Questions and Answers #1 Abstract Algebra

Questions and Answers Sheet 1 Abstract Algebra Question #1 Let F be a field. Suppose that a polynomial p(x) = a 0 + a1 x + ⋯ + a n x n is reducible in 𝐹[𝑥] . Prove that 1 the polynomial 𝑞(𝑥) == 𝑥 𝑛 𝑝 ( ) = 𝑎𝑛 + 𝑎𝑛−1 𝑥 + ⋯ 𝑎0 𝑥 𝑛 is also reducible. 𝑥 Answer: Step 1 Definition: Let (𝐹, +,⋅) be a field and let 𝑓 ∈ 𝐹[𝑥] . Then f is said to be Irreducible over F if f cannot be factored into a product of polynomials all of which having lower degree than f. If f is not irreducible over F then we say that f is Reducible over F. Step 2 We will prove with the help of principal of mathematical equation. For 𝑛 = 1 𝑃(𝑥) = 𝑎0 + 𝑎1 𝑥 is reducible in field ⇒ 𝑎0 𝑎1 ∈ 𝐹(𝑥) and 𝑞(𝑥) = 𝑥 (𝑎0 + 𝑎1 𝑥 ) = 𝑎0 𝑥 + 𝑞1 is also reducible in field as 𝑎0 𝑎1 ∈ 𝐹(𝑥) Now let for n=k is true 𝑃(𝑥) = 𝑎0 + 𝑎1 𝑥 + ⋯ 𝑎𝑘 𝑥 𝑘 is reducible in fidd So any polynomial of degree dess than k with 𝑎0 𝑎1 𝑎𝑘 ∈ 𝐹(𝑥) is reducible ⇒ 𝑞(𝑥) = 𝑎0 𝑥 𝑘 + 𝑎1 𝑥 𝑘−1 ⋯ + 𝑎𝑘 is reducible in field Now for 𝑛 = 𝑘 + 1 𝑃(𝑥) = 𝑎0 + 𝑎1 𝑥 + ⋯ 𝑎𝑘 𝑥 𝑘 + 𝑎𝑘+1 𝑥 𝑘+1 in reducible ⇒ 𝑎0 ⋅ 𝑎𝑘+1 ∈ 𝑓(𝑥) with polynomial of degree less or equal to k is reducible Step 3 Now 𝑘 = 1 𝑞(𝑥) = 𝑎𝑘+1 + 𝑞𝑘 𝑥 + ⋯ 𝑞0 𝑥 𝑘+1 𝑞(𝑥) = 𝑎𝑘+1 + 𝑥(𝑞𝑘 + ⋯ 𝑎0 𝑥 𝑘 ) Since 𝑎𝑘 + ⋯ 𝑎0 𝑥 𝑘 is reducible in f(x) and 𝑘 + 1 degree polinomial with 𝑞𝑘+1 ∈ 𝑓(𝑥) can also reducible ⇒ 𝑞(𝑥) is reducible. Question #2 Prove 𝑥 6 − 6𝑥 4 + 12𝑥 2 − 11 is irreducible over \mathbb{Q} Answer: Step 1 Let 𝑝(𝑥) = 𝑥 6 − 6𝑥 4 + 12𝑥 2 − 11 , whith we can transform into a polynomial in 𝑍𝟛 [𝑥]: 𝑥6 + 1 Since none of the three elements 0,1,2 in 𝑍𝟛 is a root of the polynomial, the polynomial has no factor of degree 1 in 𝑍𝟛 [𝑥] . So the only possible factorings into non constant polynomials are 𝑥 6 + 1 = (𝑥 3 + 𝑎𝑥 2 + 𝑏𝑥 + 𝑐)(𝑥 3 + 𝑑𝑥 2 + 𝑒𝑥 + 𝑓) or 𝑥 6 + 1 = (𝑥 4 + 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑)(𝑥 2 + 𝑒𝑥 + 𝑓) From the first equation, since corresponding coefficients are equal, we have 1) 𝑥 0 : 𝑐𝑓 = 1 2) 𝑥 1 : 𝑏𝑓 + 𝑐𝑒 = 0 3) 𝑥 2 : 𝑎𝑓 + 𝑏𝑒 + 𝑐𝑑 = 0 4) 𝑥 3 : 𝑐 + 𝑓 + 𝑏𝑑 + 𝑎𝑒 = 0 5) 𝑥 5 : 𝑎 + 𝑑 = 0 From (1), 𝑥 = 𝑓 = ±1 , and from (5), 𝑎 + 𝑑 = 0 . Consequently, 𝑎𝑓 + 𝑐𝑑 = 𝑐(𝑎 + 𝑑) = 0 , and by (3), 𝑒𝑏 = 0 . But from (2) (since 𝑐 = 𝑓 ), 𝑏 + 𝑒 = 0 , and therefore 𝑏 = 𝑒 = 0 . It follows from (4) that 𝑐 + 𝑓 = 0 which is impossible since c=f=\pm1 . We have just shown that 𝑥 6 + 1 cannot be factored into