Questions and Answers #4 Abstract Algebra

Questions and Answers Sheet 4 Abstract Algebra Question #1 The set of all even integers 2𝑍 is a subgroup of (𝑍, +). Then the right coset −4 + 2𝑍 contains the element 1) 3 2) 5 3) 10 4) 1 Answer: Given the set of all even integers Ififis a subgroup of (𝑍, +) The right coset is −4 + 2𝑍 If 𝑎 ∈ 2𝑍 then −4 + 𝑎 ∈ −4 + 2𝑍 Hence the correct option is 10. Question #2 Let 𝐺 = 𝑆_{𝑛} "{𝑎𝑛𝑑 𝑙𝑒𝑡} 𝐻 = {𝛼 ∈ 𝐺|𝛼(1) = 1} . Show that H is a subgroup of G. Answer: (𝐻,∘) is a subgroup of (𝐺,∘) iff 1) 𝐻 ≠ 𝜙 2) 𝛼, 𝛽 ∈ 𝐻 ⇒ 𝛼 ⋅ 𝛽 ∈ 𝐻 3) 𝛼 ∈ 𝐻 ⇒ 𝛼 −1 ∈ 𝐻 G = Sn H = {  Sn |  (1) = 1} identify permutation 𝑒 ∈ 𝐻 , so 𝐻 ≠ 𝜙. Let 𝛼, 𝛽 ∈ 𝐻 then 𝛼(1) = 1and𝛽(1) = 1 ⇒ 𝛼 ⋅ 𝛽(1) = 𝛼(𝛽(1)) = 𝛼(1) = 1 ⇒ (𝛼𝛽)(𝑖) = 1 ⇒ 𝛼𝛽 ∈ 𝐻 Let 𝛼 ∈ 𝐻 ⇒ 𝛼(1) = 1 ⇒ 𝛼 −1 (1) = 1 ⇒ 𝛼 −1 ∈ 𝐻 Thus, H is a subgroup of 𝐺 = 𝑆𝑛 Question #3 Julio is studying for his final exams in chemistry and algebra. He knows he only has 12 hours to study and it will take him at least 2 more hours to study for algebra than chemistry. The following system of equations represents the situation, where x represents algebra and y represents chemistry: 𝑥 + 𝑦 ≤ 12 𝑥 ≥𝑦+2 Which of the following can Julio do? Study algebra for 7 hours and chemistry for 5 hours Study algebra for 8 hours and chemistry for 4 hours Study algebra for 7 or 8 hours and chemistry for 4 or 5 hours None of the above Answer: Let 𝑥 → algebra 𝑦 → chemistry The inequalities are: 𝑥 + 𝑦 ≤ 12 (1) 𝑥 ≥ 𝑦 + 2 (2) From equation (2) we have −𝑥 + 𝑦 ≤ −2 (3) Now salving (1) and (3) we get by taking the equality sign (1) + (3) (𝑥 + 𝑦) + (−𝑥 + 𝑦) = 12 + (−2) ⇒ 2𝑦 = 10 ⇒𝑦=5 When 𝑦 = 5 then using 𝑥 + 𝑦 = 12 we have 𝑥 = 7 Question #4 Consider the following permutations in 𝑆9 : 1 2 3 4 5 6 7 8 9 ) and 3 5 2 1 4 6 9 7 8  =( 1 2 3 4 5 6 7 8 9 ) 5 2 6 7 1 3 8 9 4  =( Find 𝛼 by solving 𝜎𝛼 = 𝜏 . Write your answer in cycle notation. Answer: 1 2 3 4 5 6 7 8 9 =( ), 3 5 2 1 4 6 9 7 8 Let  1 2 3 4 5 6 7 8 9 ) 5 2 6 7 1 3 8 9 4  =( 3 5 2 1 4 6 9 7 8 ) 1 2 3 4 5 6 7 8 9  −1 = ( 𝜎𝛼 = 𝜏 ⇒ 𝛼 = 𝛿 −1 1 2 3 4 5 6 7 8 9  = ( ) 7 6 5 1 2 3 9 4 8 𝛼 = (17984)(2635) Let 1 2 3 4 5 6 7 8 9 ), 3 5 2 1 4 6 9 7 8  =( 1 2 3 5 2 6 3 5  −1 = ( 1 2  =( 4 5 6 7 1 3 2 1 4 3 4 5 7 8 6 6 8 9 9 7 9 ) 4 7 8 ) 8 9 