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


 Sharon Collins
 3 years ago
 Views:
Transcription
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
Computations/Applications
Computations/Applications 1. Find the inverse of x + 1 in the ring F 5 [x]/(x 3 1). Solution: We use the Euclidean Algorithm: x 3 1 (x + 1)(x + 4x + 1) + 3 (x + 1) 3(x + ) + 0. Thus 3 (x 3 1) + (x + 1)(4x
More informationMATH RING ISOMORPHISM THEOREMS
MATH 371  RING ISOMORPHISM THEOREMS DR. ZACHARY SCHERR 1. Theory In this note we prove all four isomorphism theorems for rings, and provide several examples on how they get used to describe quotient rings.
More informationRINGS: SUMMARY OF MATERIAL
RINGS: SUMMARY OF MATERIAL BRIAN OSSERMAN This is a summary of terms used and main results proved in the subject of rings, from Chapters 1113 of Artin. Definitions not included here may be considered
More information2a 2 4ac), provided there is an element r in our
MTH 310002 Test II Review Spring 2012 Absractions versus examples The purpose of abstraction is to reduce ideas to their essentials, uncluttered by the details of a specific situation Our lectures built
More informationHomework 10 M 373K by Mark Lindberg (mal4549)
Homework 10 M 373K by Mark Lindberg (mal4549) 1. Artin, Chapter 11, Exercise 1.1. Prove that 7 + 3 2 and 3 + 5 are algebraic numbers. To do this, we must provide a polynomial with integer coefficients
More informationHonors Algebra 4, MATH 371 Winter 2010 Assignment 3 Due Friday, February 5 at 08:35
Honors Algebra 4, MATH 371 Winter 2010 Assignment 3 Due Friday, February 5 at 08:35 1. Let R 0 be a commutative ring with 1 and let S R be the subset of nonzero elements which are not zero divisors. (a)
More informationCHAPTER I. Rings. Definition A ring R is a set with two binary operations, addition + and
CHAPTER I Rings 1.1 Definitions and Examples Definition 1.1.1. A ring R is a set with two binary operations, addition + and multiplication satisfying the following conditions for all a, b, c in R : (i)
More informationRings. Chapter 1. Definition 1.2. A commutative ring R is a ring in which multiplication is commutative. That is, ab = ba for all a, b R.
Chapter 1 Rings We have spent the term studying groups. A group is a set with a binary operation that satisfies certain properties. But many algebraic structures such as R, Z, and Z n come with two binary
More informationPRACTICE FINAL MATH , MIT, SPRING 13. You have three hours. This test is closed book, closed notes, no calculators.
PRACTICE FINAL MATH 18.703, MIT, SPRING 13 You have three hours. This test is closed book, closed notes, no calculators. There are 11 problems, and the total number of points is 180. Show all your work.
More informationA few exercises. 1. Show that f(x) = x 4 x 2 +1 is irreducible in Q[x]. Find its irreducible factorization in
A few exercises 1. Show that f(x) = x 4 x 2 +1 is irreducible in Q[x]. Find its irreducible factorization in F 2 [x]. solution. Since f(x) is a primitive polynomial in Z[x], by Gauss lemma it is enough
More informationCHAPTER 14. Ideals and Factor Rings
CHAPTER 14 Ideals and Factor Rings Ideals Definition (Ideal). A subring A of a ring R is called a (twosided) ideal of R if for every r 2 R and every a 2 A, ra 2 A and ar 2 A. Note. (1) A absorbs elements
More informationMath 547, Exam 1 Information.
Math 547, Exam 1 Information. 2/10/10, LC 303B, 10:1011:00. Exam 1 will be based on: Sections 5.1, 5.2, 5.3, 9.1; The corresponding assigned homework problems (see http://www.math.sc.edu/ boylan/sccourses/547sp10/547.html)
More informationEighth Homework Solutions
Math 4124 Wednesday, April 20 Eighth Homework Solutions 1. Exercise 5.2.1(e). Determine the number of nonisomorphic abelian groups of order 2704. First we write 2704 as a product of prime powers, namely
More informationAlgebraic structures I
MTH5100 Assignment 110 Algebraic structures I For handing in on various dates January March 2011 1 FUNCTIONS. Say which of the following rules successfully define functions, giving reasons. For each one
More informationCSIR  Algebra Problems
CSIR  Algebra Problems N. Annamalai DST  INSPIRE Fellow (SRF) Department of Mathematics Bharathidasan University Tiruchirappalli 620024 Email: algebra.annamalai@gmail.com Website: https://annamalaimaths.wordpress.com
More informationSolutions for Some Ring Theory Problems. 1. Suppose that I and J are ideals in a ring R. Assume that I J is an ideal of R. Prove that I J or J I.
