Finite fields / Galois Fields
... This sequence must begin to repeat and the first element to repeat is 0. Proof. Since the field is finite, this sequence must begin to repeat at some point. If j (1) is the first repeated element, being equal to k (1) for 0 ≤ k < j , it follows that k must be zero; otherwise ( j − k ) (1) = 0 is an ...
... This sequence must begin to repeat and the first element to repeat is 0. Proof. Since the field is finite, this sequence must begin to repeat at some point. If j (1) is the first repeated element, being equal to k (1) for 0 ≤ k < j , it follows that k must be zero; otherwise ( j − k ) (1) = 0 is an ...
School of Mathematics and Statistics The University of Sydney
... 2. Z2 ⊕ Z2 - not isomorphic to ring 1 because every element of Z2 ⊕ Z2 added to itself is zero, while 1 + 1 = 2 6= 0 ∈ Z4 . 3. Z2 [x]x2 - not isomorphic to ring 1 for the same reason as ring 2 is not isomorphic to 1. It is not isomorphix to ring 2 because there is an element of ring 3, namely x whic ...
... 2. Z2 ⊕ Z2 - not isomorphic to ring 1 because every element of Z2 ⊕ Z2 added to itself is zero, while 1 + 1 = 2 6= 0 ∈ Z4 . 3. Z2 [x]x2 - not isomorphic to ring 1 for the same reason as ring 2 is not isomorphic to 1. It is not isomorphix to ring 2 because there is an element of ring 3, namely x whic ...
Trigonometric functions, elliptic functions, elliptic modular forms 1
... [4.0.2] Claim: An entire doubly-periodic function is constant. Proof: Let ω1 , ω2 be Z-generators for Λ. Since the ωi are linearly independent over R, every z ∈ C is an R-linear combination of them. Given z = aω1 +bω2 with a, b ∈ R, let m, n be integers such that 0 ≤ a−m < 1 and 0 ≤ b − n < 1. Then ...
... [4.0.2] Claim: An entire doubly-periodic function is constant. Proof: Let ω1 , ω2 be Z-generators for Λ. Since the ωi are linearly independent over R, every z ∈ C is an R-linear combination of them. Given z = aω1 +bω2 with a, b ∈ R, let m, n be integers such that 0 ≤ a−m < 1 and 0 ≤ b − n < 1. Then ...
congruent numbers and elliptic curves
... Of course, we want to deal with triangles with rational sides as well. Suppose we have a right triangle with sides X, Y, Z ∈ Q and area N . It is easy to see that we can clear denominators and obtain a right triangle with integers sides and congruent number a2 N where a is the least common multiple ...
... Of course, we want to deal with triangles with rational sides as well. Suppose we have a right triangle with sides X, Y, Z ∈ Q and area N . It is easy to see that we can clear denominators and obtain a right triangle with integers sides and congruent number a2 N where a is the least common multiple ...
Semisimple Varieties of Modal Algebras
... Thus, m b ≡β 0 and therefore b ≡β 0. It follows that β ≥ α, contradicting the choice of β as a lower cover of α. a By Lemma 17 all semisimple varieties satisfy (?). Theorem 23 (κ < ω.) All semisimple varieties of modal algebras are weakly transitive. a It follows that semisimple varieties have a de ...
... Thus, m b ≡β 0 and therefore b ≡β 0. It follows that β ≥ α, contradicting the choice of β as a lower cover of α. a By Lemma 17 all semisimple varieties satisfy (?). Theorem 23 (κ < ω.) All semisimple varieties of modal algebras are weakly transitive. a It follows that semisimple varieties have a de ...
8 The Gelfond-Schneider Theorem and Some Related Results
... otherwise wj 6= 0. Let k = d1 + · · · + dn + n. We do induction on k. If k = 1, then n = 1 and d1 = 0, and the lemma easily follows. Let ` ≥ 2 be such that the lemma holds whenever k < `, and suppose k = `. Let N be the number of real roots of F (t). By Rolle’s Theorem, the number of real roots of F ...
... otherwise wj 6= 0. Let k = d1 + · · · + dn + n. We do induction on k. If k = 1, then n = 1 and d1 = 0, and the lemma easily follows. Let ` ≥ 2 be such that the lemma holds whenever k < `, and suppose k = `. Let N be the number of real roots of F (t). By Rolle’s Theorem, the number of real roots of F ...