𝜎𝛼 = 𝜏 ⇒ 𝛼 = 𝜎 −1 1 2 3 4 5 6 7 8 9  = ( ) 7 6 5 1 2 3 9 4 8 𝛼 = (17984)(2635) Question #5 Simple question about finitely generated algebras I have two finitely generated algebras A and B over a field K such that 𝐵[𝑎1 , ⋯ , 𝑎𝑛 ] for some 𝑎1 , ⋯ , 𝑎𝑛 ∈ 𝐴 ? B  A . Is it true that 𝐴 = Answer: If A is finitely generated over K then we have 𝐴 = 𝐾[𝑎1 , ⋯ , 𝑎𝑛 ] for some elements 𝑎1 , ⋯ , 𝑎𝑛 A . Then: 𝐴 = 𝐾[𝑎1 , ⋯ , 𝑎𝑛 ] ⊆ 𝐵[𝑎1 , ⋯ , 𝑎𝑛 ] ⊆ 𝐴 Question #6 Suppose that c is transcendental over 𝑄 . Show that √𝑐 and 𝑐 + √𝑐 are also transcendental. My solution (a bit rough): (a) A polynomial that √𝑐 satisfies is 𝑥 2 − 𝑐 . However, since c is not algebraic over 𝑄 , c is not in 𝑄 , 𝑥 − 𝑐 would be a polynomial that c satisfies in the rationals. So 𝑥 2 − 𝑐 is not in 𝑄|𝑥| and so c is not algebraic over 𝑄 (not sure about that statement... why could there not be another polynomial that is satisfies, is it because it is the minimal polynomial?). (b) A polynomial that 𝑐 + √𝑐 satisfies is 𝑝(𝑥) = 𝑥 3 − 𝑐𝑥 2 − 𝑐𝑥 = 2𝑐 . For reasons mentioned above, this is not in 𝑄[𝑥] either, so c is not algebraic over 𝑄 . (Also because it is the minimal polynomial and the minimal polynomial must divide everything that c satisfies? Really not sure about how to finish up here). Answer: a) Let 𝑓 ∈ 𝑄[𝑥] . Then 𝑓(√𝑐) = 𝑎(𝑐) + 𝑏(𝑐)√𝑐 with 𝑎(𝑐), 𝑏(𝑐) ∈ 𝑄[𝑐] . If 𝑓(√𝑐) = 0 and 𝑓 ≠ 0 it follows that a(c) and b(c) are not both zero, and 𝑎2 (𝑐) = 𝑏 2 (𝑐)𝑐 . But the last relation is not possible for degree reasons. (b) Try a similar approach. Question #7 Very basic question about localizations Let A be a ring, S a multiplicatively closed subset. Is it true that if 𝑎 ∈ 𝐴 is? Answer: 𝑎 1 ∈ 𝑆 −1 𝐴 is a non zero-divisor if and only Set A = 6 and S = A3 6 . Then 2̂ is a zerodivisor in A and ̂ 2 ̂ 1 is not zerodivisor in 𝑆 −1 𝐴 . (Actually it is invertible in 𝑆 −1 𝐴 .) Question #8 Is there a name for the set E ( x, n) = {x | p  N ?  