Solutions for Some Ring Theory Problems 1. Suppose that I and J are ideals in a ring R. Assume that I J is an ideal of R. Prove that I J or J I. SOLUTION. Assume to the contrary that I is not a subset
More informationMath Introduction to Modern Algebra
Math 343  Introduction to Modern Algebra Notes Rings and Special Kinds of Rings Let R be a (nonempty) set. R is a ring if there are two binary operations + and such that (A) (R, +) is an abelian group.
More informationAN INTRODUCTION TO THE THEORY OF FIELD EXTENSIONS
AN INTRODUCTION TO THE THEORY OF FIELD EXTENSIONS SAMUEL MOY Abstract. Assuming some basic knowledge of groups, rings, and fields, the following investigation will introduce the reader to the theory of
More informationJohns Hopkins University, Department of Mathematics Abstract Algebra  Spring 2013 Midterm Exam Solution
Johns Hopkins University, Department of Mathematics 110.40 Abstract Algebra  Spring 013 Midterm Exam Solution Instructions: This exam has 6 pages. No calculators, books or notes allowed. You must answer
More informationPolynomial Rings. i=0. i=0. n+m. i=0. k=0
Polynomial Rings 1. Definitions and Basic Properties For convenience, the ring will always be a commutative ring with identity. Basic Properties The polynomial ring R[x] in the indeterminate x with coefficients
More informationHomework problems from Chapters IVVI: answers and solutions
Homework problems from Chapters IVVI: answers and solutions IV.21.1. In this problem we have to describe the field F of quotients of the domain D. Note that by definition, F is the set of equivalence
More informationNOTES ON FINITE FIELDS
NOTES ON FINITE FIELDS AARON LANDESMAN CONTENTS 1. Introduction to finite fields 2 2. Definition and constructions of fields 3 2.1. The definition of a field 3 2.2. Constructing field extensions by adjoining
More information4.5 Hilbert s Nullstellensatz (Zeros Theorem)
4.5 Hilbert s Nullstellensatz (Zeros Theorem) We develop a deep result of Hilbert s, relating solutions of polynomial equations to ideals of polynomial rings in many variables. Notation: Put A = F[x 1,...,x
More informationbe any ring homomorphism and let s S be any element of S. Then there is a unique ring homomorphism
21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UFD. Therefore
More informationM2P4. Rings and Fields. Mathematics Imperial College London
M2P4 Rings and Fields Mathematics Imperial College London ii As lectured by Professor Alexei Skorobogatov and humbly typed by as1005@ic.ac.uk. CONTENTS iii Contents 1 Basic Properties Of Rings 1 2 Factorizing
More informationφ(a + b) = φ(a) + φ(b) φ(a b) = φ(a) φ(b),
16. Ring Homomorphisms and Ideals efinition 16.1. Let φ: R S be a function between two rings. We say that φ is a ring homomorphism if for every a and b R, and in addition φ(1) = 1. φ(a + b) = φ(a) + φ(b)
More informationTHE JOHNS HOPKINS UNIVERSITY Faculty of Arts and Sciences FINAL EXAM  FALL SESSION ADVANCED ALGEBRA I.
THE JOHNS HOPKINS UNIVERSITY Faculty of Arts and Sciences FINAL EXAM  FALL SESSION 2006 110.401  ADVANCED ALGEBRA I. Examiner: Professor C. Consani Duration: take home final. No calculators allowed.
