Questions and Answers #3 Abstract Algebra

Questions and Answers Sheet 3 Abstract Algebra Question #1 Find the order 5 + 𝐻 Answer: Step 1 Here is an alternative answer, trying to bring structure in the game. We have the short exact sequence: 6𝑍 𝑍 𝑍 0→ ⏟ → ⏟ → ⏟ →0 18𝑍 𝐻 18𝑍 𝐺 6𝑍 𝐺 𝐻 The image of that complicated element 5 + 18𝑍 ∈ 𝐺 inside (5 + 18𝑍) + 𝐻 = (5 + 18𝑍) + (6𝑍 + 18𝑍) = (5 + 6𝑍) ∈ 𝐺 𝐺 𝐻 , so 𝐻 is simply 5 + 6𝑍 , an element of order six Question #2 Factor 𝑥 4 − 2 into irreducible polynomials. Answer: Step 1 You had a wrong sign in the factorization in 𝑅[𝑥] (typo): 4 4 (𝑥 2 + √2)(𝑥 − √2)(𝑥 + √2) is the factorization: the quadratic factor has no roots in 𝑅, so it's irreducible. The polynomial is indeed irreducible in 𝑄[𝑥] : it has no rational root so it would be a product of two quadratic factors, but such a decomposition would also lead to a splitting in 𝑅[𝑥] and the only way you can combine the previous factorization in the required way would be (𝑥 2 + √2)(𝑥 2 − √2) and the factors are not in 𝑄[𝑥] . Or you can use Eisenstein's criterion, which applies to 𝑥 4 − 2 for the ' 2. Sorry, but the factorization in 𝐶[𝑥] is wrong: it should be a product of linear factors and you get them from the factorization in 𝑅[𝑥] I'll do the case 𝑍𝟛 [𝑥]. The polynomial has no roots in 𝑍𝟛 , but it could be a product of two (monic) quadratic factors: 𝑥 4 − 2 = (𝑥 2 + 𝑎𝑥 + 𝑏)(𝑥 2 + 𝑐𝑥 + 𝑑) Since 𝑏𝑑 = −2 = 1 , you can have 𝑏 = 1 and 𝑑 = 1 or 𝑏 = 2 and 𝑑 = 2 . In the first case we have. (𝑥 2 + 𝑎𝑥 + 1)(𝑥 2 + 𝑐𝑥 + 1) = 𝑥 4 + (𝑎 + 𝑐)𝑥 3 +(2 + 𝑎𝑐)𝑋 2 + (𝑎 + 𝑐)𝑥 + 1 and so you need 𝑎 + 𝑐 = 0 and 𝑎𝑐 + 2 = 0 . Hence 𝑐 = −𝑎 and 𝑎2 = 2 , which is impossible. In the second case we have (𝑥 2 + 𝑎𝑥 + 2)(𝑥 2 + 𝑐𝑥 + 2) = 𝑥 4 + (𝑎 + 𝑐)𝑥 3 + (1 + 𝑎𝑐)𝑥 2 + (2𝑎 + 2𝑐)𝑥 + 1 Again 𝑎 + 𝑐 = 0 and 1 + 𝑎𝑐 = 0 . Since 𝑐 = −𝑎 , we obtain 𝑎2 = 1 . Well, this is indeed possible: we have 𝑎 = 1 or 𝑎 = 2 . Thus the factorization is (𝑥 2 + 𝑥 + 2)(𝑥 2 + 2𝑥 + 2) Question #3 Group theory exercise from Judson text 𝑆= 𝑅 −1 and define a binary operation on S by 𝑎 × 𝑏 = 𝑎 + 𝑏 + 𝑎𝑏 . Prove that (𝑆,×) is an abelian group. Answer: Step 1 Let 𝑓(𝑥) = 𝑥 + 1 and 𝑔(𝑥) = 𝑥 − 1 . Then what we have is 𝑥 × 𝑦 = (𝑥 + 1)(𝑦 + 1) − 1 = 𝑔(𝑓(𝑥)𝑓(𝑦)) Or, to write this in a way that might be easier to catch what is going on: 𝑔(𝑥) × 𝑔(𝑦) = 𝑔(𝑥𝑦) 𝑅 So S is isomorphic to ( ,×) 0 Question #4 ⋆ Find subgroup ≪ 𝑎, 𝑏 ≫ of 𝑍𝟚𝟘 which is not cyclic. Answer: Step 1 ⋆ Since you already know 𝐺 = 𝑍𝟚𝟘 is not cyclic, and every group is a subgroup of itself, you could