two polynomials each of degree 3. For the second equation, however, 𝑥 6 + 1 = (𝑥 2 + 1)3 in 𝑍𝟛 [𝑥] . So we cnnot say 𝑝(𝑥) is irreducible over 𝑄 because 𝑥 6 + 1 is irreducible over 𝑍𝟛 . Question #3 We're trying to understand the field 3 2 𝑍𝟚 [𝑥] ≪𝑥 3 +𝑥 2 +1≫ , and in particular why it has eight elements. I know that 𝑥 + 𝑥 + 1 is irreducible in 𝑍𝟚 and so factor is going to be a field. The issue I'm having is that the notes claim that (𝑥 + 𝐼) has an inverse in the field, 𝑤ℎ𝑒𝑟𝑒𝐼 =≪ 𝑥 3 + 𝑥 2 + 1 ≫ . They write (𝑥 + 𝐼)(𝑥 2 + 𝑥 + 𝐼) = 𝑥 3 + 𝑥 2 + 𝐼 = 1 + 𝐼 I don't understand the last line, I know that multiplication of cosets is defined that way, I just don't see why we get 1 at the end. I think it's possible I didn't fully understand cosets and ideals but I can't fully articulate what the issue is. Any help towards filling in the gaps of my understanding would be greatly appreciated. Answer: Step 1 The might know that 𝑎 + 𝐼 = 𝑏 + 𝐼 ⇔ 𝑎 − 𝑏 ∈ 𝐼𝑍𝐾 So (𝑥 3 + 𝑥 2 ) − 1 = (𝑥 3 + 𝑥 2 + 1) ∈ 𝐼 So (𝑥 3 + 𝑥 2 ) + 𝐼 = 1 + 𝐼 also any element in 𝑍𝟚 [𝑥] (𝑥 3 +𝑥 2 +1) is of the form (𝑎𝑥 2 + 𝑏𝑥 + 𝑐) + 𝐼 . There are 2 choices for each a, b, c. So there are 8 distinct elements in 𝑍𝟚 [𝑥] by a cubic polynomial, you get a remainder which is a quadratic polynomial of the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 So if 𝑝(𝑥) ∈ 𝑍𝟚 [𝑥] then 𝑝(𝑥) = 𝑞(𝑥)(𝑥 3 + 𝑥 2 + 1) + 𝑟(𝑥) where 𝑟(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 . For some 𝑎, 𝑏, 𝑐 ∈ 𝑍𝟚 So 𝑝(𝑥) + 𝐼 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 + 𝐼 as 𝑞(𝑥)(𝑥 3 + 𝑥 2 + 1) ∈≪ 𝑥 3 + 𝑥 2 + 1 ≫ Question #4 Let V be the variety of algebras (𝐴,⋅) satisfying the identitie 𝑥 ⋅ 𝑥 ≈ 𝑥 and (𝑥 ⋅ 𝑦) ⋅ 𝑧 ≈ (𝑧 ⋅ 𝑦) ⋅ 𝑥 Let W be the subvariety of V defined by the additional identity 𝑦 ⋅ (𝑥 ⋅ 𝑦) Determine 𝐹𝑊 (𝑥, 𝑦) . Write out a Cayley table. Answer: Step 1 Bergman asks to derive a number of identities from the defining identities for W , and these identities help to show that the universe of 𝐹𝑊 (𝑥, 𝑦) is {𝑥, 𝑦, (𝑥 ⋅ 𝑦), (𝑦 ⋅ 𝑥)} . More precisely, Bergman's identities help to show that every word in x and y reduces to one of 𝑥, 𝑦, (𝑥 ⋅ 𝑦) or (𝑦 ⋅ 𝑥) To show that no further reductions are possible, it suffices to observe that if 𝐹𝟜 = 𝐹𝟚 [α] for α satisfying 𝑥 2 + 𝑥 + 1 = 0 , then the operation 𝑥 ⋅ 𝑦 ≔ α𝑥 + (1 − α)𝑦 acting on the set 𝐹𝟜 satisfies all the defining identities of