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Provide a justification for each step in the proof outlined below showing that: A finite commutative ring (R, +, *) with identity that has no divisors of zero will be a field. Proof: We will be assuming that (R, +, *) is a commutative ring with identity, is finite, and that (R, +, *) has no divisors of zero. To show that (R, +, *) is, in fact, a field we need to show that each non-zero element in R has a multiplicative inverse in R. Since R is finite, let N be the number of distinct elements in R, listed as a, a, ... an i. One of the elements from the list must be the multiplicative identity. Now, let r be some non-zero element of R; we want to show that one of the elements in the list is a multiplicative inverse of r. ii. That is, we want to show that there is an element in the list, a, for which r*a = 1. Let's look at all the products of r with each element in the list: r*a₁, r*a, ..., r*an ii. All of these products must be distinct: A. For, suppose there exists two of the products that are equal, that is, suppose r*a = r*ax with j not equal to k B. Now, a has an additive inverse, (-ai) C. Then, r*aj + r*(-a) = r*a+r*(-a) D. On the left: r*aj + r*(-a) = r*(aj + (-ax))

Provide a justification for each step in the proof outlined below showing that: A finite commutative ring (R, +, *) with identity that has no divisors of zero will be a field. Proof: We will be assuming that (R, +, *) is a commutative ring with identity, is finite, and that (R, +, *) has no divisors of zero. To show that (R, +, *) is, in fact, a field we need to show that each non-zero element in R has a multiplicative inverse in R. Since R is finite, let N be the number of distinct elements in R, listed as a, a, ... an i. One of the elements from the list must be the multiplicative identity. Now, let r be some non-zero element of R; we want to show that one of the elements in the list is a multiplicative inverse of r. ii. That is, we want to show that there is an element in the list, a, for which r*a = 1. Let's look at all the products of r with each element in the list: r*a₁, r*a, ..., r*an ii. All of these products must be distinct: A. For, suppose there exists two of the products that are equal, that is, suppose r*a = r*ax with j not equal to k B. Now, a has an additive inverse, (-ai) C. Then, r*aj + r*(-a) = r*a+r*(-a) D. On the left: r*aj + r*(-a) = r*(aj + (-ax))