take the ⋆ subgroup to be 𝑍𝟚𝟘 Step 2 Since 3 and 11 in 𝐺, 3 × 11 = 11 × 3 3  11 = {1,11} and 11  3 = {1,3,9,7} we have  3,11  x, y | x 2 , y 4 , xy = yx  ≅ 𝑍𝟚 × 𝑍𝟜 which is not cyclic. But |𝑍𝟚 × 𝑍𝟜 | = 8 = 𝜙(20) = |𝐺| A proper subgroup can be found similarly by considering ≪ 9,11 ≫ Question #5 Find a polynomial 𝑃𝑛 ∈ 𝐶[𝑋] such as 𝑆𝑛 = 𝑘𝑒𝑟𝑃𝑛 (𝐷) With 𝑆𝑛 (𝑛 ≥ 1) all of the functions such as 𝑦 (𝑛) = 𝑦 (where 𝑦 (𝑛) is the nth derivative of y) D the endomorphism sending the functions 𝐶 ∞ on their derivatives 𝐷: 𝑦 → 𝑦 ′ I already proved that D is an endomorphism and 𝑆𝑛 a vector space containing the functions 𝑘𝑒 𝑥 . But my probleme is that i dont really understand the meaning of 𝑃𝑛 (𝐷) , and so dont see what 𝑆𝑛 = 𝑘𝑒𝑟 𝑛 (𝐷) could be. It probably have to do with eigenvalues, eigenvectors...But really I dont see anything. Answer: Let 𝑃𝑛 = 𝑋 𝑛 − 1 Saying 𝑥 ∈ 𝑘𝑒𝑟(𝑃𝑛 (𝐷)) is precisely saying 𝑃𝑛 (𝐷)(𝑥) = 0 . Then, we have 𝐷𝑛 (𝑥) − 𝐼𝑑(𝑥) = 𝑥 (𝑥) − 𝑥 = 0 , which exactly 𝑥 (𝑥) = 𝑥 . Also, 𝑃𝑛 (𝐷) is a polynomial evaluated at an endomorphism. If f is an endomorphism of a K-vector space E and 𝑃 = 𝑎𝑛 𝑋 𝑛 + ⋯ + 𝑎1 𝑋 + 𝑎0 ∈ 𝐾[𝑋] a polynomial, then P(f) is the endomorphism 𝑎𝑛 𝑓 𝑛 + ⋯ + 𝑎1 𝑓 + 𝑎0 where 𝑓 𝑖 is f composed with itself i times. Question #6 Is it true that for a Group G with Normal Group 𝑁: have to write: [𝐺, 𝐺] 𝑁 𝑁 𝐺 𝑁 here instead of just [𝐺, 𝐺]/𝑁? = 𝐺𝑁 𝑁 ? I think the statement is correct. But why do we Answer: (1) N is normal subgroup of G; so it is normal subgroup of every subgroup between N and G. (2) When talking of (some subgroup)/N, this (some subgroup) should be taken containing N. (3) We ask: given a normal subgroup N of G, does it always contain [𝐺, 𝐺]? You can easily see negative answer by taking simple examples of finite order groups G. (4) Thus, when we want to factor [𝐺, 𝐺] by N, first look smallest subgroup above 𝑁(= containing 𝑁) which contains [𝐺, 𝐺]; it is precisely [𝐺, 𝐺]𝑁. Question #7 Number of elements in 𝑆13 conjugate to both (12)(34) and (123)(45) then is the answer 0? Because (12)(34) and (123)(45) have not got the same cycle structure. Answer: Step 1 That is correct. For this example, also notice that the two elements you mention have different orders. (12)(34) has order 2 while (123)(45) has order 6. Given a group G suppose that 𝑎, 𝑏 ∈ 𝐺 are conjugate. Then there exists 𝑐 ∈ 𝐺 such that 𝑐𝑎𝑐 −1 = 𝑏 If a has order n, that is if 𝑎𝑛 = 𝑒 , then 𝑏 𝑛 = (𝑐𝑎𝑐 −1 ) = 𝑐𝑎𝑐 −1 𝑐𝑎𝑐 −1 ⋯ 𝑐𝑎𝑐 −1 = 𝑐𝑎𝑛 𝑐 −1 = 𝑐𝑐 −1 =𝑒 So that 𝑜𝑟𝑑(𝑏) ≤ 𝑜𝑟𝑑(𝑎). By summetry, we also have that 𝑜𝑟𝑑(𝑎) ≤ 