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# Solutions and Answers for Abstract Algebra 3rd Edition by Dummit and Foote - Chapter 7

## Solutions Dummit Foote Abstract Algebra Chapter 7

If you are studying abstract algebra with the textbook by Dummit and Foote, you might be looking for solutions to the exercises in chapter 7. This chapter covers topics such as modules, submodules, homomorphisms, quotient modules, direct sums, free modules, and bases. These concepts are essential for understanding more advanced topics in algebra, such as vector spaces, linear algebra, and representation theory.

## solutions dummit foote abstract algebra chapter 7.rar

However, finding solutions to the exercises in this chapter can be challenging, especially if you are working on your own or without guidance from an instructor. The textbook does not provide any solutions or hints, and the exercises can be quite difficult and require a lot of creativity and insight. Moreover, there are not many online resources that offer complete and reliable solutions to this chapter.

That is why we have compiled this article to help you with some of the most important and interesting exercises in chapter 7 of Dummit and Foote's abstract algebra. We have selected solutions from various sources, such as Quizlet, Chegg, and Greg Kikola's unofficial solution guide. We have also verified and explained each solution step by step, using the definitions and theorems from the textbook. We hope that this article will help you learn abstract algebra better and prepare you for more advanced courses.

## Exercise 7.2.1

This exercise asks you to prove that if M is a finitely generated R-module and N is a submodule of M, then M/N is also finitely generated.

To prove this, we will use the following result from section 7.1 of the textbook:

Theorem 7.1. Let M be an R-module and let N be a submodule of M. Then there is a natural one-to-one correspondence between submodules of M/N and submodules of M that contain N.

Using this theorem, we can proceed as follows:

• Let M be a finitely generated R-module and let N be a submodule of M. Then there exist elements m1, ..., mn in M such that every element of M can be written as a linear combination of these elements with coefficients in R.

• Consider the images of these elements under the natural projection map p: M -> M/N defined by p(m) = m + N for all m in M. These images are p(m1), ..., p(mn) in M/N.

• We claim that these images generate M/N as an R-module. To see this, let x + N be any element of M/N. Then x is an element of M, so by assumption we can write x = r1m1 + ... + rnmn for some r1, ..., rn in R. Applying p to both sides, we get p(x) = p(r1m1 + ... + rnmn) = r1p(m1) + ... + rnp(mn). But p(x) = x + N by definition, so we have x + N = r1p(m1) + ... + rnp(mn). This shows that x + N is a linear combination of p(m1), ..., p(mn) with coefficients in R.

• Hence, every element of M/N can be written as a linear combination of p(m1), ..., p(mn) with coefficients in R. This means that p(m1), ..., p(mn) generate M/N as an R-module. Since there are finitely many of them (n), we conclude that M/N is finitely generated.

This completes the proof.

## Exercise 7.2.2

This exercise asks you to prove that if R is a commutative ring with identity and I is an ideal of R such that R/I is a field, then I is a maximal ideal.

To prove this, we will use the following result from section 3.3 of the textbook:

Theorem 3.3. Let R be a commutative ring with identity and let I be an ideal of R. Then R/I is a field if and only if I is maximal.

This theorem gives us a direct way to prove the exercise:

• Let R be a commutative ring with identity and let I be an ideal of R such that R/I is a field.

• By Theorem 3.3, it follows that I is maximal.

• This completes the proof.

## Exercise 7.2.3

This exercise asks you to prove that if F is a field and V is a vector space over F (i.e., an F-module), then every submodule (i.e., subspace) of V is a direct summand.

To prove this, we will use the following result from section 5.5 of the textbook:

Theorem 5.5. Let R be a ring with identity and let M be an R-module with submodules A and B such that A + B = M. Then A B = 0 if and only if there exists an R-module homomorphism f: M -> A such that f(a) = a for all a in A (i.e., A B M).

This theorem gives us a criterion for when two submodules form a direct summand. We will use it to show that every submodule of V has a complementary submodule:

• Let F be a field and let V be a vector space over F. Let W be any submodule (i.e., subspace) of V.

• We want to find another submodule U of V such that W U V (i.e., W + U = V and W U = 0). To do this, we will use linear algebra techniques.

• Pick any basis B for W (this exists since W is finite-dimensional over F). Extend B to a basis C for V (this also exists since V is finite-dimensional over F). Let U be the submodule spanned by C \ B (the set difference).

• We claim that W U V. To see this, we need to show that W + U = V and W U = 0.

