Could you explain how to show 9.48 in great detail? I also included a list of theorems and definitions in my textbook as a reference. Really appreciate your help! Definition 9.9. Let R be a ring. An ideal M of R is said to be maximal if1. M + R; and2. if I is an ideal of R containing M, then I = M or I = R.Example 9.27. Let R = Z and let n be a nonnegative integer. Then we claim that (n)is a maximal ideal of Rif and only if n is prime. Indeed, (0) is certainly not maximal,as (0) Ç (2) Ç R. Also, (1) is not maximal, since (1) = R. If n is composite, sayn = kl, with 1 2. If n is composite, say n = kl with1

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Could you explain how to show 9.48 in great detail? I also included a list of theorems and definitions in my textbook as a reference. Really appreciate your help! Definition 9.9. Let R be a ring. An ideal M of R is said to be maximal if1. M + R; and2. if I is an ideal of R containing M, then I = M or I = R.Example 9.27. Let R = Z and let n be a nonnegative integer. Then we claim that (n)is a maximal ideal of Rif and only if n is prime. Indeed, (0) is certainly not maximal,as (0) Ç (2) Ç R. Also, (1) is not maximal, since (1) = R. If n is composite, sayn = kl, with 1 < k,l < n, then we note that (n) Ç (k) Ç R, so (n) is not maximal.Finally, let n be prime. Suppose that I is an ideal of R with (n) Ç IÇ R. Takea e I\(n). Since a is not divisible by n, and n is prime, we know that (a, n) = 1.Thus, by Corollary 2.1, we can find integers u and v such that au + nv =a, n ɛ I, this implies that 1 e I, hence I = R, giving us a contradiction. (As weshall see shortly, there is another way to prove this.)1. But asExample 9.28. In any field, the ideal {0} is maximal! Remember, by Corollary 9.1,a field only has two ideals.In a commutative ring with identity, there is a nice test for maximality of ideals.Theorem 9.20. Let R be a commutative ring with identity, and M an ideal of R.Then M is maximal if and only if R/M is a field.Example 9.29. This gives us another way to deal with Example 9.27. If n = 0, thenwe note that Z/{0} is simply Z, which is not a field. Thus, (0) is not maximal. Ifn = 1, then observe that Z/Z is the ring with one element, which is not a field;hence, (1) is not maximal. For any n > 2, we see from Example 9.24 that Z/nZ isisomorphic to Z„. But by Theorem 8.11, Z, is a field if and only if n is prime. Thus,(n) is maximal if and only if n is prime.Example 9.30. By Example 9.25, R[x]/(x) is isomorphic to R, which is a field.Thus, (x) is a maximal ideal of R[x].Example 9.31. In the same manner as Example 9.30, we see that Z[x]/(x) is iso-morphic to Z. But Z is not a field, and hence (x) is not maximal. In fact, we cansee this by noting that (x) is properly contained in the ideal M consisting of thosepolynomials whose constant terms are multiples of 5. We can use Theorem 9.20 toshow that M is maximal. See Exercise 9.43.Definition 9.10. Let R be a commutative ring and P an ideal of R. Then we saythat P is a prime ideal? if1. Р 2 R; and2. if a, b e R and ab e P, then either a e P or b e P.Example 9.32. In any integral domain, {0} is a prime ideal. If ab = 0, then a = 0or b = 0.Example 9.33. Let us consider R = Z. By the preceding example, {0} is prime, sowe know immediately that maximal and prime are not the same thing. Of course,(1) = R, so (1) is not prime. Suppose that n > 2. If n is composite, say n = kl with1 < k,l < n, we see that kl e (n) but neither k nor l lies in (n). Thus, (n) is notprime. But if n is prime, then (n) is a prime ideal. Indeed, if ab e (n), then n|ab.Thus, by Theorem 2.7, n|a or n|b, and hence a or b is in (n).Theorem 9.21. Let R be a commutative ring with identity and P an ideal. Then Pis prime if and only if R/P is an integral domain.Example 9.34. Let us look at Example 9.33 again. We know that Z/(0) is just Z,which is an integral domain, and hence (0) is a prime ideal. If n > 2, then Z/(n) isisomorphic to Zn, by Example 9.24, and Theorem 8.11 tells us that this is an integraldomain if and only if n is prime. Thus, for a nonnegative integer n, (n) is a primeideal of Z if and only if n is 0 or prime.Example 9.35. Refer to Example 9.31. We see that (x) is a prime ideal in Z[x], sinceZ[x]/(x) is isomorphic to Z.Example 9.36. Naturally, (x) is prime in R[x], because we saw in Example 9.30that R[x]/(x) is isomorphic to R, which is a field, hence an integral domain.Theorem 9.22. Let R be a commutative ring with identity. Then every maximal idealof R is also a prime ideal. 9.48. Find all prime ideals in each of the following rings.1. Z102. Z50

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Find all prime ideals in each of the following rings:1 ℤ102 ℤ50

9.48. Find all prime ideals in each of the following rings. 1. Z10 2. Z50