W , and the four words 𝑥, 𝑦, 𝑥 ⋅ 𝑦 , and 𝑦 ⋅ 𝑥 have distinct interpretations on 𝐹𝟜 . To show that every word in x and y reduces to some word in the set {𝑥, 𝑦, (𝑥 ⋅ 𝑦), (𝑦 ⋅ 𝑥)} one must show that the product of any two words in this set reduces to a word in this set. For example, the product of x and (𝑥 ⋅ 𝑦) (I mean the product 𝑥 ⋅ (𝑥 ⋅ 𝑦)) reduces to (𝑦 ⋅ 𝑥) . To see this, substitute 𝑌 = 𝑥, 𝑋 = (𝑦 ⋅ 𝑥) in the defining identity 𝑌 ⋅ (𝑋 ⋅ 𝑌) = 𝑋 to obtain E) 𝑥 ⋅ ((▁{𝑦} ⋅ 𝑥) ⋅ ▁{𝑥} = (𝑦 ⋅ 𝑥) Now swap the underlined characters using the defining identity (𝑋 ⋅ 𝑌) ⋅ 𝑍 = (𝑍 ⋅ 𝑌) ⋅ 𝑋 to obtain from (E) that 𝑥 ⋅ ((𝑥 ⋅ 𝑥) ⋅ 𝑦) = (𝑦 ⋅ 𝑥) holds. Now apply the defining identity 𝑋⋅𝑋 =𝑋 to reduce this to 𝑥 ⋅ (𝑥 ⋅ 𝑦) = (𝑦 ⋅ 𝑥) This verifies the first sentence of this paragraph. Question #5 Is there an algorithm for working out the best way (i.e. fewest multiplications) of calculating 𝐴𝑛 in a structure where multiplication is associative? Answer: Step 1 All you can think of are minimum and maximum values, and you can get stuck doing this by eye. You can check that example 510 can be done in 11 multiplications. 1) 𝑥 × 𝑥 = 𝑥 2 2) 𝑥 2 × 𝑥 = 𝑥 3 3) 𝑥 3 × 𝑥 3 = 𝑥 6 4) 𝑥 6 × 𝑥 6 = 𝑥 12 5) 𝑥 12 × 𝑥 3 = 𝑥 15 6) 𝑥 15 × 𝑥 15 = 𝑥 30 7) 𝑥 30 × 𝑥 30 = 𝑥 60 8) 𝑥 60 × 𝑥 60 = 𝑥 120 9) 𝑥 120 × 𝑥 120 = 𝑥 240 10) 𝑥 240 × 𝑥 15 = 𝑥 255 11) 𝑥 255 × 𝑥 255 = 𝑥 510 We can guarantee that this is the minimum. Unfortunately, this was a simple example and we don't know how to generalize it into a program. With this speed, you can solve half of the problem manually ... Not sure how much to add because this is a project of Euler's problem, but if you express these metrics in binary you can see this strategy, why we thought the 510 was a simple example, and why we were so sure it couldn't be rolled up to fewer multiplications. And it might be possible to spot some cases in which you cannot achieve more than a binary method. Question #6 Can you find a domain where 𝑎𝑥 + 𝑏𝑦 = 1 has a solution for all a and b relatively ', but which is not a PID? Answer: Step 1 Define a domain 𝑅0 as follows. Take a field K, adjoin an indeterminate 𝑥0 , and localize at (𝑥0 ) (that is, adjoin inverses to everything not a multiple of 𝑥0 ). 