The Irrationality of Pi and Various Trigonometric Values
... Recall in (3) that F 0 (α) = 0. Observe that f (k) (0) = 0 if 0 ≤ k < n − 1 and, from Lemma 1, f (k) (0) is divisible by n if k > n − 1. Also, f (n−1) (0) = a2n (2a)n−1 . We choose n = p where p is a sufficiently large prime so that the integral in (3) is 0 as above and so that p > max{2a, d}. Then ...
... Recall in (3) that F 0 (α) = 0. Observe that f (k) (0) = 0 if 0 ≤ k < n − 1 and, from Lemma 1, f (k) (0) is divisible by n if k > n − 1. Also, f (n−1) (0) = a2n (2a)n−1 . We choose n = p where p is a sufficiently large prime so that the integral in (3) is 0 as above and so that p > max{2a, d}. Then ...
4.2 Every PID is a UFD
... a = p1 p2 . . . pr and a = q1 q2 . . . qs where each pi and qj is irreducible in R, and s ≥ r. Then p1 divides the product q1 . . . qs , and so p1 divides qj for some j, as p1 is prime. After reordering the qj if necessary we can suppose p1 |q1 . Then q1 = u1 p1 for some unit u1 of R, since q1 and p ...
... a = p1 p2 . . . pr and a = q1 q2 . . . qs where each pi and qj is irreducible in R, and s ≥ r. Then p1 divides the product q1 . . . qs , and so p1 divides qj for some j, as p1 is prime. After reordering the qj if necessary we can suppose p1 |q1 . Then q1 = u1 p1 for some unit u1 of R, since q1 and p ...
Solutions — Ark 1
... Solution: We shall show that (XY − ZW ) is irreducible, and then appeal to the previous exercise. Assume therefore that P Q = (XY − ZW ). We may write P = P0 + P1 + · · · + Pn and Q = Q0 + Q1 + · · · + Qm where the Pi ’s and the Qi ’s are homogenous polynomials of degree i. If Pn and Qm both are not ...
... Solution: We shall show that (XY − ZW ) is irreducible, and then appeal to the previous exercise. Assume therefore that P Q = (XY − ZW ). We may write P = P0 + P1 + · · · + Pn and Q = Q0 + Q1 + · · · + Qm where the Pi ’s and the Qi ’s are homogenous polynomials of degree i. If Pn and Qm both are not ...
78 Topics in Discrete Mathematics Example 3.5. According to these
... (1 − z)k+1 n=0 (You would normally change the summation sothat it started from n = k, but in this case it makes no difference because nk = 0 for n = 0, 1, · · · , k−1.) Since we can express nm as a linear combination of binomial coefficients nk for various k by Theorem 1.86, we can use (3.9) to ge ...
... (1 − z)k+1 n=0 (You would normally change the summation sothat it started from n = k, but in this case it makes no difference because nk = 0 for n = 0, 1, · · · , k−1.) Since we can express nm as a linear combination of binomial coefficients nk for various k by Theorem 1.86, we can use (3.9) to ge ...
Solutions to Homework 7 27. (Dummit
... clearly non-zero. Since K is a field it has no non-zero ideals and thus our map is injective. Since it is obviously surjective, we are done. (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 ...
... clearly non-zero. Since K is a field it has no non-zero ideals and thus our map is injective. Since it is obviously surjective, we are done. (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 ...
Finite Fields
... We now need to link the additive structure of a finite field coming from the vector space interpretation and the multiplicative structure coming from the representation of all nonzero elements by the powers of a primitive element. Theorem 1.14. Let p be a prime and P be an irreducible polynomial of ...
... We now need to link the additive structure of a finite field coming from the vector space interpretation and the multiplicative structure coming from the representation of all nonzero elements by the powers of a primitive element. Theorem 1.14. Let p be a prime and P be an irreducible polynomial of ...