ALGEBRA AND NUMBER THEORY II: Solutions 3 (Michaelmas term 2008)

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1 ALGEBRA AND NUMBER THEORY II: Solutions 3 Michaelmas term 28 A A C B B D 61 i If ϕ : R R is the indicated map, then ϕf + g = f + ga = fa + ga = ϕf + ϕg, and ϕfg = f ga = faga = ϕfϕg. ii Clearly g lies in Kerϕ, so g Kerϕ. If f lies in Kerϕ, define h by hx = fx/x a for x a, and by ha =. Then hg = f because fa =, so f g. So in this case indeed Kerϕ = g. This fails for continuous functions: if again f is in Kerϕ, and f were to equal hg for some h, then for any x a we must have hx = fx/x a. But h cannot always be extended to a continuous function on R, e.g., if fx = x a, so hx = x a/ x a for x a. 62 i We check that I A is an ideal: the zero function is clearly in I A. If f and g are in I A, then for all a in A, f ga = fa ga =. And if f is in I A and g is in R, then f ga = faga = and g fa = gafa = so f g and g f are in I A. [Of course R is commutative, so f g = g f.] ii Let J = {f in R such that fx = for x large enough}. Then J is an ideal of R: clearly the zero function is in J. If fx and gx are zero for x large enough, then the same holds for fx gx, so f g is in J. And if fx is zero for x large enough, and g is in R, then f gx = fxgx is zero for x large enough. As R is commutative, this shows that J is an ideal of R. To show that J is not equal to any I A, first assume A, and let a be in A. We can always find a function f a in J that is nonzero at a take a function with a bump around a, zero everywhere else. Then f a is in J, but not in I A so that J and I A cannot be the same. To show that J I, note that I = R as there is no condition in this case. But not all functions in R are zero for x large enough, so we cannot have J = I either. a c b b 63 If d is in M 2 R, then I will also have to contain d 1 = a c and 1 A B C D = for all a, b, c, d, A, B, C and D in R. In particular I must contain 1, so it must contain 1 = A C, for all A, B, C and D in R. So for any a, b, c and d in R, putting A = b and C = d,, hence their sum. I must contain a c and b d a c b d 64 i Let K = ϕ 1 J. We check that K is an ideal of R: R is in K because ϕ R = S, and S is in J because it is an ideal of S. If k 1 and k 2 are in K, so ϕk 1 and ϕk 2 are in J, then k 1 k 2 is in K because ϕk 1 k 2 = ϕk 1 ϕk 2 is in J as J is a subgroup of S under addition. If k is in K and r is in R, then ϕrk = ϕrϕk is in J because ϕk is in J and J is an ideal of S. Similarly kr is in K. ii In general the image of an ideal is not an ideal. For example, take R = Z and S = Q, with the inclusion map a ring homomorphism because Z is a subring of Q. Then Z is an ideal of Z, but Z is not an ideal of Q, as Q is a field so it has only the ideals {} and Q. If ϕ : R S is surjective, and I is an ideal of R, then ϕi is an ideal of S: ϕ is a homomorphism of groups with respect to addition. That doesn t use the surjectivity, but the last bit does: if j is in ϕi, and s is in S, then there exist i in I with ϕi = j by definition of ϕi and r in R with ϕr = s because ϕ is surjective. Then sj = ϕrϕi = ϕri is in ϕi as ri is in I: I is an ideal of R. Similarly js is in ϕi. 65 Certainly, if A I, then a = 1 a lies in A I for all a A. On the other hand, assume that a I for all a A. An element in A is of the form a r aa for some r a in R and the sum is finite. Each 1 a c

