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 fieldif and only if R is simple.Proof. () Assume that R is a field. Let J be an ideal of R, and assume thatJ contains at least one non-zero element r. Since R is a field, r is a unit, andso r-l E R. Now when we multiply any element of the ring with an element inthe ideal, we get something in the ideal. Hence 1 = r-1r E J. Now if s is anyarbitrary element of R, then sl € J-again something in R times something in J isin 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 associativeand 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 twonon-zero elements was zero (in other words, if we had zero-divisors). But if ry = 0and x is invertible, then we have y = xxy = 0. Hence, all wthat every non-zero element in R has a (multiplicative) inverse in R. Now let r bea non-zero element of R. Consider the ideal (r) = {rs sE R}. Since R is simpleand (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 isa field.-1 Theorem 16.45 (Homomorphism theorems). Let o: R S be a ring homomor-phism. Let K = ker(ø). LetI= {ideals of R containing K} and J = {ideals of o(R)}.34016. Homomorphisms, Ideals, and Quotient RingsThen the map o extends to a map o: I–→ I byI2, 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 anideal 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).

Recall that in a ring A not necessarily commutative and with an identity, an ideal M⊂A is a maximal…

Answered: Let R be a commutative ring with…

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

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Chapter2: Second-order Linear Odes

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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

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).

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