𝑅0 has all its ideals principal and linearly ordered: (𝑥0 ) contains (𝑥02 ) contains (𝑥03 ) Now given 𝑅𝑖 define 𝑅𝑖+1 inductively: Adjoin an indeterminate 𝑥𝑖+1 , so we have 𝑅𝑖 [𝑥𝑖+1 ] . Quotient by 2 (𝑥𝑖+1 − 𝑥𝑖 ) . Finally, localize at the ' ideal (𝑥𝑖+1 ) This effectively just gives us one more principal ideal containing all the principal ideals from 𝑅𝑖 : (𝑥𝑖+1 ) 2 ) = (𝑥𝑖 ) contains (𝑥𝑖2 ) contains (𝑥𝑖+1 Now let R be the union of all the 𝑅𝑖 , and it's obvious that any finitely generated ideal is principal, but there's a non-fg one generated by all the 𝑥𝑖 . Question #7 Are 𝑥 ⋅ 0 = 0, 𝑥 ⋅ 1 = 𝑥, and −(−𝑥) = 𝑥 axioms? Answer: Step 1 The question is more profound than is initially seems, and is really about algebraic structures. The first question you have to ask yourself is where you're working: In general, addition and multiplication are defined on a structure, which in this case is a set (basically a collection of ''things'') with two operators we call addition (marked + ) and multiplication (marked ⋅ or × or or whatever). If this structure holds some properties, which are sometimes called axioms, then it is called a unit ring. The properties are: 1) The set is closed under the operator + . That is, if a and b are in R, then 𝑎 + 𝑏 is also in R. 2) The set has a member which we mark as 0. It has the properties that for every a in R, 𝑎 + 0 = 0 and 0+𝑎 =0 . 3) The operation + is commutative: 𝑎 + 𝑏 = 𝑏 + 𝑎 . 4) The operation + is associative: (𝑎 + 𝑏) + 𝑐 = 𝑎 + (𝑏 + 𝑐). 5) Every member has an additive inverse: for every a in R there is some b in R such that 𝑎 + 𝑏 = 0 (we mark b as -a). 6) The set is closed under the operator × . That is, if a and b are in R, then 𝑎 × 𝑏 is also in R. 7) The set has a member which we mark as 1. It has the properties that for every a in R, 𝑎 × 1 = 𝑎 and 1×𝑎 =𝑎. 8) The operation \times is associative: (𝑎 × 𝑏) × 𝑐 = 𝑎 × (𝑏 × 𝑐) . 9) Multiplication is distributive over addition: 𝑎 × (𝑏 + 𝑐) = 𝑎 × 𝑏 + 𝑎 × 𝑐 and (𝑎 + 𝑏) × 𝑐 = 𝑎 × 𝑏 + 𝑎 × 𝑐 . Step 2 While this is a long list, and introduces the operator + which is not even explicitly mentioned in the question, these properties are quite natural. For example, the integers {⋯ , −2, −1, 0, 1, 2, ⋯ } we all know and love indeed form a ring. The real numbers also form a ring (in fact they form a field, which means they hold even more properties). In regard to your question, the identity 𝑥 × 1 = 𝑥 (I assume that's what you meant) is in fact an axiom - it is axiom 7. However, the other two identities are