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
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Note: Not all rings without divisors of zero will necessarily be a field, e.g., (Z, +, *) is a commutative ring with identity and no divisors of zero (no two non-zero integers multiplied together can result in zero), but non-zero integers other than 1 or -1 do not have multiplicative inverses in Z, so the system is not a field. But, if the underlying set in the ring is finite, and the ring has no divisors of zero, then it will, in fact, be a field. Justifying the proof is your next task. b. Provide a justification for each step in the proof outlined below showing that: A finite commutative ring (R, +, *) with identity that has no divisors of zero will be a field. Proof: We will be assuming that (R, +, *) is a commutative ring with identity, is finite, and that (R, +, *) has no divisors of zero. To show that (R, +, *) is, in fact, a field we need to show that each non-zero element in R has a multiplicative inverse in R. Since R is finite, let N be the number of distinct elements in R, listed as a, a, ... an. i. One of the elements from the list must be the multiplicative identity. Now, let r be some non-zero element of R; we want to show that one of the elements in the list is a multiplicative inverse of r. ii. That is, we want to show that there is an element in the list, a; for which r*a = 1. Let's look at all the products of r with each element in the list: r*a₁, r*a₂, ..., r*a iii. All of these products must be distinct: A. For, suppose there exists two of the products that are equal, that is, suppose r*aj = r*a with j not equal to k B. Now, ax has an additive inverse, (-a) C. Then, r*a;+r*(-a)=r*a+r*(-a) D. On the left: r*aj + r*(-as) = r*(a) + (-ax.)) E. On the right: a. b. r*(a +(-a))=r*0 c. r*0=0 r*a+r*(-a)=r*(a +(-a)) F. So, r* (aj + (-a)) = 0 G. Then, (aj + (-a)) = 0 H. Then, (aj + (-ax))+ ax=0+ ax On the left: I. a. (aj + (-a))+ ax= aj + ((-a) + a) b. aj + ((-a)+ ax) = aj + 0 c. aj +0=aj J. On the right: 0 + a = ax K. So, a, a, which is a contradiction iv. So, one of the products must equal 1, that is, r*a = 1 for some element a; in the set R v. Also, a*r = 1 vi. So, r has a multiplicative inverse in R vii. So, (R, +, *) is a field ring is c. From HW#2, (M:(R#), +, *) is a ring with identity, but the system is not a field (the commutative, and some non-zero elements of M (R#) do not have a multiplicative inverse. Determine 2x2 matrices A and B for which A*B= 0 (the zero matrix), but no entry of A or B is 0 (the real number 0). This shows that (M:(R#), +, *) has divisors of zero.
Transcribed Image Text:Note: Not all rings without divisors of zero will necessarily be a field, e.g., (Z, +, *) is a commutative ring with identity and no divisors of zero (no two non-zero integers multiplied together can result in zero), but non-zero integers other than 1 or -1 do not have multiplicative inverses in Z, so the system is not a field. But, if the underlying set in the ring is finite, and the ring has no divisors of zero, then it will, in fact, be a field. Justifying the proof is your next task. b. Provide a justification for each step in the proof outlined below showing that: A finite commutative ring (R, +, *) with identity that has no divisors of zero will be a field. Proof: We will be assuming that (R, +, *) is a commutative ring with identity, is finite, and that (R, +, *) has no divisors of zero. To show that (R, +, *) is, in fact, a field we need to show that each non-zero element in R has a multiplicative inverse in R. Since R is finite, let N be the number of distinct elements in R, listed as a, a, ... an. i. One of the elements from the list must be the multiplicative identity. Now, let r be some non-zero element of R; we want to show that one of the elements in the list is a multiplicative inverse of r. ii. That is, we want to show that there is an element in the list, a; for which r*a = 1. Let's look at all the products of r with each element in the list: r*a₁, r*a₂, ..., r*a iii. All of these products must be distinct: A. For, suppose there exists two of the products that are equal, that is, suppose r*aj = r*a with j not equal to k B. Now, ax has an additive inverse, (-a) C. Then, r*a;+r*(-a)=r*a+r*(-a) D. On the left: r*aj + r*(-as) = r*(a) + (-ax.)) E. On the right: a. b. r*(a +(-a))=r*0 c. r*0=0 r*a+r*(-a)=r*(a +(-a)) F. So, r* (aj + (-a)) = 0 G. Then, (aj + (-a)) = 0 H. Then, (aj + (-ax))+ ax=0+ ax On the left: I. a. (aj + (-a))+ ax= aj + ((-a) + a) b. aj + ((-a)+ ax) = aj + 0 c. aj +0=aj J. On the right: 0 + a = ax K. So, a, a, which is a contradiction iv. So, one of the products must equal 1, that is, r*a = 1 for some element a; in the set R v. Also, a*r = 1 vi. So, r has a multiplicative inverse in R vii. So, (R, +, *) is a field ring is c. From HW#2, (M:(R#), +, *) is a ring with identity, but the system is not a field (the commutative, and some non-zero elements of M (R#) do not have a multiplicative inverse. Determine 2x2 matrices A and B for which A*B= 0 (the zero matrix), but no entry of A or B is 0 (the real number 0). This shows that (M:(R#), +, *) has divisors of zero.
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E. On the right:

a. r*ak + r*(-ak) = r*(ak + (-ak))

b. r*(ak + (-ak)) = r*0

c. r*0 = 0

F. So, r*(aj + (-ak)) = 0

G. Then, (aj + (-ak)) = 0

H. Then, (aj + (-ak)) + ak = 0 + ak

II. On the left:

      1. (aj + (-ak)) + ak = aj + ((-ak) + ak)
      2. aj + ((-ak) + ak) = aj + 0
      3. aj + 0 = aj
        1. On the right: 0 + ak = ak
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