \mathbb{Z} \ | \ [r] = [0] \} \tag{Definition of } f $$ $$ = \{ r \in \mathbb{Z} \ | \ r \equiv 0 \pmod{6} \} \tag{Theorem 2.3} $$ $$ = \{ r \in \mathbb{Z} \ | \ 6 | r \} \tag{Definition of congruence mod 6} $$ $$ = \text{all multiples of 6} \tag{$6 | r$ means $r$ is a multiple of 6}. $$ So the kernel $K$ is the principal ideal $(6)$ in $\mathbb{Z}$. ### Example 5 The function $\theta : \mathbb{R}[x] \rightarrow \mathbb{R}$ that sends each polynomial in $\mathbb{R}[x]$ to its constant term in $\mathbb{R}$ is a ring homomorphism (Exercise 1). Its kernel consists of all polynomials with constant term 0. But every polynomial with 0 constant term is divisible by $x$. So the kernel is the principal ideal $(x)$ in $\mathbb{R}[x]$. Examples 4 and 5 provide examples of the following theorem. ### Theorem 6.10 Let $f: R \rightarrow S$ be a homomorphism of rings. Then the kernel $K$ of $f$ is an ideal in the ring $R$. #### Proof We shall use Theorem 6.1 to show that $K = \{ r \in R \ | \ f(r) = 0_S \}$ is an ideal. We must verify that it is a nonempty subset of $R$ that is closed under subtraction and absorbs products. First, $K$ is nonempty because $0_R \in K$ as noted before Example 4. To prove that $K$ is closed under subtraction, we must show that for $a, b \in K$, the element $a - b$ is also in $K$. To show $a - b \in K$, we must show that $f(a - b) = 0_S$. This follows from the fact that $f$ is a homomorphism and that $f(a) = 0_S$ and $f(b) = 0_S$ (because $a, b \in K$): $$ f(a - b) = f(a) - f(b) = 0_S - 0_S = 0_S. $$ To prove that $K$ absorbs products we must first verify that $ra \in K$ for any $r \in R$ and $a \in K$, that is, that $f(ra) = 0_S$; here's the proof: $$ f(ra) = f(r)f(a) = f(r) \cdot 0_S = 0_S. $$ A similar argument shows that $ar \in K$. Therefore $K$ is an ideal by Theorem 6.1. In Examples 4 and 5, the kernel of the homomorphism contained many nonzero elements. Sometimes, however, the kernel of a homomorphism contains only $0_R$, in which case we have an interesting result. ### Theorem 6.11 Let $f: R \rightarrow S$ be a homomorphism of rings with kernel $K$. Then $K = (0_R)$ if and only if $f$ is injective. #### Proof Suppose that $K = (0_R)$. We must show that $f$ is injective, so assume that $a, b \in R$ and $f(a) = f(b)$. Because $f$ is a homomorphism, $$ f(a - b) = f(a) - f(b) = 0_S. $$ Hence, $a - b$ is in the kernel $K = (0_R)$, which means that $a - b = 0_R$ and $a = b$. Therefore $f$ is injective. Conversely, suppose $f$ is injective. If $c \in K$, we must show that $c = 0_R$. By the definition of the kernel, $f(c) = 0_S$. By Theorem 3.10, $f(0_R) = 0_S = f(c)$. Therefore, $c = 0_R$ because $f$ is injective. Hence, the kernel consists of the single element $0_R$, that is, $K = (0_R)$. ### Example 6 In Example 7 of Section 3.3, we saw that the function $g: \mathbb{R} \rightarrow M(R)$ given by $$ g(r) = \begin{pmatrix} 0 & r \\ -r & 0 \end{pmatrix} $$ is a ring homomorphism. Its kernel of $g$ consists of all real numbers $r$ such that $g(r) = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$, that is, such that $$ \begin{pmatrix} 0 & r \\ -r & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}. $$ This can only occur when $r = 0$. So the kernel is the zero ideal $(0)$. Hence, $g$ is injective by Theorem 6.11. Theorem 6.10 states that every kernel is an ideal. Conversely, every ideal is the kernel of a homomorphism: ### Theorem 6.12 Let $I$ be an ideal in a ring $R$. Then the map $\pi: R \rightarrow R/I$ given by $\pi(r) = r + I$ is a surjective homomorphism with kernel $I$. The map $\pi$ is called the **natural homomorphism** from $R$ to $R/I$. #### Proof of Theorem 6.12 The map $\pi$ is surjective because given any coset $r + I$ in $R/I$, $\pi(r) = r + I$. The definition of addition and multiplication in $R/I$ shows that $\pi$ is a homomorphism: $$ \pi(r + s) = (r + s) + I = (r + I) + (s + I) = \pi(r) + \pi(s); $$ $$ \pi(rs) = rs + I = (r + I)(s + I) = \pi(r) \pi(s). $$ The kernel of $\pi$ is the set of elements $r \in R$ such that $\pi(r) = 0_{R/I} + I$ (the zero element in $R/I$). However, $\pi(r) = 0_{R/I} = 0 + I$ if and only if $r + I = 0_R + I$, which occurs if and only if $r = 0_R \ (\text{mod } I)$, that is, if and only if $r \in I$. Therefore, $I$ is the kernel of $\pi$. The natural homomorphism $\pi$ in Theorem 6.12 is a special case of a more general situation. If $f: R \rightarrow S$ is a surjective homomorphism of rings, we say that $S$ is a **homomorphic image** of $R$. If $f$ is actually an isomorphism (so that $S$ is an isomorphic image of $R$), then we know that $R$ and $S$ have identical structure. Whenever one of them has a particular algebraic property, the other one has it too. If $f$ is not an isomorphism, then properties of one ring may not hold in the other. However, the properties of $S$ and the homomorphism $f$ often give us useful information about $R$. An analogy with sculpture and photography may be helpful: If $f: R \rightarrow S$ is an isomorphism, then $S$ is an exact, three-dimensional replica of $R$. If $f$ is only a surjective homomorphism, then $S$ is a two-dimensional photographic image of $R$ in which some features of $R$ are accurately reflected but others are distorted or missing. The next theorem tells us precisely how $R$, $S$, and the kernel of $f$ are related in these circumstances. ### Theorem 6.13: First Isomorphism Theorem Let $f: R \rightarrow S$ be a surjective homomorphism of rings with kernel $K$. Then the quotient ring $R/K$ is isomorphic to $S$. The theorem states that every homomorphic image of a ring $R$ is isomorphic to a quotient ring $R/K$ for some ideal $K$. Thus if you know all the quotient rings of $R$, then you know all the possible homomorphic images of $R$. The ideal $K$ measures how much information is lost in passing from the ring $R$ to the homomorphic image $R/K$. When $K = (0_R)$, then $f$ is an isomorphism by Theorem 6.11, and no information is lost. But when $K$ is large, quite a bit may be lost. #### Proof of Theorem 6.13 We shall define a function $\varphi$ from $R/K$ to $S$ and then show that it is an isomorphism. To define $\varphi$, we must associate with each coset $r + K$ of $R/K$ an element of $S$. A natural choice for such an element would be $f(r) \in S$; in other words, we would like to define $\varphi: R/K \rightarrow S$ by the rule $\varphi(r + K) = f(r)$. The only possible problem is that a coset can be labeled by many different elements of $R$. So we must show that the value of $\varphi$ depends only on the coset and not on the particular representative $r$ chosen to name it. If $r + K = t + K$, then $r \equiv t \ (\text{mod } K)$ by Theorem 6.6, which means that $r - t \in K$ by the definition of congruence. Consequently, since $f$ is a homomorphism, $f(r) - f(t) = f(r - t) = 0_S$. Therefore, $r + K = t + K$ implies that $f(r) = f(t)$. It follows that the map $\varphi: R/K \rightarrow S$ given by the rule $\varphi(r + K) = f(r)$ is a well-defined function, independent of how the coset is written. If $s \in S$, then $s = f(r)$ for some $r \in R$ because $f$ is surjective. Thus $s = f(r) = \varphi(r + K)$, and $\varphi$ is surjective. To show that $\varphi$ is injective, we assume that $\varphi(r + K) = \varphi(c + K)$ and show that $r + K = c + K$, as follows: $$ \varphi(r + K) = \varphi(c + K) $$ $$ f(r) = f(c) \tag{Definition of } \varphi $$ $$ f(r) - f(c) = 0_S $$ $$ f(r - c) = 0_S. \tag{$f$ is a homomorphism.