𝑜𝑟𝑑(𝑏) and hence 𝑜𝑟𝑑(𝑎) = 𝑜𝑟𝑑(𝑏) Question #8 How can one show (hopefully in an elementary manner) that there exist irreducible polynomials of arbitrary degree and number of variables over arbitrary field? Does ∀𝑛, 𝑑 ∈ 𝑁 field 𝐹 exist an irreducible 𝑓 ∈ 𝐹[𝑥1 , ⋯ , 𝑥𝑛 ] of degree d? Answer: Step 1 This is false. Any irreducible polynomial in one variable over an algebraically closed field (such as 𝐶 ) has degree 1. For 𝑛 ≥ 2 , the answer is yes: just take the irreducible polynomial 𝑦 − 𝑥 𝑑 Proof: If 𝑥1 − 𝑥2𝑑 = 𝑓(𝑥)𝑔(𝑥) = (∑𝑎 𝛼𝑎 𝑥 𝑎 )(∑𝑏 𝛽𝑏 𝑥 𝑏 ) = ∑𝑎,𝑏 𝛼𝑎 𝛽𝑏 𝑥 𝑎+𝑏 , then fg would include a monomial with 𝑥1 𝑥2 a contradiction. Question #9 What reduced quadratic equation has solutions √2 and √3 ? Answer: Step 1 A polynomial with coefficients in 𝑄 that has √2 as a root must also have −√2 as a root; likewise, if it has √3 as a root, then it must also have −√3 as a root. So any polynomial with coefficients in 𝑄 that has both √2 and √3 as roots must be of degree at least 4 or equal to 0, and must be a multiple of (𝑥 2 − 2)(𝑥 2 − 3) . So there is no quadratic polynomial with coefficients in 𝑄 that will do it. Step 2 To see this, suppose that p(x) is a polynomial with coefficients in 𝑄 that has √2 as a root. Dividing with remainder by 𝑥 2 − 2 we get 𝑝(𝑥) = (𝑥 2 − 2)𝑞(𝑥) + 𝑟(𝑥) , where q(x) and r(x) have rational coefficients, and 𝑟(𝑥) = 0 or else r(x) is of degree 0 or 1. Plugging in √−2 , we get 0 = 0𝑞(√2) + 𝑟(√2) = 𝑟(√2) But there are no polynomials of degree 0 or 1 with coefficients in 𝑄 that are 0 at √2 (that would yield that √2 is rational), so 𝑟(𝑥) = 0 . Therefore, 𝑟(𝑥) = 0 , so p(x) is a multiple of 𝑥 2 − 2 ; and since 𝑥 2 − 2 is 0 at −√2 , then so is p(x). The same argument shows that it is also a multiple of 𝑥 2 − 3 , and so p(x) will also have −√3 a a root. Step 3 Since 𝑥 2 − 3 and 𝑥 2 − 2 are relatively ' in 𝑄[𝑥] , any polynomial that is a multiple of both is a multiple of their product. The smallest degree polynomial that doe is (𝑥 2 − 2)(𝑥 2 − 3) = 𝑥 4 − 5𝑥 2 + 6 If you allow other coefficients, then the only quadratics that work are the form 𝑐(𝑥 − √2)(𝑥 − √3) = 𝑐𝑥 2 − 𝑥(√2 + √3)𝑥 + √6 and its multiples. Question #10 Let k be a field and R be the exterior ring over 𝑘 𝑑 , that is, k-algebra genarated by elements 𝑥1 , ⋯ . 