results of the other axioms. Step 3 First identity: We use axioms 2 and 9 to get 0 × 𝑥 = (0 + 0) × 𝑥 = 0 × 𝑥 + 0 × 𝑥 and then by adding −(0 × 𝑥) (the additive inverse of 0 × 𝑥 , from axiom 5) to both sides, 0=0×𝑥 . Step 4 Second identity: As stated in axiom 5, −(−𝑥) is just a notation used which means ''the additive inverse of -x''. To show that −(−𝑥) = 𝑥 we need to show that x is in fact the additive inverse of -x, or in other words that 𝑥±𝑥 = 0 and −𝑥 + 𝑥 = 0 . But that's just what axiom 5 says, so we're done. Step 5 Last point: You might be wondering why did we have to go and introduce addition to answer a question about multiplication? Well, it so happens that without addition the other two identities are simply not true. For example, if we look at the positive integers { 1, 2, 3, ⋯ } with only multiplication, then there is no 0 there! Simply put, this is because the positive integers do not form a ring. Question #8 When does the produt of two polynomials = 𝑥 𝑘 ? Suppose f and g are are two polynomials with complex coefficents (i.e 𝑓, 𝑔 ∈ 𝐶[𝑥] ). Let m be the order of f and let n be the order of g. Are there some general conditions where 𝑓𝑔 = α𝑥 𝑛+𝑚 for some non-zero α ∈ 𝐶 Answer: Step 1 We don't need the strong property of UFD. If D is a domain D then x is ' in 𝐷[𝑥] (by 𝐷[𝑥] 𝑥 ≅ 𝐷 a domain), and products of 's factor uniquely in every domain (same simple proof as in 𝑍 ). In particular, the only factorizations of the ' power 𝑥 𝑖 are 𝑥 𝑗 𝑥 𝑘 , 𝑖 = 𝑗 + 𝑘 (up to associates as usual). This fails over non-domains, e.g. 𝑍 𝑥 = (2𝑥 + 3)(3𝑥 + 2) ∈ 6[𝑥] Question #9 Finding the Units in the Ring 𝑍[𝑡][√𝑡 2 − 1] Answer: Step 1 Given: 𝑍[𝑡][√𝑡 2 − 1] Write: 𝑥 = 𝑡 + √𝑡 2 − 1 and 𝑦 = 𝑡 − √𝑡 2 − 1 Then 𝑥𝑦 = 1, i.e.𝑦 = 𝑥 −1 , and 𝑡 = √𝑡 2 −1= (𝑥−𝑦) 2 Thus we see that (𝑥+𝑦) 2 1 𝑍[𝑡] [√𝑡 2 − 1] ⊂ 𝑍 [ ] [𝑥, 𝑥 −1 ] 2 Now it is easy to check that if A is an integral domain, then the only units in 𝐴[𝑥, 𝑥 −1 ] are of the form 𝑎𝑥 𝑛 , where 𝑎∈𝐴 1 is a unit and n is an integer. Since the units in 𝑍 [ ] are precisely the elements ±2𝑚 (for some m) we 2 see that the units in 1 𝑍 [ ] [𝑥, 𝑥 −1 ] 2 are of the form 𝑛 ±2𝑚 𝑥 𝑛 = ±2𝑚 (𝑡 + √𝑡 2 − 1) It is easy to check that if such an element and its inverse both actually lie in 𝑍[𝑡][√𝑡 2 − 1] , then necessarily 𝑚 = 0 (e.g. for norm reasons, or just looking explicitly at their denominators), and so we get the answer that the units are precisely the elements of the form ± (𝑡 + √𝑡 2 − 1) 𝑛 𝑛 (Here n runs over 𝑍 ; this is the same set as ±(𝑡 ± √𝑡 2 − 1) , where n is now non-negative.) Question #10 Why does 𝐾 → 𝐾(𝑋) preserve the degree of field extensions? Answer: Step 1 𝐾 Let be a field extension. 