p  n − 1}with x  G, G a group? p Answer: A common name for 𝐸(𝑥, 𝑛), considered as a group, is 𝐶𝑛 , the cyclic group of order n. So 𝐶𝑛 = {𝑒, 𝑥, 𝑥 2 , … , 𝑥 𝑛−1 } , where 𝑥 𝑛 = 𝑥 0 = 𝑒 . In the context of your question, 𝐶𝑛 ≅≪ 𝑥 ≫ , where x has order n, e.g., 𝑥 𝑛 = 𝑒, but 𝑥 𝑚 ≠ 𝑒 for 1 ≤ 𝑚 < 𝑛. It is also common to write this by the presentation Cn = x | xn = e  . The "C" stands for "cyclic". Another example of a group presentation is, for the dihedral group 𝐷𝑛 , Dn = {r, s | r n = s 2 = e, srs = r −1}, for 𝑛 ≥ 3 . This is not a cyclic group in general. Question #9 Order of field extension I'm working on some algebra exercises and I'm really struggling with finite fields. I'm currently working on the following: find the splitting field let's say F of a cubic polynomial over 𝑍𝟙𝟙 and to find the degree of the extension [𝐹: 𝑍𝟙𝟙 ] . I've determined that the splitting field is 𝐹 = 𝑍𝟙𝟙 /(𝑓) where (f) is the ideal generated by f. Now i'm stumped because I'm not sure how to figure out the order of the extension. My guess would be 3 because cubic polynomial but it's just a guess. Any hints/help will be appreciated Answer: Let 𝑓 = 𝑥 3 + 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 be an irreducible polynomial over 𝑍𝟙𝟙 . The extension field 𝐸 = 𝑍𝟙𝟙 [𝑥]/≪ 𝑓 ≫ contains a zero of f, namely the residue class 𝛼 = 𝑥+≪ 𝑓 ≫ , This gives 𝛼 3 + 𝑎𝛼 2 + 𝑏𝛼 + 𝑐 = ,so 𝛼 3 = −𝑎𝛼 2 − 𝑏𝛼 − 𝑐 , and the extension field is E = {u  2 + v  + w | u, v, w  } with degree [𝐸: 𝑍𝟙𝟙 ] = 3 . In particular, if f is primitive, the powers of 11 𝛼 are exactly the nonzero elements of E. NB: 𝑓(𝑥) = 𝑓(𝑥+≪ 𝑓 ≫) = 𝑓(𝑥)+≪ 𝑓 ≫=≪ 𝑓 ≫= 0 . Question #10 Do we have [𝐸: 𝐾]𝑖 = [𝐸𝑟 : 𝐾] where 𝐸𝑟 is purely inseparable closure? Let 𝐸 𝐾 be an algebraic extension. Let 𝐸𝑠 (resp. 𝐸𝑟 ) be the set of elements of E which are separable [𝐸:𝐾] (resp. purely inseparable) over K. Let [𝐸: 𝐾]𝑖 = [𝐸: 𝐸𝑠 ] = [𝐸:𝐾] , be the inseparable degree. Do we have 𝑠 [𝐸: 𝐾]𝑖 = [𝐸𝑟 : 𝐾]? Answer: 𝑘 = 𝐹𝟛 , 𝐾 = 𝑘(𝑥 6 , 𝑦 6 ), 𝑅 = 𝐾(𝑥 + 𝑦 2 ), 𝐿 = 𝐾(𝑥 3 ) = 𝐾(𝑥 3 + 𝑦 6 ) 𝐿 𝐸 is a separable quadratic extension, is purely inseparable of degree 3 and [𝐸: 𝐾] = 6 . 𝐾 𝐿 If there is 𝐸/𝐹/𝐾 with F/K purely inseparable then [𝐹: 𝐾] = 3, [𝐸: 𝐹] = 2 . Let 𝑓 ∈ 𝐹[𝑇] be the degree 2 minimal polynomial of 𝑥 + 𝑦 2 . f stays irreducible over 𝑘(𝑥 2 , 𝑦 2 ) ⊂ 𝐹 so in fact 𝑓 = (𝑇 − 𝑥 − 𝑦 2 )(𝑇 + 𝑥 − 𝑦 2 ) = 𝑇 2 − 2𝑦 2 𝑇 − 𝑥 2 + 𝑦 4 when 𝑦 2 , 𝑥 2 ∈ 𝐹 which is a contradiction as 𝑘(𝑥 2 , 𝑦 2 ) has degree 32 over K. Question #11 Fundamental question about field extensions Let k be an algebraically closed