
2. EUCLIDEAN RINGS
... cancellation law breaks down. This is a large and important subject but, so far, I haven’t written any notes on it. An important sub-department of this is the theory of integral domains and Euclidean rings where there is a divisibility theory, just like we have for the integers – where there are pri ...
... cancellation law breaks down. This is a large and important subject but, so far, I haven’t written any notes on it. An important sub-department of this is the theory of integral domains and Euclidean rings where there is a divisibility theory, just like we have for the integers – where there are pri ...
Factorization of C-finite Sequences - Institute for Algebra
... Let us now drop the condition that r ∈ k[x] is square free. Write r∗ for the square free part of r. It is clear from equation (1) that when p, q ∈ k[x] are such that r = p ⊗ q, then r∗ = p∗ ⊗ q ∗ , where p∗ , q ∗ denote the square free parts of p and q, respectively. It is therefore natural to first ...
... Let us now drop the condition that r ∈ k[x] is square free. Write r∗ for the square free part of r. It is clear from equation (1) that when p, q ∈ k[x] are such that r = p ⊗ q, then r∗ = p∗ ⊗ q ∗ , where p∗ , q ∗ denote the square free parts of p and q, respectively. It is therefore natural to first ...
Homework # 7 Solutions
... 7. Let a ∈ Z. If (a + 1)2 − 1 is even, then a is even. Solution: Proof. (contrapositive) Suppose that a is an odd integer. Then a = 2k + 1 for some integer k. So (a + 1)2 − 1 = (2k + 2)2 − 1 = 4k 2 + 8k + 3 = 4k 2 + 8k + 2 + 1 = 2(2k 2 + 4k + 1) + 1. Since 2k 2 + 4k + 1 is an integer, (a + 1)2 − 1 ...
... 7. Let a ∈ Z. If (a + 1)2 − 1 is even, then a is even. Solution: Proof. (contrapositive) Suppose that a is an odd integer. Then a = 2k + 1 for some integer k. So (a + 1)2 − 1 = (2k + 2)2 − 1 = 4k 2 + 8k + 3 = 4k 2 + 8k + 2 + 1 = 2(2k 2 + 4k + 1) + 1. Since 2k 2 + 4k + 1 is an integer, (a + 1)2 − 1 ...
Finite Fields - (AKA Galois Fields)
... The degree of a constant polynomial is 0, except that by special agreement we define the degree of the zero polynomial to be −1 or sometimes −∞. ...
... The degree of a constant polynomial is 0, except that by special agreement we define the degree of the zero polynomial to be −1 or sometimes −∞. ...