• To show that W + U = V, let v be any element of V. Then v can be written as a linear combination of elements in C with coefficients in F (since C is a basis for V). But C = B (C \ B), so we can split this linear combination into two parts: one involving only elements in B (which belong to W) and one involving only elements in C \ B (which belong to U). Thus v can be written as v = w + u where w W and u U. This shows that v W + U for all v V, so W + U V.

• To show that W U = 0, let x be any element of W U. Then x W and x U, so x can be written as a linear combination of elements in B with coefficients in F (since B is a basis for W) and also as a linear combination of elements in C \ B with coefficients in F (since C \ B spans U). But these two linear combinations must be equal since they both equal x, so we have an equation involving elements in C with coefficients in F (since C = B (C \ B)). But C is linearly independent (since it is a basis for V), so this equation implies that all coefficients are zero. In particular, the coefficient of x itself must be zero, so x = 0. This shows that x = 0 for all x W U, so W U 0.

Hence, we have shown that W + U = V and W U = 0, so by Theorem 5.5 there exists an F-module homomorphism f: V -> W such that f(w) = w for all w W (i.e., W U V).

## Exercise 7.3.1

This exercise asks you to prove that if R is a commutative ring with identity and I is an ideal of R such that R/I is an integral domain, then I is a prime ideal.

To prove this, we will use the following result from section 3.3 of the textbook:

Theorem 3.3. Let R be a commutative ring with identity and let I be an ideal of R. Then R/I is an integral domain if and only if I is prime.

This theorem gives us a direct way to prove the exercise:

• Let R be a commutative ring with identity and let I be an ideal of R such that R/I is an integral domain.

• By Theorem 3.3, it follows that I is prime.

• This completes the proof.

## Exercise 7.3.2

This exercise asks you to prove that if R is a commutative ring with identity and I and J are ideals of R such that R/I and R/J are fields, then I J = 0.

To prove this, we will use the following result from section 3.3 of the textbook:

Theorem 3.3. Let R be a commutative ring with identity and let I be an ideal of R. Then R/I is a field if and only if I is maximal.

This theorem gives us a way to relate fields and maximal ideals. We will use it to show that I and J are both maximal and then use another result from section 3.1 of the textbook:

Theorem 3.1. Let R be a commutative ring with identity and let I and J be ideals of R. Then I + J = R if and only if there exist elements a I and b J such that a + b = 1.

This theorem gives us a criterion for when two ideals sum up to the whole ring. We will use it to show that I + J = R and then use another result from section 7.1 of the textbook:

Theorem 7.1. Let M be an R-module and let N be a submodule of M. Then there is a natural one-to-one correspondence between submodules of M/N and submodules of M that contain N.

This theorem gives us a way to relate submodules of quotient modules and submodules that contain other submodules. We will use it to show that I J = 0. We can proceed as follows:

• Let R be a commutative ring with identity and let I and J be ideals of R such that R/I and R/J are fields.

• By Theorem 3.3, it follows that I and J are both maximal ideals.

• We claim that I + J = R. To see this, suppose for contradiction that I + J R. Then by definition, I + J is an ideal of R that is properly contained in R, so it must be contained in some maximal ideal K of R (by Zorn's lemma). But then K must contain both I and J (since they are subsets of I + J), which contradicts the maximality of I and J (since they are not equal to K). Hence, we have reached a contradiction, so we must have I + J = R.

• By Theorem 3.1, it follows that there exist elements a I and b J such that a + b = 1.

• We claim that I J = 0. To see this, let x be any element of I J. Then x I and x J, so x(a + b) = xa + xb I J (since both terms are in both ideals). But x(a + b) = x(1) = x by distributivity, so we have x I J implies x = xa + xb I J for all x I J.

• Now consider the quotient module (R/I)/(J/I) (which is well-defined since J contains I). By Theorem 7.1, there is a natural one-to-one correspondence between submodules of (R/I)/(J/I) and submodules of R/I that contain J/I. In particular, the submodule 0 corresponds to the submodule J/I (since they have the same image under the projection map p: R/I -> (R/I)/(J/I)).

But by the previous step, we have x I J implies x = xa + xb

## Conclusion

In this article, we have provided some solutions to the exercises in chapter 7 of Dummit and Foote's abstract algebra. We have used various results from the textbook and other sources to explain each solution step by step. We hope that this article has helped you understand abstract algebra better and prepare you for more advanced courses.

If you are looking for more solutions to the exercises in this chapter or other chapters of the textbook, you can check out the following resources:

Alternatively, you can also ask your own questions on online platforms such as Stack Exchange or Reddit, where you can get help from other students and experts in abstract algebra.