2 r a a lies in I as a lies in A I and I is Because ideals are additive subgroups a r aa lies in I as well, so we get A I. 66 Showing that ϕ is a ring homomorphism is straightforward using that 5 2 = 25 = 1 in Z 13. To show the kernel is of the required shape, note that if a + bi is in Kerϕ, then a + bi = a + 5b + b 5 + i = 13k+b 5+i for some k in Z, so a+bi lies in 5+i, 13. Conversely, it is easy to check that 13 and 5+i are in the kernel, so any element of the form α13+β 5+i will go to as ϕ is a ring homomorphism. That shows that 5 + i, 13 Kerϕ as well, so 5 + i, 13 = Kerϕ. To see that 5 + i, 13 = 3 + 2i, we verify that 3 + 2i is in Kerϕ, so 3 + 2i Kerϕ = 5 + i, 13. We also note that 13 = 3 2i3 + 2i and 5 + i = 1 + i3 + 2i are in 3 + 2i, so that 5 + i, i as well by Q That ϕ is a homomorphism of rings is straightforward using that 7 2 = 1 in Z 25. It is surjective because Z 25 = {, 1,...,24} and, 1,..., 24 map to those elements under ϕ. In order to find the kernel, we see that any element in the kernel is of the form a + bi = a + 7b + b 7 + i = 25k + b 7 + i for some k in Z, so that Kerϕ 25, 7 + i. Note that 25 and 7 + i are in the kernel of ϕ, so any element of the form α25 + β 7 + i lies in Kerϕ: ϕα25 + β 7 + i = ϕαϕ25 + ϕβϕ 7 + i =. So Kerϕ = 25, 7 + i. If 25, 7 + i = γ for some γ in Z[i], we must have that γ divides both 25 and 7+i in Z[i]. Because N25 = 25 2 and N 7+i = = 5, it could only be an element c with Nγ dividing those numbers, so that Nγ = 1, 5 or 25. Trying gives that 3 4i divides both 25 = 3 4i3+4i and 7+i = 3 4i 1 i. Then any element of the form α25+β 7+i = [α3+4i+β 1 i]3 4i is in 3 4i, so that Kerϕ = 25, 7+i 3 4i. On the other hand, any element of the form δ3 4i is in the kernel, as ϕδ3 4i = ϕδϕ3 4i = ϕδ =. Therefore Kerϕ = 3 4i. 68 If I is in R, then 1 R is in both I and R the units of R. If I contains some element u in R, let v in R be such that vu = 1. Then r = r 1 = rvu lies in I for all r in R, so R I, and they must be equal as I R by definition. 69 If the ideal were of the form a, a would divide both 3 and 1+ 5, so Na would divide N3 = 9 and N1 + 5 = 6, so Na would equal 1 or 3. There are no elements with norm three, and the only elements with norm 1 are ±1, in which case the ideal would equal R. If this were the case, we could write 1 = a + b 53 + c + d = 3a + c 5d + 3b + c + d 5 for some a, b, c and d in Z. This means that 3a + c 5d = 1 and 3b + c + d =. The last tells us that c + d = in Z 3, but the first says that c + d = 1 in Z 3, so that is impossible. Hence the ideal is not principal. 7 Write d = xa + yb for some x and y in Z, so for any integer k, kd = kxa + kyb is in a, b, and hence d a, b. Conversely, if a = md and b = nd, then the elements in a, b are of the form sa + tb = smd + tnd = sm + tnd for s and t in Z, so a, b d as well. [We could also refer to Q.65, noticing that a and b are multiples of d, so they lie in d, and that d = xa + by for some integers x and y, so d lies in a, b.] 71 Clearly, {} = is principal, so assume I {}. Let α, α I, have lowest norm among the nonzero elements of I. Such α exists because the norms of the nonzero elements in I form a subset of N. According to Q.47, if β lies in I, we can write β = qα + r with q and r in Z[i], and Nr < Nα. Because r = β qα, r is in I. Because α has lowest norm among the nonzero elements of i, we must have r =. This shows that I α. But as α lies in I, we also have α = Z[i]α I so that α = I. 72 Suppose that X, 2 = fx for some fx in Z[X]. Then 2 = gxfx for some gx in Z[X], so working in Q[X], we see that fx and gx must have degree, and are elements of Z. Then fx = ±1 or ±2. If fx = ±2, because X lies in X, 2 = fx, X = hx±2 for some hx in Z[X], which cannot happen because the coefficients in hx±2 are even. So fx = ±1. Then 1 lies in fx = X, 2, so 1 = axx + bx2 for some ax and bx in Z[X]. Substituting for X this is nothing but to apply the specialisation homomorphism Z[X] Z, we find that 1 = a + b2 = b2 in Z, which is impossible. This gives a contradiction, from which we can conclude that X, 2 cannot be a principal ideal in Z[X]. 2