field. Let 𝐾 ⊂ 𝑘 be a subfield, and suppose that it has an algebraic extension 𝐾 ⊂ 𝐿 . Is it true (or at least well defined, because I'm not even sure about this) that 𝐿 ⊆ 𝑘 ? One slightly different point of view is: given injections 𝑖: 𝐾 → 𝑘, 𝑗: 𝐾 → 𝐿 , the last one being algebraic, is there an injection 𝑙: 𝐿 → 𝑘 such that 𝑙 ∘ 𝑗 = 𝑖 (i.e. l is a homomorphism of K-algebras)? Answer: You cannot prove that 𝐿 ⊆ 𝑘 , but you're right that there exists an embedding 𝐿 → 𝑘 that's the identity o K. Consider the set 𝜉 of pairs (𝐹, 𝑗𝐹 ) , where 𝐾 ⊆ 𝐹 ⊆ 𝐿 (as subfields) and 𝑗𝐹 : 𝐹 → 𝑘 is the identity on K. We can partially order 𝜉 by declaring (𝐹1 , 𝑗𝐹1 ) ≤ (𝐹2 , 𝑗𝐹2 ) if (and only if) 𝐹1 ⊆ 𝐹2 and the restriction of 𝑗𝐹2 to 𝐹1 is 𝑗𝐹1 . Clearly 𝜉 is not empty, because (𝐾, 𝑖) ∈ 𝜉 . Moreover any chain in 𝜉 has an upper bound: take the union of the first components of the pair and define the homomorphism accordingly. By Zorn's lemma, there is a maximal element (𝐹, 𝑗𝐹 ) and we want to prove that 𝐹 = 𝐿 . Here the fact that L is algebraic over K is needed (up to now we didn't use it). 𝐿 If 𝑎 ∈ , the element a is algebraic over K and all it takes is to prove that we can extend 𝑗𝐹 to a 𝐹 homomorphism 𝑗: 𝐹(𝑎) → 𝑘 . This is standard. But now (F(a), j) contradicts the maximality of (𝐹, 𝑗𝐹 ) and we're done. Question #12 Convoluted definition of a set: 𝐻(𝑆) = {𝑥 ∈ 𝐺| ∃𝑛 ∈ 𝑁, ∃{𝑥_{1}, 𝑥_{2}, ⋯ , 𝑥_{𝑛}} ⊆ 𝑆 ∪ 𝑆^{−1}, 𝑥 = 𝑥_{1} ⋯ 𝑥_{𝑛}} . Answer: No. H(S) is the set of all elements g of G such that g can be expressed as a finite product of elements of 𝑆 ∪ 𝑆 −1 . Maybe lets have a look at example. Consider 𝐺 = {0,1,2,3} with addition modulo 4 as group operation, i.e. 𝐺 = 𝑍𝟜 . Now let 𝑆 = {2} . Since we are dealing with addition, then 𝑆 −1 = {2} because 2 + 2 = 0 in our group. Therefore 𝑆 ∪ 𝑆 −1 = {2} . Next H(S) is the set of all elements g of G such that 𝑔 = 𝑥1 + ⋯ + 𝑥𝑛 for some 𝑥1 , ⋯ , 𝑥𝑛 ∈ 𝑆 ∪ 𝑆 −1 . Since our last set has exactly one element then only possibilities for g are: 2,2 + 2,2 + 2 + 2 , and which we can write in a compact way as 𝑛 ⋅ 2 for some 𝑛 ∈ 𝑁. By removing duplicates (remember that 2 + 2 = 0 ) this gives us 𝐻(𝑆) = {0,2} Question #13 Generalization of Rational Root Theorem [closed] Can the rational