More informationSchool of Mathematics and Statistics. MT5836 Galois Theory. Handout 0: Course Information
MRQ 2017 School of Mathematics and Statistics MT5836 Galois Theory Handout 0: Course Information Lecturer: Martyn Quick, Room 326. Prerequisite: MT3505 (or MT4517) Rings & Fields Lectures: Tutorials: Mon
More informationAbstract Algebra: Chapters 16 and 17
Study polynomials, their factorization, and the construction of fields. Chapter 16 Polynomial Rings Notation Let R be a commutative ring. The ring of polynomials over R in the indeterminate x is the set
More informationA connection between number theory and linear algebra
A connection between number theory and linear algebra Mark Steinberger Contents 1. Some basics 1 2. Rational canonical form 2 3. Prime factorization in F[x] 4 4. Units and order 5 5. Finite fields 7 6.
More informationSolutions to oddnumbered exercises Peter J. Cameron, Introduction to Algebra, Chapter 2
Solutions to oddnumbered exercises Peter J Cameron, Introduction to Algebra, Chapter 1 The answers are a No; b No; c Yes; d Yes; e No; f Yes; g Yes; h No; i Yes; j No a No: The inverse law for addition
More informationMATH 581 FIRST MIDTERM EXAM
NAME: Solutions MATH 581 FIRST MIDTERM EXAM April 21, 2006 1. Do not open this exam until you are told to begin. 2. This exam has 9 pages including this cover. There are 10 problems. 3. Do not separate
More informationPrime Rational Functions and Integral Polynomials. Jesse Larone, Bachelor of Science. Mathematics and Statistics
Prime Rational Functions and Integral Polynomials Jesse Larone, Bachelor of Science Mathematics and Statistics Submitted in partial fulfillment of the requirements for the degree of Master of Science Faculty
More informationLecture 7.3: Ring homomorphisms
Lecture 7.3: Ring homomorphisms Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 4120, Modern Algebra M. Macauley (Clemson) Lecture 7.3:
More informationMT5836 Galois Theory MRQ
MT5836 Galois Theory MRQ May 3, 2017 Contents Introduction 3 Structure of the lecture course............................... 4 Recommended texts..................................... 4 1 Rings, Fields and
More informationInformal Notes on Algebra
Informal Notes on Algebra R. Boyer Contents 1 Rings 2 1.1 Examples and Definitions................................. 2 1.2 Integral Domains...................................... 3 1.3 Fields............................................
More informationEuclidean Domains. Kevin James
Suppose that R is an integral domain. Any function N : R N {0} with N(0) = 0 is a norm. If N(a) > 0, a R \ {0 R }, then N is called a positive norm. Suppose that R is an integral domain. Any function N
More informationMath Introduction to Modern Algebra
Math 343  Introduction to Modern Algebra Notes Field Theory Basics Let R be a ring. M is called a maximal ideal of R if M is a proper ideal of R and there is no proper ideal of R that properly contains
More informationφ(xy) = (xy) n = x n y n = φ(x)φ(y)
Groups 1. (Algebra Comp S03) Let A, B and C be normal subgroups of a group G with A B. If A C = B C and AC = BC then prove that A = B. Let b B. Since b = b1 BC = AC, there are a A and c C such that b =
More informationMTH310 EXAM 2 REVIEW
MTH310 EXAM 2 REVIEW SA LI 4.1 Polynomial Arithmetic and the Division Algorithm A. Polynomial Arithmetic *Polynomial Rings If R is a ring, then there exists a ring T containing an element x that is not
More informationMATH 3030, Abstract Algebra Winter 2012 Toby Kenney Sample Midterm Examination Model Solutions
MATH 3030, Abstract Algebra Winter 2012 Toby Kenney Sample Midterm Examination Model Solutions Basic Questions 1. Give an example of a prime ideal which is not maximal. In the ring Z Z, the ideal {(0,
More information1. Group Theory Permutations.