3 73 i If α = a + bi is in R, then we can write a = 2k + a and b = 2l + b with k and l in Z, and a = or 1, b = or 1. Then α = a + b i + k + li2, so α = a + b i, showing that there are at most the four given elements in R/I. In order to check that they are all different, assume a + b i = c + d i with a, b, c and d in {, 1}. Then a c + b d i =, so a c + b d i is in I, i.e., a c + b d i = e + fi2 = 2e + 2fi for some e and f in Z. But a c = or ±1, so we must have a = c. Similarly b = d, and therefore all four given elements in R/I are different. ii + 1 i 1 + i 1 i 1 + i 1 i 1 + i i i i i 1 + i i 1 + i i i 1 + i i i i 1 + i 1 + i 1 + i Note that both addition and multiplication are commutative in this ring, so we do not have to indicate how to read the tables. In the tables, we sometimes have to identify which of the four elements we get. E.g., i i = i 2 = 1 = 1 as 1 1 = 2 is in I. R/I is not a field: 1 + i R/I =, but from the multiplication table we see that there is no element a in R/I such that a 1 + i = 1 R/I = 1. Alternatively, 1 + i R/I, but 1 + i 2 = = R/I, so that R/I is not even an integral domain. iii Note that the diagonal of the addition table for R/I tells us that every element in R/I has order 1 or 2. As Z 4 has two elements of order 4, this means that R/I and Z 4 are not even isomorphic as groups under addition, hence certainly not as rings. This argument does not work for Z 2 Z 2. In fact, R/I and Z 2 Z 2 are isomorphic as groups under addition. But every element x in Z 2 Z 2 satisfies x 2 = x. This property would be preserved under a ring isomorphism, but in R/I this fails for i and 1 + i. So R/I and Z 2 Z 2 cannot be isomorphic as rings. 74 We have to remember that, if I is an ideal in a ring R, then in R/I, a b = ab, and c = d if and only if c d lies in I. Here we have to check if 3 + 2i4 + 3i 7 + 6i = i lies in 5 + 2i in Z[i]. As 13+11i 5+2i = 3 + i this is the case, so the answer is yes. 75 As in a general quotient ring R/I, α β = γ if and only if in R αβ γ lies in I, we have to check if any of 2+i1+i 2, 2+i1+i 1 i and 2+i1+i 1 i lie in 3+i = {α3+i with α in Z[i]}. We can do this by simply dividing each of those expressions by 3 + i and see if the quotient is in Z[i]. We find this is the case for the first and the last, but not the middle. So 2 + i 1 + i = 2 = 1 i but it is not equal to 1 i. 76 Let ϕ : R S be a homomorphism of rings, I an ideal of R, and π : R R/I the canonical projection. Show that if there exists a ring homomorphism ϕ : R/I S such that ϕ π = ϕ, then I Kerϕ. [This shows that the condition I Kerϕ in Proposition 3.14 is necessary.] 77 i ϕa 1 +b 1 i+a 2 +b 2 i = ϕa 1 +a 2 +b 1 +b 2 i = a 1 + a 2 + 4b 1 + 4b 2. This equals ϕa 1 +b 1 i+ ϕa 2 + b 2 i = a 1 + 4b 1 + a 2 + 4b 2. ϕa 1 + b 1 ia 2 + b 2 i = ϕa 1 a 2 b 1 b 2 + a 1 b 2 + a 2 b 1 i = a 1 a 2 b 1 b 2 + 4a 1 b 2 + a 2 b 1. On the other hand, ϕa 1 +b 1 iϕa 2 +b 2 i = a 1 + 4b 1 a 2 + 4b 2 = a 1 a 2 + 4a 1 b 2 + a 2 b b 1 b 2. The two are equal as 16 = 1 in Z 17. The homomorphism is surjective because, 1,...,16 map to all the elements, 1,..., 16 in Z 17. ii Elements in 4+i are of the form a+bi 4+i, and then ϕa+bi 4+i = ϕa+biϕ 4+i = ϕa + bi =. iii Let a + bi lie in the kernel of ϕ. Writing a + bi = a + 4b + b 4 + i, so ϕa + bi = a + 4b = means that 17 divides a + 4b in Z, say a + 4b = k17. As 17 = 4 i 4 + i in Z[i], a + bi = k 4 i 4 + i + b 4 + i = 4k + b ki 4 + i lies in 4 + i as desired. 3

