Could you explain how to show 9.6 in detail? I included list of theorems and definitions from the textbook. 9.6. Let R be a commutative ring. If I = {a e R : a" = 0 for some n e N}, show%3Dthat I is an ideal of R. Definition 9.1. Let R be a ring. Then a subring I of R is said to be an ideal ifir, ri e I for all i e I and r e R. We call this the absorption property.Theorem 9.1. Let R be a ring and I a subset of R. Then I is an ideal if and only if1. 0 e I;2. i – je I for all i, j e I; and3. ir, ri e I for all i e I, r e R.Example 9.1. Let n be any integer. Then nZ is an ideal of Z. Indeed, we already knowthat it is a subring. But also, if nk e nZ, then for any integerr,r(nk) = n(rk) E nZ.Example 9.2. Let I be the set of all polynomials f (x) e R[x] such that f (0) = 0.We claim that I is an ideal in R[x]. Certainly I contains the zero polynomial. Also, iff (0) = g(0) = 0, then (f – )(0) = ƒ (0) – g(0) = 0, hence f (x)–g(x) e I.Also,if f(0) = 0 and h(x) e R[x], then h(0) f (0) = h(0)0 = 0. Hence, h(x)f (x) e I.|Example 9.3. Let I be the set of all polynomials in Z[x] whose constant term is amultiple of 5. Then I is an ideal. See Exercise 9.2.Example 9.4. For any ring R, {0} and R are ideals of R.Example 9.5. In Q, the ring of integers is a subring but not an ideal. Indeed, 3 e Z,but 3(1/5) ¢ Z. Thus, Z does not have the absorption property.Theorem 9.2. Let R be a ring with identity. If an ideal I of R contains a unit, thenI = R.Corollary 9.1. Let F be a field. Then the only ideals of F are {0} and F.Example 9.6. In Z, if we let I = 4Z and J = 6Z, then I + J = 2Z. Indeed, ifm e I +J, then m = 4a+6b = 2(2a+3b), for some integers a and b. In particular,I+JC 2Z. On the other hand, for any c e Z, 2c = 4(-c)+6c e I + J, and hence2Z CI+ J. We can say something more general here. See Exercise 9.4.Theorem 9.3. If I and J are ideals of a ring R, then so is I + J.Theorem 9.4. Let I and J be ideals in a ring R. Then I J is also an ideal.Definition 9.2. Let R be a commutative ring with identity and a e R. Then theprincipal ideal generated by a, denoted (a), is the set {ra :r e R}.Example 9.7. In Z, we have (n) = nZ for any n e Z.Example 9.8. The ideal from Example 9.2 is (x). Indeed, if f(x) € R[x], thenf (0) = 0 if and only if the constant term is 0; that is, if and only if the polynomialis a multiple of x.Theorem 9.5. If R is a commutative ring with identity, and a e R, then (a) is anideal of R; indeed, it is the intersection of all ideals of R containing a.

Name: Could you explain how to show 9.6 in detail? I included list of theore
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Description: Could you explain how to show 9.6 in detail? I included list of theorems and definitions from the textbook. 9.6. Let R be a commutative ring. If I = {a e R : a" = 0 for some n e N}, show%3Dthat I is an ideal of R. Definition 9.1. Let R be a ring. The

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Could you explain how to show 9.6 in detail? I included list of theorems and definitions from the textbook. 9.6. Let R be a commutative ring. If I = {a e R : a&#34; = 0 for some n e N}, show%3Dthat I is an ideal of R. Definition 9.1. Let R be a ring. Then a subring I of R is said to be an ideal ifir, ri e I for all i e I and r e R. We call this the absorption property.Theorem 9.1. Let R be a ring and I a subset of R. Then I is an ideal if and only if1. 0 e I;2. i – je I for all i, j e I; and3. ir, ri e I for all i e I, r e R.Example 9.1. Let n be any integer. Then nZ is an ideal of Z. Indeed, we already knowthat it is a subring. But also, if nk e nZ, then for any integerr,r(nk) = n(rk) E nZ.Example 9.2. Let I be the set of all polynomials f (x) e R[x] such that f (0) = 0.We claim that I is an ideal in R[x]. Certainly I contains the zero polynomial. Also, iff (0) = g(0) = 0, then (f – )(0) = ƒ (0) – g(0) = 0, hence f (x)–g(x) e I.Also,if f(0) = 0 and h(x) e R[x], then h(0) f (0) = h(0)0 = 0. Hence, h(x)f (x) e I.|Example 9.3. Let I be the set of all polynomials in Z[x] whose constant term is amultiple of 5. Then I is an ideal. See Exercise 9.2.Example 9.4. For any ring R, {0} and R are ideals of R.Example 9.5. In Q, the ring of integers is a subring but not an ideal. Indeed, 3 e Z,but 3(1/5) ¢ Z. Thus, Z does not have the absorption property.Theorem 9.2. Let R be a ring with identity. If an ideal I of R contains a unit, thenI = R.Corollary 9.1. Let F be a field. Then the only ideals of F are {0} and F.Example 9.6. In Z, if we let I = 4Z and J = 6Z, then I + J = 2Z. Indeed, ifm e I +J, then m = 4a+6b = 2(2a+3b), for some integers a and b. In particular,I+JC 2Z. On the other hand, for any c e Z, 2c = 4(-c)+6c e I + J, and hence2Z CI+ J. We can say something more general here. See Exercise 9.4.Theorem 9.3. If I and J are ideals of a ring R, then so is I + J.Theorem 9.4. Let I and J be ideals in a ring R. Then I J is also an ideal.Definition 9.2. Let R be a commutative ring with identity and a e R. Then theprincipal ideal generated by a, denoted (a), is the set {ra :r e R}.Example 9.7. In Z, we have (n) = nZ for any n e Z.Example 9.8. The ideal from Example 9.2 is (x). Indeed, if f(x) € R[x], thenf (0) = 0 if and only if the constant term is 0; that is, if and only if the polynomialis a multiple of x.Theorem 9.5. If R is a commutative ring with identity, and a e R, then (a) is anideal of R; indeed, it is the intersection of all ideals of R containing a.

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9.6. Let R be a commutative ring. If I = {a e R : a" = 0 for some n e N}, show that I is an ideal of R.