1.1. Permutations. 1. Group Theory Problem 1.1. Let G be a subgroup of S n of index 2. Show that G = A n. Problem 1.2. Find two elements of S 7 that have the same order but are not conjugate. Let π S 7
More informationMath 120 HW 9 Solutions
Math 120 HW 9 Solutions June 8, 2018 Question 1 Write down a ring homomorphism (no proof required) f from R = Z[ 11] = {a + b 11 a, b Z} to S = Z/35Z. The main difficulty is to find an element x Z/35Z
More informationTHROUGH THE FIELDS AND FAR AWAY
THROUGH THE FIELDS AND FAR AWAY JONATHAN TAYLOR I d like to thank Prof. Stephen Donkin for helping me come up with the topic of my project and also guiding me through its various complications. Contents
More informationPolynomial Rings. (Last Updated: December 8, 2017)
Polynomial Rings (Last Updated: December 8, 2017) These notes are derived primarily from Abstract Algebra, Theory and Applications by Thomas Judson (16ed). Most of this material is drawn from Chapters
More informationFinite Fields. Sophie Huczynska. Semester 2, Academic Year
Finite Fields Sophie Huczynska Semester 2, Academic Year 200506 2 Chapter 1. Introduction Finite fields is a branch of mathematics which has come to the fore in the last 50 years due to its numerous applications,
More informationHomework 8 Solutions to Selected Problems
Homework 8 Solutions to Selected Problems June 7, 01 1 Chapter 17, Problem Let f(x D[x] and suppose f(x is reducible in D[x]. That is, there exist polynomials g(x and h(x in D[x] such that g(x and h(x
More informationAlgebra Homework, Edition 2 9 September 2010
Algebra Homework, Edition 2 9 September 2010 Problem 6. (1) Let I and J be ideals of a commutative ring R with I + J = R. Prove that IJ = I J. (2) Let I, J, and K be ideals of a principal ideal domain.
More informationName: Solutions Final Exam
Instructions. Answer each of the questions on your own paper. Be sure to show your work so that partial credit can be adequately assessed. Put your name on each page of your paper. 1. [10 Points] All of
More informationTotal 100
Math 542 Midterm Exam, Spring 2016 Prof: Paul Terwilliger Your Name (please print) SOLUTIONS NO CALCULATORS/ELECTRONIC DEVICES ALLOWED. MAKE SURE YOUR CELL PHONE IS OFF. Problem Value 1 10 2 10 3 10 4
More informationMATH 3030, Abstract Algebra FALL 2012 Toby Kenney Midyear Examination Friday 7th December: 7:0010:00 PM
MATH 3030, Abstract Algebra FALL 2012 Toby Kenney Midyear Examination Friday 7th December: 7:0010:00 PM Basic Questions 1. Compute the factor group Z 3 Z 9 / (1, 6). The subgroup generated by (1, 6) is
More informationMath 547, Exam 2 Information.
Math 547, Exam 2 Information. 3/19/10, LC 303B, 10:1011:00. Exam 2 will be based on: Homework and textbook sections covered by lectures 2/33/5. (see http://www.math.sc.edu/ boylan/sccourses/547sp10/547.html)
More informationHomework 6 Solution. Math 113 Summer 2016.
Homework 6 Solution. Math 113 Summer 2016. 1. For each of the following ideals, say whether they are prime, maximal (hence also prime), or neither (a) (x 4 + 2x 2 + 1) C[x] (b) (x 5 + 24x 3 54x 2 + 6x
More informationBackground Material in Algebra and Number Theory. Groups
PRELIMINARY READING FOR ALGEBRAIC NUMBER THEORY. HT 2016/17. Section 0. Background Material in Algebra and Number Theory The following gives a summary of the main ideas you need to know as prerequisites
More informationSome practice problems for midterm 2
Some practice problems for midterm 2 Kiumars Kaveh November 14, 2011 Problem: Let Z = {a G ax = xa, x G} be the center of a group G. Prove that Z is a normal subgroup of G. Solution: First we prove Z is
More informationALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ.