4 iv We can apply the first isomorphism theorem: we know ϕ is a surjective ring homomorphism with kernel 4 + i, so Z[i]/ 4 + i = Z i If α = a+b 5 and β = c+d 5 are in R, then ϕα+β = ϕa+c+b+d 5 = a + c + b + d = a + b + c + d and ϕα + ϕβ = a + b + c + d = a + b + c + d as well. Also, ϕα β = ϕac + 5bd+ad+bc 5 = ac + 5bd + ad + bc = ac + ad + bc + bd as 5 = 1 in Z 2. On the other hand, ϕα ϕβ = a + b c + d = a + bc + d = ac + ad + bc + bd as well, so ϕ is a homomorphism of rings. Note that ϕ2 = 2 + = and ϕ1 + 5 = = so that both 2 and lie in Kerϕ. By Q.65 we get that 2, Kerϕ because Kerϕ is an ideal of R. ii We now only have to show that Kerϕ 2, as we know the reverse inclusion from i. If α = a + b 5 is in Kerϕ, then a + b =, so that a + b = 2k for some k in Z. Then α = a + b 5 = a b + b1 + 5 = k b2 + b1 + 5 lies in 2, iii We of course apply the first isomorphism theorem for rings, but before doing so we still have to check that ϕ is surjective. Clearly, ϕ = and ϕ1 = 1 so ϕ is surjective. As we now know that ϕ is a surjective ring homomorphism with Kerϕ = 2, 1 + 5, we get that R/2, = Z i If α = a+b 5 and β = c+d 5 are in R, then ϕα+β = ϕa+c+b+d 5 = a + c + b + d = a + b + c + d and ϕα + ϕβ = a + b + c + d = a + b + c + d so they agree. Also, ϕα β = ϕac + 7bd + ad + bc 5 = ac + 7bd + 4ad + bc = ac + 4ad + 4bc + 7bd, and ϕαϕβ = a + 4b c + 4d = ac + 4ad + 4bc + 16bd and the two agree as 16 = 7 in Z 9. Note that ϕ9 = 9 + = and ϕ4 5 = = so that both 9 and 4 5 lie in the ideal Kerϕ. By Q.65 we get that 9, 4 5 Kerϕ as Kerϕ is an ideal of R. ii We now only have to show that Kerϕ 9, 4 5 as we know the reverse inclusion from i. If α = a + b 5 is in Kerϕ, then a + 4b =, so that a + 4b = 9k for some k in Z. Then α = a + b 5 = a + 4b b4 5 = 9k b4 5 lies in 9, 4 5. iii We apply the first isomorphism theorem, but before doing so we still have to check that ϕ is surjective. Clearly, ϕ maps, 1,..., 8 to, 1,...,8 so ϕ is surjective. Then ϕ is a surjective ring homomorphism with Kerϕ = 9, 4 5, so we get that R/9, 4 5 = Z 9. 8 Define a map ϕ : Z 3 [X] Z 3 by mapping fx to f2, so that X+1 is in the kernel. This is a ring homomorphism as it is a specialisation homomorphism. If fx is in X+1, fx = gxx+1 for some gx in Z 3 [X], and ϕfx = ϕgxϕx +1 =. If fx lies in Kerϕ, write fx = qxx +1+c for some qx in Z 3 [X] and c a polynomial of degree at most, i.e., an element of Z 3. Then ϕfx = c, so c =, and fx is in X + 1. This shows that Kerϕ = X + 1. ϕ is surjective because it maps, 1 and 2 in Z 3 [X] to, 1 and 2 in Z 3. So by the first isomorphism theorem we get an isomorphism Z 3 [X]/X + 1 = Z i ϕx 2 2 = =, so if fx lies in the ideal X 2 2, fx = gxx 2 2 for some gx in Q[X], and ϕfx = ϕgxϕx 2 2 = ϕgx =. This shows that X 2 2 Kerϕ. If hx is in Kerϕ, write hx = X 2 2qX + a + bx with a and b in Q use division with remainder. Then = ϕhx = ϕa + bx = a + b 2. 2 is irrational, so a = b =. Hence hx is in X 2 2, and Kerϕ X 2 2 as well. ii If fx is in Q[X], then we can still write fx = qxx a + bx with a and b in Q. Then ϕfx = a+b 2 lies in Q[ 2]. ϕ is surjective onto Q[ 2] as a+bx maps to a+b 2 for a and b in Q. Because ϕ is a surjective ring homomorphism with kernel X 2 2 and image Q[ 2], the first isomorphism theorem gives us an isomorphism of rings Q[X]/X 2 2 = Q[ 2]. 82 i Note that for a in Z, A = a be in R. Then ϕ a1 b 1 c 1 + a2 b 1 c 1 and a2 a1 lies in R and ϕa = a, so ϕ is surjective. Let b 2 c 2 = ϕ a1+a2 b 1+b 2 c 1+c 2 = a 1 + a 2 = ϕ a1 b 1 c 1 + ϕ a2 4 b 2 c 2 b 2 c 2 and

5 b 1 ϕ a1 c 1 a2 b 2 c 2 = ϕ a1a2 a 1b 2+b 1c 2 c 1c 2 = a 1 a 2 = ϕ a1 b 1 c 1 ϕ a2 b 2 c 2 so that ϕ is a homomorphism of rings. ii a b c is in Kerϕ if and only if a =, so that Kerϕ = { b c with b and c in Z}. By the first isomorphism theorem we now get that R/Kerϕ = Z as ϕ is a surjective homomorphism of rings. The elements in R/Kerϕ are the cosets with respect to addition of Kerϕ, so the elements of R/Kerϕ are of the form a b c + Kerϕ = a b c + { d e with d and e in Z} = { a B C with B and C in Z}. R/Kerϕ = Z is given by picking out the upper left entry a: ϕx = ϕx in the first isomorphism theorem here means that if x = { a B C