Solutions for Practice Problems for the Math 403 Midterm

Transcription

1 Solutions for Practice Problems for the Math 403 Midterm 1. This is a short answer question. No explanations are needed. However, your examples should be described accurately and precisely. (a) Given an example of a prime ideal of Z which is not a maximal ideal of Z. SOLUTION. There is just one such example, namely the ideal (0) of Z. This ideal is {0}. (b) Give an example of a maximal ideal of Z. SOLUTION. There are many examples. Suppose that p is a prime number. Then (p) is a maximal ideal of Z. In particular, (2) = 2Z, the set of even integers, is one example of a maximal ideal of Z. (c) Give an example of a maximal ideal in the ring Z[i]. SOLUTION. There are many examples, as we will soon discuss in class. One example is the principal ideal I = (3) of Z[i]. This is a nonzero ideal and we have I Z[i]. As explained in class, Z[i] is a PID. Suppose that J is an ideal of Z[i]] and that I J. Now J must be a principal ideal. Suppose that β is a generator of J. Thus, J = (β). Since I J, and 3 I, it follows that 3 J. Therefore, 3 = βγ for some γ Z[i]. We will use the norm map for Q[i]. This is defined by N(r+si) = (r+si)(r si) = r 2 +s 2 for all r, s Q. We have N(3) = 9 and N(βγ) = N(β)N(γ). That is, N(β)N(γ) = 9 and therefore, N(β) is an integer which divides 9. We have β = a + bi, where a, b Z. We have N(β) = a 2 + b 2. Thus, N(β) is a positive integer which divides 9. Thus, N(β) {1, 3, 9}. But the equation a 2 + b 2 = 3 has no solutions where a, b Z. Thus, either N(β) = 1 or N(β) = 9. If N(β) = 1, then β is a unit in Z[i]. It then follows that J = Z[i]. On the othet hand, if N(β) = 9, then N(γ) = 1. It then follows that γ is a unit of Z[i]. Since 3 = βγ and γ is a unit of Z[i], it follows that (3) = (β). Therefore, we have I = (3) = (β) = J. It follows that if J is any ideal of Z[i] containing I, then either J = Z[i] or J = I. We also have I Z[i]. It follows that I is a maximal ideal of Z[i].

2 (d) Give an example of a maximal ideal in the ring R R. SOLUTION. There are two examples. One is the ideal I = { (a, 0) a R }. This is the principal ideal of R R generated by (1, 0). (e) Give an example of an injective ring homomorphism ϕ : Z/5Z Z/10Z. SOLUTION. There are four idempotents e in the ring Z/10Z. As explained in class, for each of those idempotents, there exists a ring homomorphism from Z to Z/10Z which sends 1 to e. Take e to be the idempotent Z. We then get a ring homomorphism χ from Z to Z/10Z. For any m Z, we have χ(m) = me = 6m + 10Z. One finds that Ker(χ) = 5Z. Therefore, by the first isomorphism theorem, we obtain an injective ring isomorphism ϕ : Z/5Z Z/10Z defined by ϕ(m + 5Z) = χ(m) = 6m + 10Z for all m Z. 2. Suppose that R and S are rings with identity and that ϕ : R S is a surjective ring homomorphism. Let 1 R and 1 S denote the multiplicative identity elements of R and S, respectively. Prove that ϕ(1 R ) = 1 S. SOLUTION: Let e = ϕ(1 R ). Thus, e S. Furthermore, suppose that s S. Since ϕ is surjective, there exists an element r R such that ϕ(r) = s. Note that es = ϕ(1 R )ϕ(r) = ϕ(1 R r) = ϕ(r) = s and se = ϕ(r)ϕ(1 R ) = ϕ(r1 R ) = ϕ(r) = s. Therefore, e S and es = s = se for all s S. Thus, e is a multiplicative identity element for the ring S. However, so is 1 S. We proved in class that the multiplicative identity element of a ring (if it exists) is unique. Therefore, we have e = 1 S. That is, we have ϕ(1 R ) = 1 S. 3. Suppose that R is a ring and that I is an ideal of R. Suppose also that R/I = Z, where Z is the ring of integers. Suppose that J is an ideal of R, that I J, and that I J. Prove that R/J is a finite ring. SOLUTION: As pointed out in class one day, there exists a surjective ring homomorphism ϕ : R R/I. By assumption, there exists a ring isomorphism ψ : R/I Z. Let χ = ψ ϕ, which is a surjective ring homomorphism from R to Z. Furthermore, Ker(χ) = Ker(ϕ) because ψ is an isomorphism. Thus, Ker(χ) = I. We will consider the surjective homomorphism χ : R Z and use the correspondence theorem. We are assuming that J is an ideal of R containing I and that J I. The