root theorem in univariate variable polynomial equations be extended to more than one variable? For instance could we apply it to 𝑓(𝑥, 𝑦) = 1 + 𝑥 + 𝑥𝑦 − 2𝑦 2 + 4𝑥 2 to find all possible rational roots? I am sure the obvious answer is no but is there any hint on how to find all possible rational roots say of this example? Answer: As said in the comments: the answer is NO in general. For your particular case, you can do the following thing. We have 𝑥 2 𝑥2 4 2 𝑓(𝑥, 𝑦) = −2 (𝑦 − ) + + 4𝑥 2 + 𝑥 + 1 𝑥 2 9 4 𝑥 2 2 9 1 2 17 4 2 9 18 = −2 (𝑦 − ) + 𝑥 2 + 𝑥 + 1 = −2 (𝑦 − ) + (𝑥 + ) + . 9 17 𝑥 2 18 4 Soving 𝑓(𝑥, 𝑦) = 0 where 𝑥, 𝑦 ∈ 𝑄 is the same as solving −2𝑢2 + 𝑣 2 = − , setting 𝑢 = 𝑦 − 𝑥+ 1 9 and 𝑉 = (it is easy to derive x,y from u, v. This is equivalent to 2 -36𝑢 + 81𝑣 2 = −17 , that is (3𝑢 − 9𝑣)(3𝑢 + 9𝑣) = 17 , which is easy to solve. For 𝑡 ∈ 𝑄 × , this is 17 equivalent to solve 3𝑢 − 9𝑣 = 𝑡, 3𝑢 + 9𝑣 = . You then get u,v as a rational fraction of t, then you get 𝑡 x,y. [The idea is that you have an affine conic, so you can reduce it to a "diagonal form", and then reduce to the easier case to a rational conic 𝑎𝑥 2 + 𝑏𝑦 2 = 𝑐 . This situation is well understood from quadratic form specialist, and you have systematic methods to solve this even if it can be a bit tricky. Here we are in a particularly nice situation.] Question #14 Abelian group order divides 1cm I want to prove the following: Let G be an abelian group and let 𝑎1 , ⋯ , 𝑎𝑟 ∈ 𝐺 then |𝑎1 ⋯ 𝑎𝑟 | (the order of 𝑎1 ⋯ 𝑎𝑟 ) divides 1𝑐𝑚(|𝑎1 |, … , |𝑎𝑟 |) So far I have shown that (𝑎1 , … , 𝑎𝑟 )1𝑐𝑚(|𝑎1|,…,|𝑎𝑟 |) = 1 but I don't think this implies the desired result. Thank you in advance Answer: It does imply the desired result. Note that if 𝑎𝑛 = 1 , then |𝑎| divides n. Simply write 𝑛 = |𝑎|𝑞 + 𝑟 for 0 ≤ 𝑟 < |𝑎| , and then 1 = 𝑎𝑛 = 𝑎 |𝑎|𝑞 ⋅ 𝑎𝑟 = 1𝑞 ⋅ 𝑎𝑟 = 𝑎𝑟 , and so since 𝑟 < |𝑎| and 𝑎𝑟 = 1, we have 𝑟 = 0 . Question #15 If N is normal in G, show 𝑍𝑖 (𝐺) 𝑁 𝑁 𝐺 ≤ 𝑍𝑖 ( ) where 𝑍𝑖 (𝐺) is the i-th term in the upper central series for G. 𝑁 Answer: As you note, if K is a group, then showing that 𝑥 ∈ 𝑍𝑖+1 (𝐾) is equivalent to showing that [𝑥, 𝑦] ∈ 𝑍𝑖 (𝐾) for all 𝑦 ∈ 𝐾 . You are assuming that 𝑍𝑖 (𝐺) 𝑁 𝑁 𝐺 𝑁 𝐺 𝑁 𝑁 𝑁 ≤ 𝑍𝑖 ( ). You want to prove that 𝑍𝑖+1 (𝐺) ≤ 𝑍𝑖+1 ( ) . That means that 𝐺 𝐺 𝑁 𝑁 you want to show that for all 𝑥 ∈ 𝑍𝑖+1 (𝐺), [𝑥𝑁, 𝑦𝑁] ∈ 𝑍𝑖 ( ) for all 𝑦𝑁 ∈ . 