with B and C in Z}, pick any element x in x, so x = a B C for some B and C in Z. Then we compute ϕx, which gives us a. This is independent of the choice of x in x, i.e., the choice of B and C in Z. 83 First we show that I J + I K I J + K. Note that I J + K is an ideal, so in particular is closed under addition. So I J + I K is contained in it if we can show that I J and I K are contained in it. But J J + K, so I J I J + K, and K J + K so I K I J + K as well. Next we show that I J + K I J + I K. An element in J + K is of the form j + k with j in J and k in K. An element in I J + K is then of the form: a finite sum of elements ij + k with i in I, j in J and k in K. As ij + k = ij + ik, we can rewrite such an element as a finite sum of terms of the form ij, plus a finite sum of terms of the form ik. But the finite sum of ij s is in I J, and the finite sum of ik s is in I K, so that the element is in I J + I K. 84 Let I = a 1,...,a m and J = b 1,..., b n, K = a 1,..., a m, b 1,...,b n so we have to check if I + J = K. Clearly a i = a i + is in I + J and similarly for the b j so by Q.65 K I + J. On the other hand, by Q.65 again, I K and J K. As K is an ideal so it is closed under addition, if i is in I K and j is in J K, then i + j is in K. This shows that I + J K as well. Now let L = a 1 b 1,...,a 1 b n,..., a m b 1,..., a m b n. Then a i b j is in I J by the definitions, so by Q.65, L I J. On the other hand, an element in I J looks like a finite sum of terms that look like m k=1 r ka k n l=1 r lb l for r k and r l in R. As L is an ideal so it is closed under addition, we only have to check that each of those terms is of the form m n i=1 j=1 r i,ja i b j with all r i,j in R. But R is commutative, so m k=1 r ka k n l=1 r lb l = m n k=1 l=1 r kr l a k b l so that is the case. This shows that I J L as well. [Note that we used that R has an identity to see that a j is in a 1,..., a m = { m k=1 r ka k with all r k in R} and similarly for b j. The commutativity of R is also used to see that { m k=1 r ka k with all r k in R} is indeed an ideal.] 85 Z[ 5] is a commutative ring with identity, so we use Q.65 to check if two ideals are equal by working at the level of generators. i We always have 2, as 1 = Z[ 5]. For the reverse inclusion, notice that 1 = is in 2, 3 5. ii , clearly as = For the reverse inclusion notice that 4 = is in 1 + 5, so = is in [You could find this by computing /1 + 5 = as well.] iii 3 5 = and 3+ 5 = so that 3 5, , 1 5. For the reverse inclusion, note that = 4 is in 3 5, 3+ 5, so that = 1 5 is in it. Also = 6 is in it, so 6 4 = 2 is in it, which shows the reverse inclusion. iv For we just have to show that 11 = a + b for some a and b in Z. Computing this using division we get that 11 = To show that we now have to show that / 11, i.e., we do not have a + b 511 = for any a and b in Z. This is clear as 11a = 3 has no solution in Z. 86 As Z[ 5] is a commutative ring with identity, we shall use Q.65 a lot to check two ideals are equal. i 1 5, , 2 as 1 5 = , and 1+ 5, 2 1 5, 2 as = ii As 1 = Z[ 5], we have 2 5, 3 1. Note that = 1 is in 2 5, 3, so = 1 is in it, hence 1 2 5, 3 as well. 5

6 iii By Q.84, 1 + 5, 3 1 5, 2 = 6, , 3 3 5, 6. Note that then the element = is in this ideal, so also = is in this ideal. This shows that , 3 3 5, 6. For the reverse inclusion we need to show that , and 6 are in Clearly = and 6 = are in it. Then = is in as well. [Alternatively, we can check this by checking if 2+2 5/1+ 5, 3 3 5/1+ 5 and 6/1 + 5 are in Z[ 5].] iv The manipulations here very similar to the previous part replacing 5 with 5 everywhere. v 1 5, 2 2 = 4 2 5, 2 2 5, 2 2 5, 4 by Q.84. As we need each generator only once, this equals the ideal 4 2 5, 2 2 5, 4. This is contained in 2 as = etc. Now we