3 corresponding ideal in Z is χ(j). Denote χ(j) by K. Since J I, K = χ(j) is not the zero ideal of the ring Z. Thus, J = mz for some positive integer m. Now we use proposition 4 on the handout about homomorphisms and ideals. With the above notation, it follows that R/J = Z/K. Now K = mz for some positive integer m and hence Z/K = Z/mZ. This ring has finitely many elements. To be precise, it has m elements. The ring R/J is isomorphic to Z/K and therefore also has a finite number of elements. Thus, R/J is indeed a finite ring. 4. Suppose that R and S are rings and that ϕ : R S is a ring homomorphism. Suppose that e is an idempotent of R. Prove that ϕ(e) is an idempotent of S. By making use of this result, determine all possible ring homomorphisms from Z to Q. Determine all possible ring homomorphisms from Z to Z/10Z. SOLUTION: Let s = ϕ(e), which is an element in S. Since e is an idempotent of R, we have ee = e. Thus, we have ss = ϕ(e)ϕ(e) = ϕ(ee) = ϕ(e) = s. This proves that ss = s and hence that s is an idempotent in the ring S. Now suppose that R = Z. The multiplicative identity element of Z is 1. This element is an idempotent in Z. Thus, if ϕ : Z S is a ring homomorphism, then s = ϕ(1) is an idempotent in the ring S. Furthermore, since ϕ is a homomomorphism of the additive group of Z to the additive group of S, we can say that ϕ(m) = ϕ(m 1) = mϕ(1) = ms for all m Z. Thus, ϕ is completely determined if one knows what ϕ(1) is. We also use the fact mentioned in class which states that if s is an idempotent in a ring S, and if one defines ϕ : Z S by ϕ(m) = ms for all m Z, then ϕ is actually a ring homomorphism. For completeness, we outline the proof of that fact. Lemma: Suppose that s is any idempotent in S. Define a map ϕ : Z S by ϕ(n) = ns for all n Z. The map ϕ is a ring homomorphism from Z to S. Furthermore, we have ϕ(1) = s.

4 Note that ns does not refer to ring multiplication. The element ns of S is defined in terms of the underlying additive group of S. (Recall that if G is a group, g G, and n Z, then one can define g n. But if the group operation is written additively, one would write ng instead of g n.) We will give most of the proof of the above lemma. The fact that ϕ(1) = s is clear. Suppose that n, m Z. Then ϕ(n + m) = (n + m)s = = ns + ms = ϕ(n) + ϕ(m). We have used the law of exponents from group theory. To prove that ϕ(nm) = ϕ(n)ϕ(m), we must use the distributive law many times. If n and m are positive, then ϕ(n)ϕ(m) = (ns)(ms) = (s + s + s... + s)(s + s + s...s), where the first parenthesis contains n terms, the second contains m terms. Using the distributive law, we obtain ss + ss ss, with nm terms. Thus, ϕ(n)ϕ(m) = (nm)(ss) = (nm)s = ϕ(nm) as we wanted to show. Note that we are using the fact that s is an idempotent of S when we replace ss by s. One must also consider the cases where n and m are not necessarily positive. We won t do every case. Suppose that n is positive and m is negative. Then, by the definitions in group theory, ns is as before and ms = m ( s) = m s. We will use the above calculation, but applied to the positive integers n and m. Thus, ϕ(n)ϕ(m) = (ns)(ms) = ( ns )( m s ) = ( (ns)( m s) ) = (n m )(ss) = ( n m )(ss) = (nm)s = ϕ(nm). We have used the elementary fact that a( b) = (ab) for a, b S in the third equality. We have used the law of exponents from group theory for the fifth equality. The above lemma is helpful in this problem. Consider first S = Q. Since Q is a field, Q is also an integral domain and hence the only idempotents in Q are 0 and 1. This was proved for any integral domain in the solution to problem 38 on page 265 of the text. Thus either ϕ(1) = 0 or ϕ(1) = 1. The above discussion and the lemma show that there are exactly two ring homomorphisms from Z to Q. However, we can exhibit them explicitly and so we don t have to use the lemma. One homomorphism is simply the inclusion map ϕ : Z Q defined by ϕ(n) = n for all n Z. This map is clearly a ring homomorphism. It has the property that ϕ(1) = 1.