𝐺 So, let 𝑥 ∈ 𝑍𝑖+1 𝐺 . We want to show that 𝑥𝑁 ∈ 𝑍𝑖+1 ( ) . 𝑁 Let 𝑦 ∈ 𝐺 be arbitrary. Then [𝑥𝑁, 𝑦𝑁] = [𝑥, 𝑦]𝑁 ∈ 𝐺 𝑁 . If 𝑥 ∈ 𝑍𝑖+1 (𝐺),then [𝑥, 𝑦] ∈ 𝑍𝑖 (𝐺) , so [𝑥𝑁, 𝑦𝑁] = [𝑥, 𝑦]𝑁 ∈ 𝑍𝑖 (𝐺)𝑁. But 𝑍𝑖 (𝐺)𝑁 . But 𝑍𝑖 (𝐺) 𝑁 𝑁 𝐺 ≤ 𝑍𝑖 ( ) by 𝑁 𝐺 the induction hypothesis, so [𝑥𝑁, 𝑦𝑁] ∈ 𝑍𝑖 ( ) . 𝑁 𝐺 𝐺 𝑁 𝑁 As this holds for all 𝑦𝑁 ∈ , it follows that 𝑥𝑁 ∈ 𝑍𝑖+1 ( ) , as required. Question #16 Localization in the 𝐷 + 𝑀 construction. Let T be an integral domain, and let K be a subfield of T and M a maximal ideal of T such that 𝑇 = 𝐾 + 𝑀 . Let 𝐷 ⊆ 𝐾 be an integral domain, and set 𝑅 = 𝐷 + 𝑀 . My question is: if 𝑚 ∈ 𝑀 such that 𝑚 ≠ 0, is 𝑅𝑚 = 𝑇𝑚 ? In this context, 𝑅𝑚 is the localization of R at the multiplicative set S = {m | k  k  {0}} . Answer: The answer is yes. Since 𝑅 ⊆ 𝑇, also 𝑅𝑚 ⊆ 𝑇𝑚 ; for the other direction, it suffices to show that 𝑇 ⊆ 𝑅𝑚 , so fix any 𝑎 ∈ 𝑇 . Then 𝑎𝑚 ∈ 𝑀 , since M is an ideal of T and 𝑚 ∈ 𝑀 . Hence, since 𝑀 ⊆ 𝑅 , we have 𝑎𝑚 𝑎𝑚 ∈ 𝑅 . Thus 𝑎 = ∈ 𝑅𝑚 , as desired. 𝑚 Question #17 Determine order of ≪ (12345), (2354) ≫ . I'm asked to prove that the order of 𝐻 ≔≪ (12345), (2354) ≫ is 20, in an Algebra exercise 2nd year maths. What I have achieved is that 20||𝐻|"{𝑎𝑛𝑑}|𝐻| = 20,120 since H can't be the alternating group and there are no subgroups of order 40 in S5. But I don't know how to prove H  S5 . Answer: Consider the integers modulo 5. Let 𝛼 be the operation 𝛼: 𝑥 → 𝑥 + 1 . Let \beta be the operation 𝛽: 𝑥 → 2𝑥 . 𝑍 Then 𝛼, 𝛽 generate all affine automorphisms of , which has size 20: 5𝑍 𝑃𝑆𝐾 𝑥 → 𝑎𝑥 + 𝑏 , with 𝑎 ∈ {1,2,3,4} and𝑏 ∈ {0,1,2,3,4,5} . Question #18 Show that two modules 𝑉𝐴 , 𝑉𝐵 are isomorphic. Let K be a field and define two matrices: 1 1 1 0 A := ( ) and B := ( ) 0 3 0 3 We then define a 𝐾[𝑋] −module on 𝑉 = 𝐾 2 such that 𝐾[𝑋] × 𝑉 → 𝑉, (𝑃, 𝑣) → 𝑃(𝐴) ⋅ 𝑣 where 𝑃(𝐴) ∈ 𝐾 2×2 is achieved by plugging in matrix A into polynomial P and 𝑃(𝐴) ⋅ 𝑣 is therefore the matrix-vector-multiplication. We call this module 𝑉𝐴 ⋅ 𝑉𝐵 can be constructed similarly. I want to show that for 𝐾 = 𝑄 the modules 𝑉𝐴 , 𝑉𝐵 are isomorphic. Answer: Actually one can define 𝜙: 𝑉𝐵 → 𝑉𝐴 by 𝜙(𝑣) = 