notice that = 2 is in 4 2 5, 2 2 5, 4 which shows that 1 5, as well. vi If 1 5, 2 = α for some α in Z[ 5], so α divides both 1 5 and 2 in Z[ 5]. Then Nα must divide N1 5 = 6 and N2 = 4 in Z so that Nα divides 2. As Nα, we must have Nα = 1 or 2. If α = a+b 5, Nα = a 2 +5b 2 so there are only two possible α s, ±1. As α and α generate the same ideal, we can assume α = 1 if necessary, and see if 1 5, 2 = 1 = Z[ 5]. But if that were the case, then 1 = 1 2 = 1 5, 2 2 = 2 from what we did before, so 1 = 2c + d 5 for some c and d in Z, which is not possible. So 1 5, 2 is not a principal ideal. [Alternatively, as generators of principal ideals in integral domains are unique up to units check: in an integral domain R, a = b if and only if a = bu for some unit u R, we could argue as follows. If 1 5, 2 = α, then α 2 = α 2 = 1 5, 2 2 = 2 so that 2 = ±α 2 as the units in Z[ 5] are ±1. Then 2 2 = N2 = N±α 2 = N±1Nα 2 = Nα 2 so that Nα = 2 as Nα. This is not possible as a 2 + 5b 2 = 2 has no solutions for a and b in Z.] 87 In order to show that I + J = Z[i], we have to show that 1 is in I + J, as Z[i] has an identity. So we have to write 1 = α + β with α in I and β in J. One way of doing that is using the Euclidean algorithm in Z[i] see Q.47, computing the greatest common divisor of 3 2i and 3 + 2i, and expressing it in the form α3 2i + β3 + 2i. This gives 3 2i = i3 + 2i i, 3 + 2i = 21 + i + 1, and so 1 = 3 + 2i 21 + i = 3 + 2i 2[3 2i + i3 + 2i] = 23 2i + 1 2i3 + 2i. [Alternatively, you can find that 1 is in I + J by playing around: 3 2i i = 6 is in I + J; 3 2i 3 + 2i = 4i is in I + J, and so is i 4i = 4; then 6 4 = 2 is in I + J, and hence 3 + 2i 1 + i2 = 1. Note that this approach only shows that 1 is in I + J, but does not express it in the form α + β with α in I and β in J, so this would not be enough for the last bit of the question.] Because Z[i] is commutative with identity, we have that I J = I J = 3 2i 3 + 2i = 3 2i3 + 2i = 13. By the Chinese remainder theorem, Z[i]/13 = Z[i]/I Z[i]/J = Z[i]/3 2i Z[i]/3 + 2i. If 1 = α + β with α in I and β in J, then the element in Z[i]/13 that maps to 1, 2 is given by 1 β + 2 α. With the α and β found before, we get 1 1 2i3 + 2i i = 5 + 4i. 88 Let I = X + 1 and J = X 2 + 2, which are ideals in Q[X]. Then I + J = Q[X]: Q[X] is commutative with identity, so we only have to write 1 = fxx gxx for some fx and gx in Q[X], and we can find such fx and gx using the Euclidean algorithm as in Theorem We get that 1 = XX X Then by the Chinese remainder theorem, because Q[X] is commutative with identity, I J = I J = X +1 X 2 +2 = X +1X 2 +2 = X 3 +X 2 +2X +2, and we get an isomorphism Q[X]/X 3 + X 2 + 2X + 2 = Q[X]/I Q[X]/J = Q[X]/X + 1 Q[X]/X The element mapping to X, 2X is given by X 1 3 X X XX + 1 = 1 3 X X. 89 We write 1 = X + X 1 so that as ideals X+X 1 = Q[X]. So by the Chinese remainder theorem we get that X X 1 = X X 1 = X 2 X as Q[X] is a commutative ring with 1, and Q[X]/X 2 X = Q[X]/X Q[X]/X 1 under the natural map. Then 3 = 3X + 3X 1 and 5 = 5X + 5X so that we should take 3X 1 + 5X = 2X + 3 in Q[X]: 2X + 3 in Q[X]/X 2 X maps to 3, 5 in Q[X]/X Q[X]/X 1. 9 The idea is to show that X 2 5 is a maximal ideal in Q[X] a commutative ring with identity 1, and that Q[X]/X 2 5 = Q[ 5] as rings, because then Q[X]/X 2 5 is a field by Theorem 3.25, 6