ALGEBRAIC GEOMETRY COURSE NOTES, LECTURE 2: HILBERT S NULLSTELLENSATZ. ANDREW SALCH 1. Hilbert s Nullstellensatz. The last lecture left off with the claim that, if J k[x 1,..., x n ] is an ideal, then
More informationMath 2070BC Term 2 Weeks 1 13 Lecture Notes
Math 2070BC 2017 18 Term 2 Weeks 1 13 Lecture Notes Keywords: group operation multiplication associative identity element inverse commutative abelian group Special Linear Group order infinite order cyclic
More informationMODEL ANSWERS TO HWK #10
MODEL ANSWERS TO HWK #10 1. (i) As x + 4 has degree one, either it divides x 3 6x + 7 or these two polynomials are coprime. But if x + 4 divides x 3 6x + 7 then x = 4 is a root of x 3 6x + 7, which it
More information(Rgs) Rings Math 683L (Summer 2003)
(Rgs) Rings Math 683L (Summer 2003) We will first summarise the general results that we will need from the theory of rings. A unital ring, R, is a set equipped with two binary operations + and such that
More informationGroups, Rings, and Finite Fields. Andreas Klappenecker. September 12, 2002
Background on Groups, Rings, and Finite Fields Andreas Klappenecker September 12, 2002 A thorough understanding of the Agrawal, Kayal, and Saxena primality test requires some tools from algebra and elementary
More informationMATH 403 MIDTERM ANSWERS WINTER 2007
MAH 403 MIDERM ANSWERS WINER 2007 COMMON ERRORS (1) A subset S of a ring R is a subring provided that x±y and xy belong to S whenever x and y do. A lot of people only said that x + y and xy must belong
More informationPh.D. Qualifying Examination in Algebra Department of Mathematics University of Louisville January 2018
Ph.D. Qualifying Examination in Algebra Department of Mathematics University of Louisville January 2018 Do 6 problems with at least 2 in each section. Group theory problems: (1) Suppose G is a group. The
More informationa b (mod m) : m b a with a,b,c,d real and ad bc 0 forms a group, again under the composition as operation.
Homework for UTK M351 Algebra I Fall 2013, Jochen Denzler, MWF 10:10 11:00 Each part separately graded on a [0/1/2] scale. Problem 1: Recalling the field axioms from class, prove for any field F (i.e.,
More informationAugust 2015 Qualifying Examination Solutions
August 2015 Qualifying Examination Solutions If you have any difficulty with the wording of the following problems please contact the supervisor immediately. All persons responsible for these problems,
More informationDefinitions. Notations. Injective, Surjective and Bijective. Divides. Cartesian Product. Relations. Equivalence Relations
Page 1 Definitions Tuesday, May 8, 2018 12:23 AM Notations " " means "equals, by definition" the set of all real numbers the set of integers Denote a function from a set to a set by Denote the image of
More informationHandout  Algebra Review
Algebraic Geometry Instructor: Mohamed Omar Handout  Algebra Review Sept 9 Math 176 Today will be a thorough review of the algebra prerequisites we will need throughout this course. Get through as much
More information1 Rings 1 RINGS 1. Theorem 1.1 (Substitution Principle). Let ϕ : R R be a ring homomorphism
1 RINGS 1 1 Rings Theorem 1.1 (Substitution Principle). Let ϕ : R R be a ring homomorphism (a) Given an element α R there is a unique homomorphism Φ : R[x] R which agrees with the map ϕ on constant polynomials
More information2 (17) Find nontrivial left and right ideals of the ring of 22 matrices over R. Show that there are no nontrivial two sided ideals. (18) State and pr
MATHEMATICS Introduction to Modern Algebra II Review. (1) Give an example of a noncommutative ring; a ring without unit; a division ring which is not a eld and a ring which is not a domain. (2) Show that
More information1 First Theme: Sums of Squares
I will try to organize the work of this semester around several classical questions. The first is, When is a prime p the sum of two squares? The question was raised by Fermat who gave the correct answer
More informationHonors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35
Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35 1. Let R be a commutative ring with 1 0. (a) Prove that the nilradical of R is equal to the intersection of the prime
More informationChapter 3. Rings. The basic commutative rings in mathematics are the integers Z, the. Examples
Chapter 3 Rings Rings are additive abelian groups with a second operation called multiplication. The connection between the two operations is provided by the distributive law. Assuming the results of Chapter
More information6]. (10) (i) Determine the units in the rings Z[i] and Z[ 10]. If n is a squarefree
Quadratic extensions Definition: Let R, S be commutative rings, R S. An extension of rings R S is said to be quadratic there is α S \R and monic polynomial f(x) R[x] of degree such that f(α) = 0 and S
More informationRings and Fields Theorems
Rings and Fields Theorems Rajesh Kumar PMATH 334 Intro to Rings and Fields Fall 2009 October 25, 2009 12 Rings and Fields 12.1 Definition Groups and Abelian Groups Let R be a nonempty set. Let + and (multiplication)
More informationALGEBRA PH.D. QUALIFYING EXAM September 27, 2008
ALGEBRA PH.D. QUALIFYING EXAM September 27, 2008 A passing paper consists of four problems solved completely plus significant progress on two other problems; moreover, the set of problems solved completely
More informationMath 4400, Spring 08, Sample problems Final Exam.