𝑆𝑣 . This map is bijective and clearly satisfies 𝜙(𝑣1 + 𝑣2 ) = 𝜙(𝑣1 ) + 𝜙(𝑣2 ). We have to show now that 𝜙(𝑃(𝑋)𝑣) = 𝑃(𝑋)𝜙(𝑣). But 𝜙(𝑃(𝑋)𝑣) = 𝜙(𝑃(𝐵)𝑣) = 𝑆(𝑃(𝐵)𝑣) = (𝑆𝑃(𝐵))𝑣 = (𝑃(𝐴)𝑆)𝑣 = 𝑃(𝐴)(𝑆𝑣) = 𝑃(𝑋)𝜙(𝑣) . Question #19 Module with matrix multiplication Define the rational 3 × 3 -matrix 1 4 1 A := (0 1 0) 0 1 2 We then define a 𝑄[𝑋] -module on 𝑉 = 𝑄𝟛 such that 𝑄[𝑋] × 𝑉 → 𝑉, (𝑃, 𝑣) → 𝑃(𝐴) ⋅ 𝑣 where 𝑃(𝐴) ∈ 𝑄𝟛×𝟛 is achieved by plugging in matrix A into polynomial P and 𝑃(𝐴) ⋅ 𝑣 is therefore the matrix-vector-multiplication. Call this module 𝑉𝐴 . 1 0 Argue if VA = [ X ](0)  [ X ]( 1 ) . 1 −1 Answer: Your attempt is wrong. In fact, 𝑉𝐴 equals the direct sum. The elementary divisors of A (that is, of 𝑋𝐼3 − 𝐴 ) are (𝑋 − 1)2 , 𝑋 − 2 , and then 𝑉𝐴 is isomorphic to 𝑄[𝑋]/((𝑋 − 1)2 ) ⊕ 𝑄[𝑋]/(𝑋 − 2). 1 0 0 2 On the other side, one can check that (( X − 2)(0) = 0, ( X − 1) ( 1 ) = 0 and ( X − 1)( 1 )  0 , so the 1 −1 −1 annihilator of the first vector is (𝑋 − 2) , and the annihilator of the second vector is ((𝑋 − 1)2 ) . Knowing all these and using the uniqueness of the elementary divisors one can conclude that 1 0 VA = [ X ](0)  [ X ]( 1 ) 1 −1 Question #20 Northcott Multilinear Algebra Universal Property Proof Northcott Multilinear Algebra poses a problem. Consider R-modules 𝑀1 , … , 𝑀𝑝 , M and N. Consider multilinear mapping 𝜓: 𝑀1 × ⋯ × 𝑀𝑝 → 𝑁 Northcott calls the universal problem as the problem to find M and multilinear mapping 𝜙: 𝑀1 × ⋯ × 𝑀𝑝 → 𝑀 such that there is exactly one R-module homomorphism ℎ: 𝑀 → 𝑁 such that ℎ ∘ 𝜙 = 𝜓 . If 𝜆 and 𝜆′ exist I understand why the equalities at the end of the sentence follow, based on the satisfaction of the universal problem. I can't see however why homomorphisms 𝜆 and 𝜆′ should exist. I did more group theory many years ago and this is my first serious foray into "modules" so I wouldn't be surprised if there is something obvious I'm missing. my thoughts: Clearly M and M' are both homomorphic to N through h and h', I'm not sure if this says anything about a relationship between M and M' though. If h' were injective I could say something like 𝜆(𝑚) = ℎ′−1 (ℎ(𝑚)) but I don't know if there is any guarantee that h' is injective.. Likewise, if 𝜙 were injective I could