7 and so Q[ 5] is a field because it is isomorphic to Q[X]/X 2 5 as rings. X 2 5 is irreducible in Q[X] as it is of degree two, but has no roots in Q: the candidates for roots in Q would be ±1 and ±5, and none of these are roots. Hence, since all maximal ideals in QX are of the form fx for some irreducible polynomial fx, X 2 5 is a maximal ideal in Q[X], and as Q[X] is a commutative ring with identity 1, the quotient Q[X]/X 2 5 is a field, again by Theorem In order to show that Q[X]/X 2 5 = Q[ 5], we try to define a surjective ring homomorphism ϕ : Q[X] 5 with Kerϕ = X 2 5, so that we can apply the first isomorphism theorem for rings. So once we ve done this, we are done. Q[ 5] = {a + b 5 with a, b in Q}, a subring of C. Define ϕ : Q[X] Q[ 5] via fx = f 5. Note that f 5 is in Q[ 5] because any term a 5 n with a in Q and n will be of the form b or b 5 for some rational number b. ϕ is a homomorphism of rings by the specialisation homomorphism as Q is clearly a subring of Q[ 5]. ϕ is surjective, as for a and b in Q, ϕbx + a = a + b 5, so we get all elements in Q[ 5]. In order to determine Kerϕ, first notice that X 2 5 Kerϕ: elements in X 2 5 are of the form fxx 2 5 for some fx in Q[X], and then ϕfxx 2 5 = ϕfxϕx 2 5 = ϕfx =. To see Kerϕ X 2 5 as well, we could go about it in two ways. The quick way is to notice that we know that X 2 5 Kerϕ Q[X], and we know that X 2 5 is a maximal ideal in Q[X]. So from the definition of maximal ideals, either Kerϕ = X 2 5 or Kerϕ = Q[X]. The last cannot happen as that would mean that ϕ1 = which is clearly not the case. The other way of seeing it is more computational: if gx is in Kerϕ, write gx = hxx 2 5+aX+b with a and b in Q by using long division in Q[X]. Because ϕ is a homomorphism or rings, and gx is in Kerϕ, = ϕgx = ϕhxϕx 2 5+ϕaX +b = b+a 5, so a = b = and gx = hxx 2 5 lies in X 2 5. So Kerϕ X 2 5 as well, and equality must hold: Kerϕ = X 2 5. Either way, we can now apply the first isomorphism theorem for rings: Q[X]/X 2 5 = Q[ 5] as rings. 91 We would be tempted to use the map fx fi 3, but that will not map X 2 + X + 1 to zero. Instead, we use one of the roots of X 2 + X + 1, like α = 1 i 3/2. So we define ϕ : Q[X] C by ϕfx = fα. This is a ring homomorphism, as it is a specialisation homomorphism. We first determine its kernel. As α is root of X 2 + X + 1, if fx = gxx 2 + X + 1 is in X 2 + X + 1, ϕfx = ϕgxϕx 2 + X + 1 = so that X 2 + X + 1 Kerϕ. Conversely, if fx is in Kerϕ, we can write fx = qxx 2 + X ax + b for some qx in Q[X] and a, b in Q use division with remainder. Then = ϕfx = aα+b. If a, we would get α = b/a would be in R or even Q, which is clearly not the case. So a = and hence b =. therefore fx = qxx + X + 1 lies in X 2 + X + 1 and Kerϕ = X 2 + X + 1. In order to determine the image of ϕ, Iϕ, write any fx in Q[X] as qxx 2 + X ax + b with qx in Q[X] and a, b in Q use again division with remainder. Then ϕfx = aα + b = b a 2 a 2 i 3 lies in Q[i 3]. Also ϕ 2bX + a b = a + bi 3 for a, b in Q so that ϕ is surjective. So by the first isomorphism theorem we get an isomorphism Q[X]/X 2 + X + 1 = Q[i 3]. Now in order to show that Q[i 3] is a field, we use Theorem 3.25 and since all maximal ideals in QX are of the form fx for some irreducible polynomial fx. By the latter, X 2 + X + 1 is a maximal ideal in Q[X] as X 2 + X + 1 is irreducible in Q[X]: it is of degree two but does not have any roots in Q or even R. Because Q[X] is a commutative ring with identity 1, Theorem 3.25 tells us that Q[X]/X 2 +X +1 must be a field. 92 We define a map Z[X] Z n by mapping fx to f. This is the composition of the maps Z[X] Z given by fx f, and Z Z n given by a a. both those maps are specialisation homomorphisms of rings, and so is their composition. Alternatively, it is easy to write it out starting with fx and gx in Z[X]. If fx lies in X, n, fx = gxx + hxn for some gx and hx in Z[X]. Then ϕfx = g + hn =, so X, n Kerϕ. To show the reverse inclusion, take fx in Kerϕ, and write it as gxx +m where m is the constant term of fx. Then f = m. As ϕfx =, m is divisble by n, say m = kn for some k in Z. Then fx = gxx + kn lies in X, n. This shows that Kerϕ = X, n. ϕ is surjective as, 1,..., n 1 map to, 1,..., n 1. So by the first isomorphism theorem, we get an isomorphism Z[X]/X, n = Z n. As Z[X] is a commutative ring with identity 1, Theorem 3.25 tells 7

