Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
Let R be a commutative ring with identity. Using the homomorphism theorem (Theorem 16.45) and Proposition 16.32, show that an ideal M of R is maximal if and only if R/M is a field.

Let R be a commutative ring with identity. Using the homomorphism theorem (Theorem 16.45) and Proposition 16.32, show that an ideal M of R is maximal if and only if R/M is a field.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
See similar textbooks
icon
Related questions
Question

Let R be a commutative ring with identity. Using the homomorphism theorem (Theorem 16.45) and Proposition 16.32, show that an ideal M of R is maximal if and only if R/M is a field.

Proposition 16.32. Let R be a commutative ring with identity. Then R is a field if and only if R is simple. Proof. () Assume that R is a field. Let J be an ideal of R, and assume that J contains at least one non-zero element r. Since R is a field, r is a unit, and so r-l E R. Now when we multiply any element of the ring with an element in the ideal, we get something in the ideal. Hence 1 = r-1r E J. Now if s is any arbitrary element of R, then sl € J-again something in R times something in J is in J-which means that every element of R is in J. Hence J = R is a trivial ideal. () Assume that R is simple. To show that R is a field, we have to show that (R- {0}, ·) is an abelian group. We already know that the product is associative and commutative and we have a 1. Hence, we only need closure and inverses. Since (R,·) is closed, the only way closure could fail in (R-{0}, ·) is if the product of two non-zero elements was zero (in other words, if we had zero-divisors). But if ry = 0 and x is invertible, then we have y = xxy = 0. Hence, all w that every non-zero element in R has a (multiplicative) inverse in R. Now let r be a non-zero element of R. Consider the ideal (r) = {rs sE R}. Since R is simple and (r) is not the zero ideal, we have (r) = R. Since 1 E R, we have 1 E (r), and, for some s E R. we have rs = 1. This means that r has an inverse, and hence R is a field. -1
Transcribed Image Text:Proposition 16.32. Let R be a commutative ring with identity. Then R is a field if and only if R is simple. Proof. () Assume that R is a field. Let J be an ideal of R, and assume that J contains at least one non-zero element r. Since R is a field, r is a unit, and so r-l E R. Now when we multiply any element of the ring with an element in the ideal, we get something in the ideal. Hence 1 = r-1r E J. Now if s is any arbitrary element of R, then sl € J-again something in R times something in J is in J-which means that every element of R is in J. Hence J = R is a trivial ideal. () Assume that R is simple. To show that R is a field, we have to show that (R- {0}, ·) is an abelian group. We already know that the product is associative and commutative and we have a 1. Hence, we only need closure and inverses. Since (R,·) is closed, the only way closure could fail in (R-{0}, ·) is if the product of two non-zero elements was zero (in other words, if we had zero-divisors). But if ry = 0 and x is invertible, then we have y = xxy = 0. Hence, all w that every non-zero element in R has a (multiplicative) inverse in R. Now let r be a non-zero element of R. Consider the ideal (r) = {rs sE R}. Since R is simple and (r) is not the zero ideal, we have (r) = R. Since 1 E R, we have 1 E (r), and, for some s E R. we have rs = 1. This means that r has an inverse, and hence R is a field. -1
Theorem 16.45 (Homomorphism theorems). Let o: R S be a ring homomor- phism. Let K = ker(ø). Let I= {ideals of R containing K} and J = {ideals of o(R)}. 340 16. Homomorphisms, Ideals, and Quotient Rings Then the map o extends to a map o: I–→ I by I2, 0(1). This map-which is also denoted by o-is a bijection from I to J. In particular, if I is an ideal of R containing K, then o(1) is an ideal of o(R), and if J is an ideal of o(R), then o(J) is an ideal of R containing K. Furthermore, if I e I and J EI with o(I) = J, then I/K=J and R/I = $(R)/J. In particular, R/K= 6(R).
Transcribed Image Text:Theorem 16.45 (Homomorphism theorems). Let o: R S be a ring homomor- phism. Let K = ker(ø). Let I= {ideals of R containing K} and J = {ideals of o(R)}. 340 16. Homomorphisms, Ideals, and Quotient Rings Then the map o extends to a map o: I–→ I by I2, 0(1). This map-which is also denoted by o-is a bijection from I to J. In particular, if I is an ideal of R containing K, then o(1) is an ideal of o(R), and if J is an ideal of o(R), then o(J) is an ideal of R containing K. Furthermore, if I e I and J EI with o(I) = J, then I/K=J and R/I = $(R)/J. In particular, R/K= 6(R).
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Ring
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Basic Technical Mathematics
Basic Technical Mathematics
Advanced Math
ISBN:
9780134437705
Author:
Washington
Publisher:
PEARSON
Topology
Topology
Advanced Math
ISBN:
9780134689517
Author:
Munkres, James R.
Publisher:
Pearson,
SEE MORE TEXTBOOKS