𝑥𝑑 , where 𝑥𝑖 𝑥𝑗 = −𝑥𝑗 𝑥𝑖 ? Is R Artinian? Answer: Step 1 Actually, I am not sure the question is true as stated if 1 = −1 in 𝑘! If 2 is invertible, then you get 𝑥𝑖2 = 0 for all i. As suggested in the comments and Mariano's answer, the key point is R has finite dimension as a k-vector space. Since the monomials span R, you want a lot of them to vanish. For example: 𝑥1 𝑥2 𝑥1 = −𝑥1 𝑥1 𝑥2 = −𝑥12 𝑥2 = 0 Question #11 This is a well known lemma that consistently appears in textbooks, either as a statement without proof, or as an exercise. If 0id AfBgCh0 is a short exact sequence of finitely generated abelian groups, then rank 𝐵 = rank 𝐴 + rank 𝐶 I've been trying to prove this unsuccessfully. What do we know? f is injective, g is surjective, 𝐼𝑚𝑓 = ker 𝑔 , 𝐼𝑚𝑔 = ker ℎ , 𝐶 ≅ 𝐵 𝐴 So I start with a maximally linearly independent subset {𝑎𝛼 } of A such that the sum (with only finite nonzero entries) ∑ 𝑛𝛼 𝑎𝛼 = 0 for 𝑛𝛼 ∈ 𝑍 , implies that 𝑛𝛼 = 0 Where to go from here is a puzzle? Answer: Step 1 If you're not familiar with tensor products, then learning them for this problem is overkill. But if you are familiar with them, then I think the following solution is nice. For a finitely generated abelian group A, note that rank(𝐴) = dim𝑄 (𝐴 ⊗𝑍𝑄 ) . One can show (basically following Arturo's answer) that given a short exact sequence of abelian groups 0→𝐴→𝐵→𝐶→0, the sequence obtained by tensoring with 𝑄 is also exact: 0 → 𝐴 ⊗𝑍𝑄 → 𝐵 ⊗𝑍𝑄 → 𝐶 ⊗𝑍𝑄 → 0 Now the statement about ranks boils down to an easy statement about how dimensions of vector spaces add in a short exact sequence. Question #12 Let A be a DVR in its field of fractions F, and 𝐹 ′ ⊂ 𝐹 a subfield. Then is it true that 𝐴 ∩ 𝐹 ′ is a DVR in 𝐹 ′ ? I can see that it is a valuation ring of 𝐹 ′ , but how do I show, for example, that 𝐴 ∩ 𝐹' is Noetherian? Answer: Step 1 This is not necessarily true. For instance, you could have 𝐴 = 𝑘[[𝑡]] for a field k and 𝐹 ′ ⊂ 𝑘((𝑡)) equal to k. Then 𝐴 ∩ 𝐹 ′ = 𝐹 ′ and a field is often not considered a DVR. But this is the only thing that can go wrong. In general, if a domain has a nontrivial valuation taking values in the nonnegative integers, such that the elements of valuation zero are units, then it is a DVR (and noetherianness follows - for instance, one checks that the only ideals occur as 𝑥: 𝑥 has valuation at least n.) Question #13 A question on Hamilton Quaternions How does one prove that ring of Hamilton Quaternions with coefficients coming from the field 𝑍/𝑝𝑍 is not a divison ring. Answer: This answer builds on damiano's comment. Because of Wedderburn's theorem, we know that there are no finite-dimensional non-commutative division algebras over 𝐹𝑝 . In particular, every quaternion algebra over 𝐹𝑝 is split, i.e., isomorphic to the algebra 𝑀2 (𝐹𝑝 ) of 2 × 2 matrices over 𝐹𝑝 . Can we see this directly by arguments particular to quaternion algebras? Yes. For instance, a quaternion algebra is split iff its associated norm form 𝑁(𝑥) = 𝑥𝑥 = 𝑞(𝑥1 , 𝑥2 , 𝑥3 , 𝑥4 ) is isotropic -- i.e., there exists (𝑥1 , … , 𝑥4 ) ≠ (0, … ,0) such that 𝑞(𝑥1 , … , 𝑥4 ) = 0 . In the case of the Hamiltonian quaternion algebra, 𝑁(𝑥) = 𝑥12 + 𝑥22 + 𝑥32 + 𝑥42 As Robin points out, one way to see that this form is isotropic over Fp is to realize that by Lagrange's Four Squares Theorem, p is a sum of four integral squares, and then reduce modulo p. This is overkill. The fact that any quadratic form in at least three variables over a finite field is isotropic is a special case of the (easy to prove) Chevalley-Warning theorem Even this is more than is necessary: it follows from a simple counting argument that any nondegenerate quadratic form in at least two variables over 𝐹𝑝 is universal, i.e., represents every nonzero element of the field. From this we get that every nondegenerate quadratic form in at least three variables over a finite field is isotropic, and in particular the norm form of any quaternion algebra over a finite field is split. In fact the universality of binary quadratic forms over finite fields is the key to this result. Assuming the characteristic is not 2, the quaternion algebra may be represented as ≪ 𝑎, 𝑏 ≫ . For a quaternion algebra 𝑄 = ≪ 𝑎, 𝑏 ≫ over any field F (of characteristic not 2), a necessary and sufficient criterion for Q to be split is that b is a norm in the extension 𝐹(√𝑎)/𝐹 . Thus the universality of the norm form of 𝐹𝑝 (√𝑎)/𝐹𝑝 implies that all quaternion algebras over 𝐹𝑝 are split. For that matter, from the perspective of Galois cohomology, Wedderburn's theorem is not especially deep or difficult to prove. Using standard properties of cohomology of cyclic groups, it comes down to showing that for any finite degree extension 𝐹𝑞𝑛 /𝐹𝑞 of finite fields, the norm map is surjective, and this may also be proved by an elementary counting argument (or simply by a straightforward calculation). Question #14 Why the ring 𝑆 −1 𝑅 is Noetherian if S is multiplicative? Answer: Step 1 Assume the contrary that the following ascending chain of ideals in 𝑆 −1 𝑅 does not stop 𝑆 −1 𝐼1 ⊆ 𝑆 −1 𝐼2 ⊆ 𝑆 −1 𝐼3 ⊆ 𝑆 −1 𝐼4 ⊆ ⋯. So there is not any integer n such that 𝑆 −1 𝐼𝑛 = 𝑆 −1 𝐼𝑛+1 . Hence we find the following ascending chain of ideals in R which does not stop. 