Math 4400, Spring 08, Sample problems Final Exam. 1. Groups (1) (a) Let a be an element of a group G. Define the notions of exponent of a and period of a. (b) Suppose a has a finite period. Prove that
More informationM3P14 LECTURE NOTES 8: QUADRATIC RINGS AND EUCLIDEAN DOMAINS
M3P14 LECTURE NOTES 8: QUADRATIC RINGS AND EUCLIDEAN DOMAINS 1. The Gaussian Integers Definition 1.1. The ring of Gaussian integers, denoted Z[i], is the subring of C consisting of all complex numbers
More informationSUMMARY ALGEBRA I LOUISPHILIPPE THIBAULT
SUMMARY ALGEBRA I LOUISPHILIPPE THIBAULT Contents 1. Group Theory 1 1.1. Basic Notions 1 1.2. Isomorphism Theorems 2 1.3. Jordan Holder Theorem 2 1.4. Symmetric Group 3 1.5. Group action on Sets 3 1.6.
More informationFactorization in Polynomial Rings
Factorization in Polynomial Rings Throughout these notes, F denotes a field. 1 Long division with remainder We begin with some basic definitions. Definition 1.1. Let f, g F [x]. We say that f divides g,
More informationCourse 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra
Course 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra D. R. Wilkins Contents 3 Topics in Commutative Algebra 2 3.1 Rings and Fields......................... 2 3.2 Ideals...............................
More informationRecall, R is an integral domain provided: R is a commutative ring If ab = 0 in R, then either a = 0 or b = 0.
Recall, R is an integral domain provided: R is a commutative ring If ab = 0 in R, then either a = 0 or b = 0. Examples: Z Q, R Polynomials over Z, Q, R, C The Gaussian Integers: Z[i] := {a + bi : a, b
More informationZachary Scherr Math 503 HW 3 Due Friday, Feb 12
Zachary Scherr Math 503 HW 3 Due Friay, Feb 1 1 Reaing 1. Rea sections 7.5, 7.6, 8.1 of Dummit an Foote Problems 1. DF 7.5. Solution: This problem is trivial knowing how to work with universal properties.
More informationLecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman
Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman October 17, 2006 TALK SLOWLY AND WRITE NEATLY!! 1 0.1 Integral Domains and Fraction Fields 0.1.1 Theorems Now what we are going
More informationMATH 113 FINAL EXAM December 14, 2012
p.1 MATH 113 FINAL EXAM December 14, 2012 This exam has 9 problems on 18 pages, including this cover sheet. The only thing you may have out during the exam is one or more writing utensils. You have 180
More informationTROPICAL SCHEME THEORY
TROPICAL SCHEME THEORY 5. Commutative algebra over idempotent semirings II Quotients of semirings When we work with rings, a quotient object is specified by an ideal. When dealing with semirings (and lattices),
More informationSupplement. Dr. Bob s Modern Algebra Glossary Based on Fraleigh s A First Course on Abstract Algebra, 7th Edition, Sections 0 through IV.
Glossary 1 Supplement. Dr. Bob s Modern Algebra Glossary Based on Fraleigh s A First Course on Abstract Algebra, 7th Edition, Sections 0 through IV.23 Abelian Group. A group G, (or just G for short) is
More informationMath 121 Homework 5: Notes on Selected Problems
Math 121 Homework 5: Notes on Selected Problems 12.1.2. Let M be a module over the integral domain R. (a) Assume that M has rank n and that x 1,..., x n is any maximal set of linearly independent elements
More information1. Factorization Divisibility in Z.