define 𝜆(𝑚) = 𝜙 ′ (𝜙 −1 (𝑚)) but again I don't know why this would be the case... I've tried replacing M and N with more familiar vector spaces and R-module homomorphisms by multilinear maps for better intuition but no luck.. I do know that if M and M' are vector spaces with the same dimension then there is an isomorphism between them. I guess more generally if M and M' have different dimensions (say dim(𝑀′) > dim(𝑀)) then there is a homomorphism from M into a subspace of M' and another homormophism from M' onto M. Maybe this carries over to modules and is in the right direction for what I need...? Answer: The existence of 𝜆 and 𝜆′ are furnished by the maps 𝜙 ′ ϕ via the universal property; that is, we are taking 𝑁 = 𝑀′ , M respectively. To spell this out further, the map 𝜙 ′ : 𝑀1 × ⋯ × 𝑀𝑝 → 𝑀′ is bilinear by assumption, and so the universal property of the pair (𝑀, 𝜙) says that there is exactly one morphism 𝜆: 𝑀 → 𝑀′ such that 𝜆 ∘ 𝜙 = 𝜙 ′ . You obtain the morphism 𝜆′ : 𝑀′ → 𝑀 analogously by switching the roles of M and M'. This is a typical situation that arises in universal problems: the data of any two solutions contains information which allows you to relate the two solutions through the universal property. In this case, the crucial piece of information is this bilinear map 𝜙: 𝑀1 × ⋯ × 𝑀𝑝 → 𝑀 which allows you to relate any two solutions. Question #21 if 𝑓: 𝑅 → 𝑅 is a homomorphism and p is an integer-coefficient polynomial, then: 𝑓(𝑝(𝑟)) = 𝑝(𝑓(𝑟))∀𝑟 ∈ 𝑅 Answer: We have 𝑓(1) = 𝑓(1 + 0) = 𝑓(1) + 𝑓(0) so 𝑓(0) = 0 . By induction on 𝑛 ∈ 𝑁 we have 𝑓(∑𝑛𝑖=1 𝑟𝑖 ) = ∑𝑛𝑖=1 𝑓(𝑟𝑖 )and𝑓(∏𝑛𝑖=1 𝑠𝑖 ) = ∏𝑛𝑖=1 𝑓(𝑠𝑖 ) . In particular when 𝑟𝑖 = 1 for each i then 𝑓(𝑛) = 𝑓(∑𝑛𝑖=0 1) = 𝑓(∑𝑛𝑖=1 𝑟𝑖 ) = ∑𝑛𝑖=1 𝑓(𝑟𝑖 ) = ∑𝑛𝑖=1 𝑓(1) = ∑𝑛𝑖=1 1 = 𝑛 . And so if 𝑛 ∈ 𝑁 then 0 = 𝑓(0) = 𝑓(𝑛 + (−𝑛)) = 𝑓(𝑛) + 𝑓(−𝑛) = 𝑛 so + 𝑓(−𝑛), 𝑓(−𝑛) = −𝑛 So 𝑓(𝑣) = 𝑣 for every integer v. And if 𝑠𝑖 = 𝑥 for each i then 𝑓(𝑥 𝑛 ) = 𝑓(∏𝑛𝑖=1 𝑥 ) = 𝑓(∏𝑛𝑖=1 𝑠𝑖 ) = = ∏𝑛𝑖=1 𝑓(𝑠𝑖 ) = ∏𝑛𝑖=1 𝑓(𝑥) = 𝑓(𝑥)𝑛 . Let 𝑝(𝑥) = 𝑣0 + ∑𝑛𝑖=1 𝑣𝑖 𝑥 𝑖 where each 𝑣𝑖 ∈ 𝑍 . For 𝑖 > 0 let each 𝑣𝑖 𝑥 𝑖 = 𝑟𝑖 and we have 𝑓(𝑝(𝑥)) = 𝑓(𝑣0 ) + 𝑓(∑𝑛𝑖=1 𝑟𝑖 ) = 𝑓(𝑣0 ) + ∑𝑛𝑖=1 𝑓(𝑟𝑖 ) = = 𝑓(𝑣0 ) + ∑𝑛𝑖=1 𝑓(𝑣𝑖 𝑥 𝑖 ) = = 𝑓(𝑣0 ) + ∑𝑛𝑖=1 𝑓(𝑣𝑖 )𝑓(𝑥 𝑖 ) = = 𝑣0 + ∑𝑛𝑖=1 𝑣𝑖 𝑓(𝑥)𝑖 = 𝑝(𝑓(𝑥)) .