} $$ Thus, $r - c \in K$ and hence, $r \equiv c \ (\text{mod } K)$. So $r + K = c + K$ by Theorem 6.6. Therefore, $\varphi$ is injective. Finally, $\varphi$ is a homomorphism because $f$ is $$ \varphi[(c + K)(d + K)] = \varphi(cd + K) = f(cd) = f(c)f(d) = \varphi(c + K)\varphi(d + K) $$ and $$ \varphi[(c + K) + (d + K)] = \varphi[(c + d) + K] = f(c + d) = f(c) + f(d) = \varphi(c + K) + \varphi(d + K). $$ Therefore, $\varphi: R/K \rightarrow S$ is an isomorphism. The First Isomorphism Theorem is a useful tool for determining the structure of quotient rings, as illustrated in the following examples. ### Example 7 In the ring $\mathbb{Z}[x]$, the principal ideal $(x)$ consists of all multiples of $x$, that is, all polynomials with constant term 0. What does the quotient ring $\mathbb{Z}[x]/(x)$ look like? We can answer the question by using the function $\theta: \mathbb{Z}[x] \rightarrow \mathbb{Z}$, which maps each polynomial to its constant term. The function $\theta$ is certainly surjective because each $k \in \mathbb{Z}$ is the image of the polynomial $k + x + k x$ in $\mathbb{Z}[x]$. Furthermore, $\theta$ is a homomorphism of rings (Exercise 1). The kernel of $\theta$ consists of all those polynomials that are mapped to 0, that is, the kernel of $\theta$ is all polynomials with constant term 0. Thus the kernel of $\theta$ is the ideal $(x)$. By Theorem 6.13 the quotient ring $\mathbb{Z}[x]/(x)$ is isomorphic to $\mathbb{Z}$. ### Example 8 Let $T$ be the ring of functions from $\mathbb{R}$ to $\mathbb{R}$ and $I$ the ideal of all functions $g$ such that $g(2) = 0$. In Example 13 of Section 6.1 we saw that $T/I$ consists of the cosets $f + I$, one for each real number $r$, where $f_r: \mathbb{R} \rightarrow \mathbb{R}$ is the constant function given by $f_r(x) = r$ for every $x$. This suggests the possibility that the quotient ring $T/I$ might be isomorphic to the field $\mathbb{R}$. We shall use Theorem 6.13 to show that this is indeed the case by constructing a surjective homomorphism from $T$ to $\mathbb{R}$ whose kernel is the ideal $I$. Let $\varphi: T \rightarrow \mathbb{R}$ be the function defined by $\varphi(f) = f(2)$. Then $\varphi$ is surjective because for every real number $r$, $r = f(2) = \varphi(f_r)$. Furthermore, $\varphi$ is a homomorphism of rings: $$ \varphi(f + h) = (f + h)(2) = f(2) + h(2) = \varphi(f) + \varphi(h) $$ $$ \varphi(fh) = (fh)(2) = f(2)h(2) = \varphi(f)\varphi(h). $$ By definition, the kernel of $\varphi$ is the set $$ \{g \in T \ | \ \varphi(g) = 0\} = \{g \in T \ | \ g(2) = 0\}. $$ Thus the kernel is precisely the ideal $I$. By Theorem 6.13, $T/I$ is isomorphic to $\mathbb{R}$. ### Example 9 What do the homomorphic images of the ring $\mathbb{Z}$ look like? To answer this question, suppose that $f: \mathbb{Z} \rightarrow S$ is a surjective homomorphism. If $f$ is actually an isomorphism, then $S$ looks exactly like $\mathbb{Z}$, of course (in terms of algebraic structure). If $f$ is surjective, but not an isomorphism (that is, not injective), then the kernel $K$ of $f$ is a nonzero ideal in $\mathbb{Z}$ by Theorem 6.11. Since $K$ is an ideal in $\mathbb{Z}$, $K$ must be a principal ideal, say $K = (n)$ for some $n \neq 0$, by Exercise 40 in Section 6.1. By Theorem 6.13, $S$ is isomorphic to $\mathbb{Z}/K = \mathbb{Z}/n\mathbb{Z} = \mathbb{Z}_n$. Thus every homomorphic image of $\mathbb{Z}$ is isomorphic either to $\mathbb{Z}$ or to $\mathbb{Z}_n$ for some $n$. ## Exercises 1. Show that the map $\theta: \mathbb{R}[x] \rightarrow \mathbb{R}$ that sends each polynomial $f(x)$ to its constant term is a surjective homomorphism. 2. Show that every homomorphic image of a field $F$ is isomorphic either to $F$ itself or to the zero ring. *[Hint: See Exercise 10 in Section 6.1 and Exercise 7 below.]* 3. If $F$ is a field, $R$ a nonzero ring, and $f: F \rightarrow R$ a surjective homomorphism, prove that $f$ is an isomorphism. 4. Let $[a]_n$ denote the congruence class of the integer $a$ modulo $n$. - (a) Show that the map $f: \mathbb{Z}_{12} \rightarrow \mathbb{Z}_4$ that sends $[a]_{12}$ to $[a]_4$ is a well-defined, surjective homomorphism. - (b) Find the kernel of $f$. 5. Let $I$ be an ideal in an integral domain $R$. Is it true that $R/I$ is also an integral domain? 6. The function $\varphi: \mathbb{R}[x] \rightarrow \mathbb{R}$ given by $\varphi(f(x)) = f(2)$ is a homomorphism of rings by Exercise 24 of Section 4.4 (with $a = 2$). Find the kernel of $\varphi$. *[Hint: Theorem 4.16.]* 7. If $R$ is a ring, show that $R / (0_R) \cong R$. 8. Let $R$ and $S$ be rings. Show that $\pi: R \times S \rightarrow S$ given by $\pi(r, s) = s$ is a surjective homomorphism whose kernel is isomorphic to $S$. 9. $R = \left\{ \begin{pmatrix} a & 0 \\ b & c \end{pmatrix} \ | \ a, b, c \in \mathbb{Z} \right\}$ is a ring with identity by Example 19 in Section 3.1. - (a) Show that the map $f: R \rightarrow \mathbb{Z}$ given by $f\left( \begin{pmatrix} a & 0 \\ b & c \end{pmatrix} \right) = a$ is a surjective homomorphism. - (b) What is the kernel of $f$? 10. (a) Let $f: R \rightarrow S$ be a surjective homomorphism of rings and let $I$ be an ideal in $R$. Prove that $f(I)$ is an ideal in $S$, where $f(I) = \{ s \in S \ | \ s = f(a) \text{ for some } a \in I \}$. - (b) Show by example that part (a) may be false if $f$ is not surjective. 11. $\mathbb{Z}[\sqrt{2}]$ is a ring by Exercise 13 of Section 3.1. Let $f: \mathbb{Z}[\sqrt{2}] \rightarrow \mathbb{Z}[\sqrt{2}]$ be the function defined by $f(a + b \sqrt{2}) = a - b \sqrt{2}$. - (a) Show that $f$ is a surjective homomorphism of rings. - (b) Use Theorem 6.11 to show that $f$ is also injective and hence is an isomorphism. *[You may assume that $\sqrt{2}$ is irrational.]* 12. Let $I$ be an ideal in a noncommutative ring $R$ such that $ab - ba \in I$ for all $a, b \in R$. Prove that $R / I$ is commutative. 13. Let $I$ be an ideal in a ring $R$. Prove that every element in $R / I$ has a square root if and only if for every $a \in R$, there exists $b \in R$ such that $a - b^2 \in I$. 14. Let $I$ be an ideal in a ring $R$. Prove that every element in $R / I$ is a solution of $x^2 = x$ if and only if for every $a \in R$, $a^2 - a \in I$. 15. Let $I$ be an ideal in a commutative ring $R$. Prove that $R / I$ has an identity if and only if there exists $e \in R$ such that $ea - a \in I$ for every $a \in R$. 16. Let $I \neq R$ be an ideal in a commutative ring $R$ with identity. Prove that $R / I$ is an integral domain if and only if whenever $ab \in I$, either $a \in I$ or $b \in I$. 