𝑘 Then 𝐾 ⊗𝑘 𝑘(𝑋) is a localization of the integral domain 𝐾 ⊗𝑘 𝑘[𝑋] = 𝐾[𝑋] thus the natural map 𝐾 ⊗𝑘 𝑘(𝑋) → 𝐾(𝑋) is injective. Now we have the following equivalence: 𝐾 a) is algebraic 𝑘 b) Every polynomial over K divides some polynomial over k. c) 𝐾 ⊗𝑘 𝑘(𝑋) = 𝐾(𝑋) 𝑎 ⇒ 𝑏 : Because of uniqueness of the division algorithm, it suffices to prove divisibilty in an algebraic extension. So take a splitting field of the polynomial and reduce to the case of linear factors: 𝑥 − 𝑎 . But they divide the minimal polynomial over k. 𝑏 ⇒ 𝑐: 𝐾 ⊗𝑘 𝑘(𝑋) is the localization of 𝐾[𝑋] at all nontrivial polynomials over k. Since we have b, we localize at all nontrivial polynomials over K, i.e. we get K(X) . 𝑐 ⇒ 𝑎: Take 𝑢 ∈ 𝐾 . 1 𝑝 Then (𝑋−𝑢) = 𝑞 for 𝑝 ∈ 𝐾[𝑋] 𝑞 ∈ 𝑘[𝑋] i.e. p(X-u) is a polynomial over k with root u. QED 𝐾 In particular, if is algebraic, we have [𝐾(𝑋): 𝑘(𝑋)] = [𝐾: 𝑘] 𝑘 Question #11 Prove that {𝑎 + 𝑏𝑖 ∶ 𝑎, 𝑏 ∈ 𝑅} is a subring of the real Hamilton Quaternions H, which is a field but it is not contained in the center of H. Answer: Step 1 Note: → 𝐻 = {𝑎0 + 𝑎1 𝑖̂ + 𝑎𝑖 𝑗̂ + 𝑎3 𝑘̂ ∣ 𝑎𝑖 ∈ 𝑅} where 𝑖 2 = 𝑗 2 = 𝑘 2 = −1 , Called Quaternions ring Now, 𝑍(𝐻) = {Set of those element of H which are Conmuty with every elements of H} 𝑍(𝐻) = {𝑎_{0}|𝑎0 ∈ 𝑅} = 𝑅 Now, let 𝑆 = {𝑎 + 𝑏𝑖, 𝑎, 𝑏 ∈ 𝑅} S\subseteq H Clearly Now ∀𝑥, 𝑦 ∈ 𝑆, 𝑥 = 𝑎 + 𝑏𝑖 𝑦 − 𝑥 + 𝑑𝑖 𝑥 − 𝑦 = 𝑎 − 𝑐 + (𝑏 − 𝑎)𝑖 ∈ 𝑆 and 𝑥𝑦 = (𝑎 + 𝑏𝑖) × (𝑐 + 𝑑𝑖) = (𝑎𝑐 − 𝑏𝑑) + (𝑎𝑑 + 𝑏𝑐)𝑖 ∈ 𝑆 ⇒ 𝑆 𝑖𝑠 𝑆𝑢𝑏𝑟𝑖𝑛𝑔 𝑜𝑓 𝐻 → (𝑎 + 𝑏𝑖) × (1 + 0 × 𝑖) = (𝑎 + 𝑏𝑖) ⇒ 1 is unity 𝑎 𝑏𝑖 → (𝑎 + 𝑏𝑖)−1 = 2 2 − 2 2 for any element 𝑎 +𝑏 𝑎 +𝑏 → 𝑆 is Commotative → 𝑆 is field any S = {a + bi | a, b  }    Z ( H ) = Question #12 Which of the following is not correct? a) 𝑈(14) is isomorphic to 𝑆3 b) 𝑈(14) is isomorphic to 𝑈(7) c) 𝑈(8) is isomorphic to 𝑈(12) d) none of these Answer: Step 1 𝑈(14) = {1,3,5,9,11,13} (𝑆3 ) = 3! = 6 1) So, the group have the same order 2) 𝑈(14) is abelian, and 𝑆3 is non abelian. But if two group are isomorphie, they are either both abelian or both non abelian So, option A is incorrect. Question #13 Let α = (1 7 3)(5 4 2 