Could you explain how to show 8.32 in detail? I also attached definitions and theorems in my textbook. 8.28. For each of the following rings, which elements are units? Which are zerodivisors?1. Z182. Z3 O Z98.32. Let R and S be rings. Under precisely what circumstances is R S an integraldomain?8.33. Let F be a field with subfields K and L. Show that K N L is a subfield of F.Extend this to show that the intersection of any collection of subfields is a subfield.8.34. Let p be a prime and F a field with p² elements. Show that F cannot havemore than one proper subfield. Definition 8.8. Let R be a commutative ring. Then a nonzero element a e R is saidto be a zero divisor if there exists a nonzero b e R such that ab = 0.Example 8.19. In Z6, we note that 4 is a zero divisor, as 4·3 = 0. On the other hand,5 is not a zero divisor.Example 8.20. The ring of integers has no zero divisors.Definition 8.9. An integral domain is a commutative ring R with identity 1 + 0having no zero divisors.Example 8.21. The rings Z, Q, R and C are all integral domains.Example 8.22. The polynomial ring R[x] is an integral domain. Indeed, we knowthat it is a commutative ring with identity. Also, if f (x) = ao + ax + . + a„x"and g(x) = bo + bịx + · . . + bmx™, with a; , b; e R and an + 0 + bm, then theunique term of highest degree in ƒ (x)g(x)is a„bmxm+n. As R is an integral domain,a„bm # 0. Thus, f(x)g(x) is not the zero polynomial.Example 8.23. The rings 2Z, Z6 and M2(R) all fail to be integral domains. The firstlacks an identity, the second has zero divisors and the third is not commutative.Theorem 8.7 (Cancellation Law). Let R be an integral domain. Suppose thata, b, c e R and ab = ac. If a +0, then b = c.Definition 8.10. Let R be a ring with identity. Then we say that an element a e Ris a unit if there exists an element b e R such that ab = ba = 1. In this case, wecall b the inverse of a and write b = a-l. We write U (R) for the set of all units ofR, and call it the unit group of R.Theorem 8.8. Let R be a ring with identity. Then U (R) is a group under multipli-cation.Example 8.24. By definition, U (M„(R)) = GL„(R).Example 8.25. The unit group of Z is {±1}.Example 8.26. The unit group of Z, is U (n). See Exercise 8.30.Example 8.27. Every element other than 0 in R is a unit. The same can be said forQ and C.This last example leads us to our next definition.Definition 8.11. Let F be a commutative ring with identity 1 7 0. Then F is saidto be a field if U (F) consists of every element of F other than 0.Example 8.28. As we noted above, Q, R and C are fields.Lemma 8.1. Let R be a commutative ring with identity. Then a unit in R cannot bea zero divisor.Theorem 8.9. Every field is an integral domain.Of course, the integers are an integral domain, but not a field. However, we cansay something for finite integral domains. As we might expect, if a e R, and n is apositive integer, we writea" = aa ·· .an timesTheorem 8.10. Let R be a finite integral domain. Then R is a field.Theorem 8.11. Let n 2 be a positive integer. Then the following are equivalent:1. Z, is an integral domain;2. Z, is a field; and3. n is prime.Definition 8.12. Let F be a field. Then a subring K of F is said to be a subfield ifit is a field using the same addition and multiplication operations.Example 8.29. Q is a subfield of R, which in turn is a subfield of C.But how do we test if a subset is a subfield?Theorem 8.12. Let F be a field. Then a subset S of F is a subfield of F if and onlyif1. 1 e S;2. if a, b e S, then a – b e S; and3. if a, b e S, and b ± 0, then ab¬l e S.

Question

Could you explain how to show 8.32 in detail? I also attached definitions and theorems in my textbook. 8.28. For each of the following rings, which elements are units? Which are zerodivisors?1. Z182. Z3 O Z98.32. Let R and S be rings. Under precisely what circumstances is R S an integraldomain?8.33. Let F be a field with subfields K and L. Show that K N L is a subfield of F.Extend this to show that the intersection of any collection of subfields is a subfield.8.34. Let p be a prime and F a field with p² elements. Show that F cannot havemore than one proper subfield. Definition 8.8. Let R be a commutative ring. Then a nonzero element a e R is saidto be a zero divisor if there exists a nonzero b e R such that ab = 0.Example 8.19. In Z6, we note that 4 is a zero divisor, as 4·3 = 0. On the other hand,5 is not a zero divisor.Example 8.20. The ring of integers has no zero divisors.Definition 8.9. An integral domain is a commutative ring R with identity 1 + 0having no zero divisors.Example 8.21. The rings Z, Q, R and C are all integral domains.Example 8.22. The polynomial ring R[x] is an integral domain. Indeed, we knowthat it is a commutative ring with identity. Also, if f (x) = ao + ax + . + a„x&#34;and g(x) = bo + bịx + · . . + bmx™, with a; , b; e R and an + 0 + bm, then theunique term of highest degree in ƒ (x)g(x)is a„bmxm+n. As R is an integral domain,a„bm # 0. Thus, f(x)g(x) is not the zero polynomial.Example 8.23. The rings 2Z, Z6 and M2(R) all fail to be integral domains. The firstlacks an identity, the second has zero divisors and the third is not commutative.Theorem 8.7 (Cancellation Law). Let R be an integral domain. Suppose thata, b, c e R and ab = ac. If a +0, then b = c.Definition 8.10. Let R be a ring with identity. Then we say that an element a e Ris a unit if there exists an element b e R such that ab = ba = 1. In this case, wecall b the inverse of a and write b = a-l. We write U (R) for the set of all units ofR, and call it the unit group of R.Theorem 8.8. Let R be a ring with identity. Then U (R) is a group under multipli-cation.Example 8.24. By definition, U (M„(R)) = GL„(R).Example 8.25. The unit group of Z is {±1}.Example 8.26. The unit group of Z, is U (n). See Exercise 8.30.Example 8.27. Every element other than 0 in R is a unit. The same can be said forQ and C.This last example leads us to our next definition.Definition 8.11. Let F be a commutative ring with identity 1 7 0. Then F is saidto be a field if U (F) consists of every element of F other than 0.Example 8.28. As we noted above, Q, R and C are fields.Lemma 8.1. Let R be a commutative ring with identity. Then a unit in R cannot bea zero divisor.Theorem 8.9. Every field is an integral domain.Of course, the integers are an integral domain, but not a field. However, we cansay something for finite integral domains. As we might expect, if a e R, and n is apositive integer, we writea&#34; = aa ·· .an timesTheorem 8.10. Let R be a finite integral domain. Then R is a field.Theorem 8.11. Let n > 2 be a positive integer. Then the following are equivalent:1. Z, is an integral domain;2. Z, is a field; and3. n is prime.Definition 8.12. Let F be a field. Then a subring K of F is said to be a subfield ifit is a field using the same addition and multiplication operations.Example 8.29. Q is a subfield of R, which in turn is a subfield of C.But how do we test if a subset is a subfield?Theorem 8.12. Let F be a field. Then a subset S of F is a subfield of F if and onlyif1. 1 e S;2. if a, b e S, then a – b e S; and3. if a, b e S, and b ± 0, then ab¬l e S.

Accepted Answer

Let R and S be rings. Find the conditions for R⊕S to be an integral domain.

8.32. Let R and S be rings. Under precisely what circumstances is ROS an integral domain?