5 Another ring homomorphism from Z to Q is defined by ϕ(n) = 0 for all n Z. This map ϕ is clearly a ring homomorphism. We have ϕ(1) = 0. Now we come to a description of all ring homomorphisms ϕ from Z to Z/10Z. As above, such a ring homomorphism is determined if one knows ϕ(1). Also, ϕ(1) must be an idempotent in the ring Z/10Z. Furthermore, by the lemma, each idempotent in Z/10Z will correspond to a homomorphism. By checking each of the elements in the ring Z/10Z, one finds four idempotents, namely Z, Z, Z, Z and the corresponding ring homomorphisms ϕ i : Z Z/10Z (where 1 i 4) are given by: ϕ 1 (m) = 0+10Z, ϕ 2 (m) = m+10z, ϕ 3 (m) = 5m+10Z, ϕ 4 (m) = 6m+10Z, for all m Z. 5. Consider the ring R = Z/256Z. Prove that every element of R is either a unit of R or a nilpotent element. Determine all maximal ideals of R. SOLUTION: In general, if m 1, we know that the units in Z/mZ are the elements a+mz, where a Z and gcd(a, m) = 1. Consider m = 256 = 2 8. If a Z, then gcd(a, 256) = 1 if and only if a is odd. Thus, the units in Z/256Z are the elements of the form a + 256Z, where a is odd. Now suppose that r R and that r is not a unit of R. Then r = a + 256Z, where a is not odd. Thus, a = 2k, where k Z. Notice that r 8 = (2k + 256Z) 8 = (2k) Z = 256k Z = Z, which is the additive identity 0 R in the ring R. Hence r is indeed a nilpotent element in R. Finally, suppose that I is an ideal of R and that I R. Then I cannot contain any units of R. To see this, suppose to the contrary that I contains a unit u of R. Since u R, there exists an element v R such that uv = 1 R. Since u I and v R, we have uv I. Hence 1 R I. Hence, for every element r R, we have r = r1 R I. This contradicts the assumption that I R. We have shown that if I is an ideal of R and I R, then I contains no units of R. Thus, the elements of I must be of the form 2k + 256Z, where k Z. Thus, they must be

6 of the form (k + 256Z)( Z), and hence must be in the principal ideal of R generated by Z. Therefore, apart from R, every ideal of R is contained in M = { 2k + 256Z k Z } = ( Z), the principal ideal of R generated by Z. It follows that this ideal M is a maximal ideal of R and that no other maximal ideals of R exist. 6. Suppose that R is a ring and that I and J are maximal ideals of R. TRUE OR FALSE: If R/I = R/J, then I = J. If true, give a proof. If false, give a specific counterexample. SOLUTION: The statement is false. As a counterexample, let R be the ring of continuous functions from the interval (0, 1) to R. Consider the homomorphisms ϕ and ψ from R to R defined as follows. For every f R, define ϕ(f) = f(1/2), ψ(f) = f(1/3). Let I = Ker(ϕ) and J = Ker(ψ). Both ϕ and ψ are surjecxtive ring homomorphisms. Therefore, by the first isomorphism theorem, we have R/I = R, R/J = R. Therefore, R/I = R/J. But, I J. To see this, let f R be the function defined by f(x) = x (1/2) for all x (0, 1). Then ϕ(f) = 0, but ψ(f) 0. Thus, f I, but f J. Hence, I J. 7. Find all of the maximal ideals in the ring R R. SOLUTION. First of all, we will determine all of the ideals in the ring R = R R. The additive identity in R is 0 R = (0, 0). Note that R is a commutative ring with identity. The multiplicative identity is 1 R = (1, 1). One ideal in R is the zero ideal Z = { (0, 0) }. Suppose that I is an ideal, but I Z. Then I contains a nonzero element (a, b). Choose one such element. At least one of the real numbers a or b is nonzero. We will consider three cases. Suppose that both a and b are nonzero. Since R is a field, both a and b are units of R. Thus, there exist c, d R such that ca = 1 and db = 1. We have (c, d)(a, b) = (1, 1), which