9) and β = (2 3)(7 4)(5 1 8) be elements of 𝑆10 . Then α2 β−1 is: a) (1 2 7 9 4)(3 8 5) b) (1 2 7 4 9)(3 5 8) c) (1 2 7 9 4)(3 5 8) d) (1 2 9 7 4)(3 5 8) Answer: Step 1 1 2 3 4 5 6 7 8 9 10  7 9 7 2 4 7 3 9 5 1  1 2 3 4 5 6 7 8 9 10  =  8 3 2 7 1 7 4 5 10 1   = 1 2 3 7 9 1  1 2   3 4 5 6 7 8 9 10 2 4 7 3 9 5 1  2 3 4 5 6 7 8 9 10 5 7 9 2 3 1 5 4 7  1 2 3 4 5 6 7 8 9 10  −1    8 3 2 7 1 7 4 5 10 1  1 2 3 4 5 6 7 8 9 10  −1 =   5 2 3 4 8 4 7 1 1 8  1 2 3 4 5 6 7 8 9 10  3 5 7 9 2 3 1 5 4 7  1 2 3 4 5 6 7 8 9 10    5 2 3 4 8 4 7 1 1 8  1 2 3 4 5 7 9 8 2 −1      (1 2 7 9 4)(3 8 5)  2 7 8 1 3 9 4 5  2  −1   Question #14 Prove ker ϕ =≪ 1 + √−5, 2 ≫ Answer: Step 1 Let ϕ: 𝑍[√−5] → 𝑍𝟚 ϕ(𝑝 + 𝑞√−5) = [𝑝 + 𝑞]2 Want to prove ker ϕ =≪ 1 + √−5, 2 ≪ So what 𝑚 = 𝑝 + 𝑞√5 ∈ ker ϕ ⇔ [𝑝 + 𝑞]2 = [0]2 ⇔ 𝑝 + 𝑞 ∈ 2𝑍 Step 2 Only need to prove that every element 𝑝 + 𝑞√5 from the kernel is in ≪ 1 + √5, 2 ≫ . Since 𝑝 + 𝑞 is even, 𝑝 − 𝑞 is also even, = 2𝑛 for some integer n. Then 𝑝 + 𝑞√5 = 𝑛 × 2 + 𝑞(1 + √5) as desired. Question #15 Finding a subgroup of 𝑆 8 isomorphic to 𝑍𝟜 × 𝑍𝟜 What might be a more clever way to go about solving this? Answer: Step 1 Well, Consider 𝑆8 Take, 𝑎 = (1234), 𝑏 = (5678) We know order of a cycle is the length of the cycle. Hence, |𝑎| = |𝑏| = 4 Again we know any two disjoint cycles commute. Now consider, 𝐻 =≪ 𝑎, 𝑏: 𝑎4 = 𝑏 4 = 𝑖𝑑, 𝑎𝑏 = 𝑏𝑎 ≫ Claim: 𝐻 ≅ 𝑍𝟜 × 𝑍𝟜 Hint: Send generators to generators. Edit: H = {a, a 2 , a 3 , a 4 , b, b 2 , b3 , ab, ab 2 , ab3 , a 2b, a 2b 2 , a 3b3 , a 3b, a 3b 2 , a 3b3 ,} Question #16 Classify via fundamental theorem of finitely generated abelian groups. 𝑍×𝑍 ≪(3,1),(8,2)≫ Answer: Step 1 Let f be as defined in the question. We get   x, y | 3x + y,8 x + 2 y, x + y = y + x  ker f  x, y | y = −3x, 8 x + 2 y, x + y = y + x  1) ≅≪ 𝑥|8𝑥 − 6𝑥, 𝑥 − 3𝑥 = −3𝑥 ≫ ≅≪ 𝑥|2𝑥, −2𝑥 = −2𝑥 ≫ ≅≪ 𝑥|2𝑥 ≫ ≅ 𝑍𝟚 where (1) holds by the Tietze transformation of eliminating y as y can be written in terms of x. Question #17 Prove that: Every 𝑄 - vector space 𝑉 ≠ 0 is not free over subring 𝑍 ⊂ 𝑄 Answer: Step 1 Dietrich Burde essentially already proved the result. Here are some more details. First we show the claim for finite vector spaces. So let V be a finite-dimensional 𝑄 vector space, 𝑛 = dim _{Ⅎo{𝑄}}𝑉 . Suppose V was free as a 𝑍 -module, then 𝑄 ⊂ 𝑉 = 𝑄𝑛 is a submodule of a free 𝑍 -module hence it is itself free. But 𝑄 