8 J. E. CREMONA 1.1. Divisibility in Z. 1. Factorization Definition 1.1.1. Let a, b Z. Then we say that a divides b and write a b if b = ac for some c Z: a b c Z : b = ac. Alternatively, we may say that
More informationLecture 7: Polynomial rings
Lecture 7: Polynomial rings Rajat Mittal IIT Kanpur You have seen polynomials many a times till now. The purpose of this lecture is to give a formal treatment to constructing polynomials and the rules
More informationChapter 4. Remember: F will always stand for a field.
Chapter 4 Remember: F will always stand for a field. 4.1 10. Take f(x) = x F [x]. Could there be a polynomial g(x) F [x] such that f(x)g(x) = 1 F? Could f(x) be a unit? 19. Compare with Problem #21(c).
More informationHomework 9 Solutions to Selected Problems
Homework 9 Solutions to Selected Problems June 11, 2012 1 Chapter 17, Problem 12 Since x 2 + x + 4 has degree 2 and Z 11 is a eld, we may use Theorem 17.1 and show that f(x) is irreducible because it has
More informationRings. Chapter Homomorphisms and ideals
Chapter 2 Rings This chapter should be at least in part a review of stuff you ve seen before. Roughly it is covered in Rotman chapter 3 and sections 6.1 and 6.2. You should *know* well all the material
More informationYale University Department of Mathematics Math 350 Introduction to Abstract Algebra Fall Midterm Exam Review Solutions
Yale University Department of Mathematics Math 350 Introduction to Abstract Algebra Fall 2015 Midterm Exam Review Solutions Practice exam questions: 1. Let V 1 R 2 be the subset of all vectors whose slope
More information(3) Let Y be a totally bounded subset of a metric space X. Then the closure Y of Y
() Consider A = { q Q : q 2 2} as a subset of the metric space (Q, d), where d(x, y) = x y. Then A is A) closed but not open in Q B) open but not closed in Q C) neither open nor closed in Q D) both open
More informationFACTORIZATION OF IDEALS
FACTORIZATION OF IDEALS 1. General strategy Recall the statement of unique factorization of ideals in Dedekind domains: Theorem 1.1. Let A be a Dedekind domain and I a nonzero ideal of A. Then there are
More informationSection III.6. Factorization in Polynomial Rings
III.6. Factorization in Polynomial Rings 1 Section III.6. Factorization in Polynomial Rings Note. We push several of the results in Section III.3 (such as divisibility, irreducibility, and unique factorization)
More information15. Polynomial rings DefinitionLemma Let R be a ring and let x be an indeterminate.
15. Polynomial rings DefinitionLemma 15.1. Let R be a ring and let x be an indeterminate. The polynomial ring R[x] is defined to be the set of all formal sums a n x n + a n 1 x n +... a 1 x + a 0 = a
More information50 Algebraic Extensions
50 Algebraic Extensions Let E/K be a field extension and let a E be algebraic over K. Then there is a nonzero polynomial f in K[x] such that f(a) = 0. Hence the subset A = {f K[x]: f(a) = 0} of K[x] does
More information12 16 = (12)(16) = 0.
Homework Assignment 5 Homework 5. Due day: 11/6/06 (5A) Do each of the following. (i) Compute the multiplication: (12)(16) in Z 24. (ii) Determine the set of units in Z 5. Can we extend our conclusion
More informationMATH 361: NUMBER THEORY TENTH LECTURE
MATH 361: NUMBER THEORY TENTH LECTURE The subject of this lecture is finite fields. 1. Root Fields Let k be any field, and let f(x) k[x] be irreducible and have positive degree. We want to construct a
More informationGEOMETRIC CONSTRUCTIONS AND ALGEBRAIC FIELD EXTENSIONS
GEOMETRIC CONSTRUCTIONS AND ALGEBRAIC FIELD EXTENSIONS JENNY WANG Abstract. In this paper, we study field extensions obtained by polynomial rings and maximal ideals in order to determine whether solutions
More information