7 is the multiplicative identity element 1 R in the ring R. Since I is an ideal of R, (a, b) I, and (c, d) R, we must have (c, d)(a, b) I. That is, we must have 1 R I. It follows that if r R, then r = r 1 R I. Therefore, I = R. Apart from the ideals Z and R, there are two other ideals, namely M = { (u, 0) u R }, N = { (0, v) v R }. It is clear that M and N are additive subgroups of R. The fact that M and N are ideals of R follows from the fact that if (c, d) R, then (c, d) (u, 0) = (c u, d 0) = (cu, 0) M, (c, d) (0, v) = (c 0, d v) = (0, dv) N Thus, the ring R has at least four distinct ideals, namely Z, R, M, and N. We will prove that no other ideals I in the ring R exist. Assume that I is an ideal of R and that I Z and that I R. Thus, if (a, b) is a nonzero element of I, then we have shown that either a = 0 or b = 0. Otherwise, we would have I = R, as explained above. Assume that a 0 and hence that b = 0. Thus, I contains (a, 0). Now a is a unit in R. There exists a c R such that ca = 1. Then I contains (c, 0)(a, 0) = (1, 0). Furthermore, if u R, it follows that I contains (u, 0)(1, 0) = (u, 0). This is true for all u R. Thus, I contains M. We claim that I = M. If this were not so, then I would be bigger than M and I would contain an element (e, f) where f 0. But then I would contain all the elements (u, 0) + (e, f) = (u + e, f) and hence I would contain an element (a, b), where both a and b are nonzero. But this would imply (as shown above) that I = R. Since we are assuming now that I R, we have shown that I = M. Now assume that b 0 and hence that a = 0. The argument in the previous paragraph can be modified to show that I = N. It is a very similar argument. Now we will find all of the maximal ideals in the ring R. As explained above, The ring R has exactly four ideals. Two of the ideals of R described in the solution to that problem are: M = { (u, 0) u R }, N = { (0, v) v R }. The other two ideals of R are R itself and the zero ideal Z = { (0, 0) }. By definition, R is not a maximal ideal. Since Z M R, M Z, and M R, it follows that Z is not a maximal ideal of R. However, the only ideals of R containing M (from among the four ideals in that ring) are M itself and R. Thus, by definition, M is a maximal ideal of R. For a similar reason, N is also a maximal ideal of R.

8 8. Suppose that R and S are fields and that ϕ : R S is a ring homomorphism. TRUE OR FALSE: If ϕ is surjective, then ϕ is injective. If true, give a proof. If false, give a specific counterexample. SOLUTION: The statement is true. Here is a proof. Let K = Ker(ϕ). Thus, K is an ideal in the ring R. Since R is a field, it has only two ideals, namely R itself or the zero ideal. Thus, either K = R or K = { 0 R }. We will show that K R. Suppose to the contrary that K = R. Then, we would have ϕ(r) = 0 S for all r R. Since S is a field, it contains an element 1 S such that 1 S 0 S. Hence 1 S is not in the image of ϕ. Therefore, ϕ cannot be surjective. This contradicts our assumption that ϕ is surjective. It follows that K = { 0 R }. This implies that ϕ is indeed injective.