is not free as a 𝑍 -module, since any two rationals are linearly dependent and clearly 𝑄 is not generated by one rational as 𝑎𝑍 -module. To deal with the infinite dimensional case it should suffice to write V as the union of its finite-dimensional subspaces. Question #18 Aluffi's proof that det(𝐴𝐵) = det(𝐴) det(𝐵) for commutative rings 1) How does 𝑍 being initial in Ring, help us when we are proving an identity that holds for 𝑍[𝑥11 , ⋯ , 𝑥𝑛𝑛 , 𝑦11 , ⋯ , 𝑦𝑛𝑛 ]? 2) My second question is about the field of fractions argument. Doesn't that immediately give a proof for all integral domains without reference 𝑍[𝑥11 , ⋯ , 𝑥𝑛𝑛 , 𝑦11 , ⋯ , 𝑦𝑛𝑛 ]? Answer: Step 1 If 𝐴 = (𝑎𝑖𝑗 ) and 𝐵 = (𝑏𝑖𝑗 ) are in 𝑀𝑛 (𝑅) , where R is a commutative unitary ring, then there is a homorphism ϕ(𝑛): 𝑍[𝑥11 , ⋯ , 𝑥𝑛𝑛 , 𝑦11 , ⋯ , 𝑦𝑛𝑛 ] → 𝑅 such that ϕ(𝑛) = 𝑛 × 1𝑅 for 𝑛 ∈ 𝑍, ϕ(𝑥𝑖𝑗 ) = 𝑎𝑖𝑗 and ϕ(𝑦𝑖𝑗 ) = 𝑏𝑖𝑗 If one proves that det(𝑋𝑌) = det 𝑋 det 𝑌 for the matrices 𝑋 = (𝑥𝑖𝑗 ), respectively 𝑌 = (𝑦𝑖𝑗 ) , then apply \phi to this relation and get det(𝐴𝐵) = det 𝐴 det 𝐵 But you know that det(𝑋𝑌) = det 𝑋 det 𝑌 holds for matrices X, Y whose entries are in a field. This is why the book suggests to look at these matrices in the field of fractions of the ring 𝑍[𝑥11 , ⋯ , 𝑥𝑛𝑛 , 𝑦11 , ⋯ , 𝑦𝑛𝑛 ] Question #19 Determine Group structure of given 2 × 2 matrix group over 𝐹𝟛 Answer: Step 1 I think that concentrating on the case 𝑝 = 3 and special properties of groups of order 12 isn't really very illuminating. So let p be an odd '. Consider the group 𝐺 = {((𝑎, 𝑏), (0, 𝑑)): 𝑎, 𝑑 ∈ 𝐹𝑝⋆ , 𝑏 ∈ 𝐹𝑝 } which is clearly of order 𝑝(𝑝 − 1)2 Then the centre of the whole matrix group 𝑍 = {((𝑎, 𝑏), (0, 𝑑)): 𝑎 ∈ 𝐹𝑝⋆ } is a cyclic subgroup of G of order (𝑝 − 1) ; and so it is a normal subgroup lying in the centre of G. Moreover G has a subgroup 𝐻 = {((𝑎, 𝑏), (0, 𝑑)): 𝑎 ∈ 𝐹𝑝⋆ , 𝑏 ∈ 𝐹𝑝 } of order 𝑝(𝑝 − 1) which intersects Z trivially, and is normalised by the central Z, and so by all of G. Hence 𝐺 = 𝑍 × 𝐻 . The group H is known as the affine group: it is clearly isomorphic to the set of mappings 𝐹𝑝 → 𝐹𝑝 of the form {𝑥 ↦ 𝑎𝑥 + 𝑏|𝑎, 𝑏 ∈ Ⅎo{𝐹}_{𝑝}, 𝑎 ≠ 0} In the case 𝑝 = 3 we clearly have 𝐺 = 𝐶2 × 𝑆3 Question #20 Let H be a subgroup of G and 𝑥, 𝑦 ∈ 𝐺 . Show that 𝑥(𝐻𝑦) = (𝑥𝐻)𝑦 Answer: Step 1 This can be done in a few lines: 𝑥(𝐻𝑦) = {𝑥ℎ′|ℎ′ ∈ 𝐻𝑦} = {𝑥(ℎ𝑦)|ℎ ∈ 𝐻} = {(𝑥ℎ)𝑦|ℎ ∈ 𝐻} = {ℎ′′𝑦|ℎ′′ ∈ 𝑥𝐻} = (𝑥𝐻)𝑦