𝐼1 ⊆ 𝐼2 ⊆ 𝐼3 ⊆ 𝐼4 ⋯. This is a contradiction. Question #15 Why require R to be a field to obtain that 𝑅[𝛼] (2𝛼 2 ) Answer: Step 1 The graded ring 𝑍[𝛼] (2𝛼 2 ) with 𝛼 in degree 1 has 2𝛼 2 = 0 but 𝛼 2 ≠ 0 is either an exterior algebra or a polynomial algebra? As Andres Mejia points out, you can only deduce that 2𝛼 2 = 0 ⇒ 𝛼 2 = 0 if 2 is invertible in the ring R. By the way, this graded ring has 𝑍 in degrees 0 and 1, 𝑍 2𝑍 in degree 2. Question #16 If H and K are subgroups of G, |𝐻| = 20 and |𝐾| = 32 then a possible value of |𝐻 ∩ 𝐾| is 1) 2 2) 8 3) 6 4) 16 Answer: Order of H ∩ 𝐾 must divide order of H and order of K i.e., |𝐻 ∩ 𝐾| must divide |𝐻| and |𝐾|. If |𝐻 ∩ 𝐾| = 8 , 8 divides 32 which is |𝐾| but 8 does not divide 20 = |𝐻| . Hence |𝐻 ∩ 𝐾| ≠ 8 . If |𝐻 ∩ 𝐾| = 6 , 6 does not divide 32 = |𝐾| and 6 does not divide 20 = |𝐻| . Hence |𝐻 ∩ 𝐾| ≠ 6 . If |𝐻 ∩ 𝐾| = 16 , 16 divides 32 = |𝐾| but 16 does not divides 20 = |𝐻| . Hence |H \cap K| \ne 16 If |𝐻 ∩ 𝐾| = 2 . 2 divides 32 = |𝐾| and 2 divides 20 = |𝐻| . Hence |𝐻 ∩ 𝐾| = 2 Question #17 Let (𝐺1 ,∘)and(𝐺2 ,∗) be two groups and 𝜙: 𝐺1 ⇒ 𝐺2 be an isomorphism. a) G2 might not be abelian even if G1 is abelian. b) G2 is abelian if and only if G1 is cyclic c) G2 is finite if G1 is finite d) G2 might be abelian even if G1 is abelian Answer: Solution: 𝐺1 and 𝐺2 two groups 𝜙: 𝐺1 → 𝐺2 be an isomorphism ⇒ 𝑓(𝑔1 , 𝑔2 ) = 𝑓(𝑔1 ) ⋅ 𝑓(𝑔2 )∀𝑔1 𝑔2 ∈ 𝐺1 , ∴ 𝐺1 , is abelian 𝑔1 𝑔2 = 𝑔2 𝑔1 ∀𝑔1 , 𝑔2 ∈ 𝐺1 , 𝑓(𝑔1 , 𝑔)𝑃2 = 𝑓(𝑔2 𝑔1 ) (∴ 𝑓 is 1 − 1) 𝑓(𝑔1 )𝑓(𝑔2 ) = 𝑓(𝑔2 )𝑓(𝑔1 )∀𝑓(𝑔1 ), 𝑓(𝑔2 ) ∈ 𝐺2 ⇒ 𝐺2 is abelian group option (D) correct Question #18 Let 𝐺 = 𝑎 . The smallest subgroup of G containing 𝑎8 and𝑎 12 is generated by 1) 2) 3) 4) Arithmetical subtraction (-) is binary relation on 1) 𝑁 2) 𝑍 + 3) 𝑄 4) 𝑍 − Answer: Smallest subgroup 𝑎𝑚 and 𝑎𝑛 is 𝑎 𝑔𝑐𝑑 (𝑚, 𝑛) = 𝑎 𝑔𝑐𝑑 (8,12) = 𝑎4 So option 2nd are correct arithmetic subtraction is a binary relation on Q so option 3 are correct Question #19 Let H be a subgroup of a group G and 𝑎, 𝑏 ∈ 𝐺 . Then 𝑎 ∈ 𝑏𝐻 if and only if 1) 𝑏𝑎 − 1 ∈ 𝐻 2) 𝑏𝑎 ∈ 𝐻 3) 𝑏 − 1𝑎 ∈ 𝐻 4) None of these Answer: Given: H be a subgroup of a group G. i.e. 𝐻 ⊂ 𝑒𝑞 𝐺 . and 𝑎, 𝑏 ∈ 𝐺 If 𝑎 ∈ 𝑏𝐻 (1) 𝑏 ∈ 𝑏𝐻 ⇒ 𝑏1 ∈ 𝑏𝐻(∴ 𝑎 ∈ 𝐺 ⇒ 𝑎1 ∈ 𝐺) (2) Now from eq-n (1) and (2) 𝑎 ∈ 𝑏𝐻 and 𝑏1 ∈ 𝑏𝐻 then 𝑏1 𝑎 ∈ 𝑏1 (𝑏𝐻) 𝑏1 𝑎 ∈ (𝑏1 𝑏)𝐻 𝑏1 𝑎 ∈ 𝐻 Question #20 The number of elements in 𝑆6 is 1) 6 2) 720 3) 36 4) 360 Answer: 𝑆6 is a symmetric group of order 6 for a symmetric group 𝑆𝑛 of order n, the number of element can be found by the formula, number of elements = 𝑛! For symmetric group 𝑆6 , 𝑛 = 6 therefore number of elements = 6! = 720 Number of elements in 𝑆6 is 720 .