Find every maximal ideal of Z, O Z7.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Could you explain how to show 9.50 in great detail? I also included a list of theorems and definitions in my textbook as a reference. Really appreciate your help!
Transcribed Image Text:9.50. Find every maximal ideal of Z7 e Z7.
![Definition 9.9. Let R be a ring. An ideal M of R is said to be maximal if
1. M + R; and
2. 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, say
n = 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. Take
a 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 we
shall see shortly, there is another way to prove this.)
1. But as
Example 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, then
we note that Z/{0} is simply Z, which is not a field. Thus, (0) is not maximal. If
n = 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 is
isomorphic 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 can
see this by noting that (x) is properly contained in the ideal M consisting of those
polynomials whose constant terms are multiples of 5. We can use Theorem 9.20 to
show 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 say
that P is a prime ideal? if
1. Р 2 R; and
2. 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 = 0
or b = 0.
Example 9.33. Let us consider R = Z. By the preceding example, {0} is prime, so
we 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 with
1 < k,l < n, we see that kl e (n) but neither k nor l lies in (n). Thus, (n) is not
prime. 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 P
is 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) is
isomorphic to Zn, by Example 9.24, and Theorem 8.11 tells us that this is an integral
domain if and only if n is prime. Thus, for a nonnegative integer n, (n) is a prime
ideal 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], since
Z[x]/(x) is isomorphic to Z.
Example 9.36. Naturally, (x) is prime in R[x], because we saw in Example 9.30
that 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 ideal
of R is also a prime ideal.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15ed467-90ec-4e60-afef-3d3f6119f74d%2Fb88c8e02-ca26-44d3-be56-81535584a590%2F49hf61_processed.png&w=3840&q=75)
Transcribed Image Text:Definition 9.9. Let R be a ring. An ideal M of R is said to be maximal if
1. M + R; and
2. 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, say
n = 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. Take
a 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 we
shall see shortly, there is another way to prove this.)
1. But as
Example 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, then
we note that Z/{0} is simply Z, which is not a field. Thus, (0) is not maximal. If
n = 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 is
isomorphic 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 can
see this by noting that (x) is properly contained in the ideal M consisting of those
polynomials whose constant terms are multiples of 5. We can use Theorem 9.20 to
show 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 say
that P is a prime ideal? if
1. Р 2 R; and
2. 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 = 0
or b = 0.
Example 9.33. Let us consider R = Z. By the preceding example, {0} is prime, so
we 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 with
1 < k,l < n, we see that kl e (n) but neither k nor l lies in (n). Thus, (n) is not
prime. 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 P
is 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) is
isomorphic to Zn, by Example 9.24, and Theorem 8.11 tells us that this is an integral
domain if and only if n is prime. Thus, for a nonnegative integer n, (n) is a prime
ideal 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], since
Z[x]/(x) is isomorphic to Z.
Example 9.36. Naturally, (x) is prime in R[x], because we saw in Example 9.30
that 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 ideal
of R is also a prime ideal.
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