8 us that X, n is a prime resp. maximal ideal if and only if Z n is an integral domain resp. a field. By Example 3.24, this happens if and only if n is a prime number. 93 For the first we ll give two answers: one directly from the definitions, and one using more theory. X consists of the polynomials with vanishing constant term. If fxgx lies in X, either fx of gx must have vanishing constant term because Z is an integral domain, so lies in X. This means X is a prime ideal. It cannot be maximal: X, 2 is certainly a larger ideal as it contains 2. It is also not the whole of Z[X]: 1 = fxx + gx2 leads to a contradiction by looking at the constant term. So we have X X, 2 Z[X] with the inclusions strict, and X is not maximal. We can get a quick answer by using the first isomorphism theorem as in Q.92. One shows that Z[X]/X = Z and Z[X]/n = Z n [X]. The maps to use are ϕ : Z[X] Z given by ϕfx = f and ψ : Z[X] Z n [X] given by ψfx = fx, i.e., reduction of the coefficients modulo n. ϕ is a ring homomorphism simply write it out or use that it is a specialisation homomorphism, and ψ is a ring homomorphism reduction of coefficients. As Z[X] is a commutative ring with identity 1 we only have to see if Z or Z n [X] are integral domains and/or fields by Theorem 3.25, Z is an integral domain but not a field, so X is a prime ideal in Z[X] but not a maximal ideal. For Z n [X] it depends on n. If n is not a prime number, Z n is not an integral domain by Example Then Z n [X] which contains Z n as the constants cannot be an integral domain, and certainly not a field, so n is neither a prime ideal nor a maximal ideal in Z[X] in this case. If n is a prime number, Z n is a field by Example 3.24, and Z n [X] is an integral domain in this case degrees add up for a product of polynomials. But it is not a field by Q.34 as not every nonzero element of Z n [X] is a unit. So now n is a prime ideal but not a maximal ideal in Z[X]. [That n is not a maximal ideal in Z[X] can be seen by hand: n, X contains n but is not equal to it as n consists of all elements in Z[X] with all coefficients divisible by n and X is not of this shape. Also n, X Z[X] because otherwise we could write 1 = fxn+gxx for some fx and gx in Z[X], and as n 2 this is impossible by looking at the constant term.] 94 All rings are commutative with nonzero identity, so we can apply Theorem 3.25 directly, as the quotient rings are identified in the Exercises. This gives us that 4 + i is a maximal ideal and hence a prime ideal in Z[i] as Z 17 is a field because 17 is prime, and Z[i]/ 4 + i = Z 17. 2, is maximal and hence prime in Z[ 5] because Z 2 is a field as 2 is prime, and Z[ 5]/2, = Z 2. 9, 4 7 is not a prime ideal in Z[ 7] because Z[ 7]/9.4 7 = Z 9 and 9 is not prime, so Z 9 is not an integral domain. Hence 9, 4 7 is not a maximal ideal either. of course, this also follows because Z 9 is not a field. X + 1 is a maximal ideal in Z 3 [X] because Z 3 [X]/X + 1 = Z 3 which is a field as 3 is prime, and hence it is also a prime ideal in Z 3 [X]. With Q[X]/X 2 2 we can argue both ways. Either we say that X 2 2 is irreducible in Q[X] because X 2 2 is of degree 2 and has no rational roots, therefore, since all maximal ideals in QX are of the form fx for some irreducible polynomial fx, we conclude that X 2 2 is maximal and conesquently prime. So then it follows that Q[ 2] is a field. We could also check by hand that Q[ 2], which is a subring of C by Q.12 is a field. See Q.23. Then we can conclude that X 2 2 is a maximal ideal hence a prime ideal in Q[X] as Q[X]/X 2 2 is isomorphic to the field Q[ 2], so must be a field itself. 8