Math 594, HW7
... c). Continue the previous procedure, creating a Kn+1 , an extension of Kn (and of F as well) in which for every polynomial f (y) ∈ Kn [y] there’s at least one root in Kn+1 [y]. Denote K = ∪∞ 1 Kn . Every polynomial f [y] ∈ K[y] has finitely many coefficient so exist N , large enough, s.t. f (y) ∈ KN ...
... c). Continue the previous procedure, creating a Kn+1 , an extension of Kn (and of F as well) in which for every polynomial f (y) ∈ Kn [y] there’s at least one root in Kn+1 [y]. Denote K = ∪∞ 1 Kn . Every polynomial f [y] ∈ K[y] has finitely many coefficient so exist N , large enough, s.t. f (y) ∈ KN ...
Finite fields - MIT Mathematics
... write q − 2 in base 2, and multiply together the a2 for k corresponding to the nonzero bits in q − 2: for example, a25 = a16 · a8 · a. This is about log2 q additional multiplications, for a total of 2 log2 q multiplications in Fq to compute a−1 . At first glance this looks comparable to the Euclidea ...
... write q − 2 in base 2, and multiply together the a2 for k corresponding to the nonzero bits in q − 2: for example, a25 = a16 · a8 · a. This is about log2 q additional multiplications, for a total of 2 log2 q multiplications in Fq to compute a−1 . At first glance this looks comparable to the Euclidea ...
Solutions to Homework 7 27. (Dummit
... (Dummit-Foote 13.2 #22) Let {αi } be a basis for K1 over F , and let {βj } be a basis for K2 over F . Then {αi ⊗ βj } is a basis for K1 ⊗F K2 over F . Define a map φ : K1 ⊗F K2 → K1 K2 by φ(αi ⊗ βj ) = αi βj , and extend it by linearity. It is easy to check that φ is an F -algebra homomorphism. The ...
... (Dummit-Foote 13.2 #22) Let {αi } be a basis for K1 over F , and let {βj } be a basis for K2 over F . Then {αi ⊗ βj } is a basis for K1 ⊗F K2 over F . Define a map φ : K1 ⊗F K2 → K1 K2 by φ(αi ⊗ βj ) = αi βj , and extend it by linearity. It is easy to check that φ is an F -algebra homomorphism. The ...
Fields - MIT Mathematics
... (9) (identity element) The multiplicative identity 1F is not equal to the additive identity 0F . (10) (multiplicative inverses) For every non-zero element a in F there is a nonzero element a−1 so that a · a−1 = a−1 · a = 1F . MULTIPLICATION IS COMMUTATIVE (11) (commutative law) For all elements a an ...
... (9) (identity element) The multiplicative identity 1F is not equal to the additive identity 0F . (10) (multiplicative inverses) For every non-zero element a in F there is a nonzero element a−1 so that a · a−1 = a−1 · a = 1F . MULTIPLICATION IS COMMUTATIVE (11) (commutative law) For all elements a an ...
Is there anything else like the complex numbers
... This remarkable fact follows from a theorem of Frobenius stated below. This fact shows that R and C are quite exceptional. You probably know the so-called vector product of vectors in R3 . This product does not turn R3 into a field, because it is not commutative and not associative. One can ask a mo ...
... This remarkable fact follows from a theorem of Frobenius stated below. This fact shows that R and C are quite exceptional. You probably know the so-called vector product of vectors in R3 . This product does not turn R3 into a field, because it is not commutative and not associative. One can ask a mo ...
Maple Lecture 4. Algebraic and Complex Numbers
... 1. Generate the multiplication table for Z12 . Which elements of Z12 do have a multiplicative inverse? 2. Use Maple to show that the polynomial p := x4 + 3x + 4 is irreducible over Z5 = {0, 1, 2, 3, 4}. Declare α to be a root of p and express α13 as a polynomial in α of degree < 4. 3. An irreducible ...
... 1. Generate the multiplication table for Z12 . Which elements of Z12 do have a multiplicative inverse? 2. Use Maple to show that the polynomial p := x4 + 3x + 4 is irreducible over Z5 = {0, 1, 2, 3, 4}. Declare α to be a root of p and express α13 as a polynomial in α of degree < 4. 3. An irreducible ...
Math 210B. Homework 4 1. (i) If X is a topological space and a
... of X admits a finite subcover) and that any subspace Y ⊂ X is noetherian. (iii) Conversely to (ii), if every subspace of a topological space X is quasi-compact then prove X is noetherian. 2. Over a field k = k with char(k) 6= 2, decompose Z(y 4 − x2 , y 4 − x2 y 2 + xy 2 − x3 ) ⊂ k 2 and Z(u2 + v 2 ...
... of X admits a finite subcover) and that any subspace Y ⊂ X is noetherian. (iii) Conversely to (ii), if every subspace of a topological space X is quasi-compact then prove X is noetherian. 2. Over a field k = k with char(k) 6= 2, decompose Z(y 4 − x2 , y 4 − x2 y 2 + xy 2 − x3 ) ⊂ k 2 and Z(u2 + v 2 ...