17. Suppose $I$ and $J$ are ideals in a ring $R$ and let $f: R \rightarrow R / I \times R / J$ be the function defined by $f(a) = (a + I, a + J)$. - (a) Prove that $f$ is a homomorphism of rings. - (b) Is $f$ surjective? *[Hint: Consider the case when $R = \mathbb{Z}$, $I = (2)$, $J = (4)$.]* - (c) What is the kernel of $f$? 18. Let $R$ be a commutative ring with identity with the property that every ideal in $R$ is principal. Prove that every homomorphic image of $R$ has the same property. 19. Let $I$ and $K$ be ideals in a ring $R$, with $K \subseteq I$. Prove that $I / K = \{ a + K \ | \ a \in I \}$ is an ideal in the quotient ring $R / K$. 20. Let $f: R \rightarrow S$ be a homomorphism of rings with kernel $K$. Let $I$ be an ideal in $R$ such that $I \subseteq K$. Show that $\overline{f}: R / I \rightarrow S$ given by $\overline{f}(r + I) = f(r)$ is a well-defined homomorphism. 21. Use the First Isomorphism Theorem to show that $\mathbb{Z}_{20} / (5) \cong \mathbb{Z}_5$. 22. Let $f: R \rightarrow S$ be a homomorphism of rings. If $J$ is an ideal in $S$ and $I = \{ r \in R \ | \ f(r) \in J \}$, prove that $I$ is an ideal in $R$ that contains the kernel of $f$. 23. (a) Let $R$ be a ring with identity. Show that the map $f: \mathbb{Z} \rightarrow R$ given by $f(k) = k \cdot 1_R$ is a homomorphism. - (b) Show that the kernel of $f$ is the ideal $(n)$, where $n$ is the characteristic of $R$. *[Hint: “Characteristic” is defined immediately before Exercise 41 of Section 3.2. Also see Exercise 40 in Section 6.1.]* 24. Find at least three idempotents in the quotient ring $\mathbb{Z}[x] / (x^4 + x^2)$. *[See Exercise 3 in Section 3.2.]* 25. Let $R$ be a commutative ring and $J$ the ideal of all nilpotent elements of $R$ (as in Exercise 30 of Section 6.1). Prove that the quotient ring $R / J$ has no nonzero nilpotent elements. 26. Let $S$ and $I$ be as in Exercise 41 of Section 6.1. Prove that $S / I \cong \mathbb{Z}_2$. 27. Let $T$ and $I$ be as in Exercise 42 of Section 6.1. Prove that $T / I \cong \mathbb{Z}_p$. 28. Let $T$ and $I$ be as in Exercise 44 of Section 6.1. Prove that $T / I \cong \mathbb{R}$. 29. Let $S$ and $I$ be as in Exercise 45 of Section 6.1. Prove that $S / I \cong \mathbb{R} \times \mathbb{R}$. 30. **(The Second Isomorphism Theorem)** Let $I$ and $J$ be ideals in a ring $R$. Then $I \cap J$ is an ideal in $I$, and $J$ is an ideal in $I + J$ by Exercises 19 and 20 of Section 6.1. Prove that $\frac{I}{I \cap J} \cong \frac{I + J}{J}$. *[Hint: Show that $f: I \rightarrow (I + J) / J$ given by $f(a) = a + J$ is a surjective homomorphism with kernel $I \cap J$.]* 31. **(The Third Isomorphism Theorem)** Let $I$ and $K$ be ideals in a ring $R$ such that $K \subseteq I$. Then $I / K$ is an ideal in $R / K$ by Exercise 19. Prove that $(R / K) / (I / K) \cong R / I$. *[Hint: Show that the map $f: R / K \rightarrow R / I$ given by $f(r + K) = r + I$ is a well-defined surjective homomorphism with kernel $I / K$.]* 32. (a) Let $K$ be an ideal in a ring $R$. Prove that every ideal in the quotient ring $R / K$ is of the form $I / K$ for some ideal $I$ in $R$. *[Hint: Exercises 19 and 22.]* - (b) If $f: R \rightarrow S$ is a surjective homomorphism of rings with kernel $K$, prove that there is a bijective function from the set of all ideals of $S$ to the set of all ideals of $R$ that contain $K$. *[Hint: Part (a) and Exercise 10.]* --- **EXCURSION:** The Chinese Remainder Theorem for Rings (Section 14.3) may be covered at this point if desired.