8.2 Closure of a Set Under an Operation
... Consider the set {0, 1, 2, 3, ...}, which are called the whole numbers. Notice that if we add or multiply any two whole numbers the result is also a whole number, but if we try subtracting two such numbers it is possible to get a number that is not in the set. We say that the whole numbers are close ...
... Consider the set {0, 1, 2, 3, ...}, which are called the whole numbers. Notice that if we add or multiply any two whole numbers the result is also a whole number, but if we try subtracting two such numbers it is possible to get a number that is not in the set. We say that the whole numbers are close ...
Solutions to Homework 1
... just showed gives 1 = xx−1 ≤ x · 0 = 0, which contradicts (xiii). Now it follows from what we showed above that (ax)x−1 ≤ (bx)x−1 =⇒ a ≤ b. The case of equality is similar. (xiv) This follows from (xii). ...
... just showed gives 1 = xx−1 ≤ x · 0 = 0, which contradicts (xiii). Now it follows from what we showed above that (ax)x−1 ≤ (bx)x−1 =⇒ a ≤ b. The case of equality is similar. (xiv) This follows from (xii). ...
Solutions - math.miami.edu
... in the sum i+j=k ai bj is zero, hence the coefficient of xk in f (x)g(x) is zero. We conclude that the degree of f (x)g(x) is m + n. For part (b), consider f, g ∈ R, both nonzero. We wish to show that f g is nonzero. If deg(f ) = deg(g) = 0 then f, g are constants and the fact that f g 6= 0 follows ...
... in the sum i+j=k ai bj is zero, hence the coefficient of xk in f (x)g(x) is zero. We conclude that the degree of f (x)g(x) is m + n. For part (b), consider f, g ∈ R, both nonzero. We wish to show that f g is nonzero. If deg(f ) = deg(g) = 0 then f, g are constants and the fact that f g 6= 0 follows ...
ALGEBRA HANDOUT 2: IDEALS AND
... is by noticing that Z[i]/(2) contains Z/2Z = {0, 1} as a subring, and is generated by adjoining to Z/2Z the single element r = 1 + i. Thus one can show that it is isomorphic to the ring Z/2Z[X]/(X 2 ), or, if you like, the ring Z[X]/(2, X 2 ). It is natural to wonder what the ring structure on the n ...
... is by noticing that Z[i]/(2) contains Z/2Z = {0, 1} as a subring, and is generated by adjoining to Z/2Z the single element r = 1 + i. Thus one can show that it is isomorphic to the ring Z/2Z[X]/(X 2 ), or, if you like, the ring Z[X]/(2, X 2 ). It is natural to wonder what the ring structure on the n ...
THE HILBERT SCHEME PARAMETERIZING FINITE LENGTH
... 2. Roots of F ϕ (x) and invertible elements in A[x]/(F (x)). 2.1. Notation. We shall use the notation of Sections (1.1) and (1.2). Let A be a ring and let P be a prime ideal. We write κ(P ) = AP /P AP for the residue field. Let k be a field and assume that A is a k–algebra. Denote by k[x](x) the lo ...
... 2. Roots of F ϕ (x) and invertible elements in A[x]/(F (x)). 2.1. Notation. We shall use the notation of Sections (1.1) and (1.2). Let A be a ring and let P be a prime ideal. We write κ(P ) = AP /P AP for the residue field. Let k be a field and assume that A is a k–algebra. Denote by k[x](x) the lo ...
enumerating polynomials over finite fields
... A polynomial is irreducible if it is of positive degree and cannot be factored into polynomials of strictly smaller degree. So for instance every polynomial of degree one is irreducible. In fact, every polynomial can be uniquely factored into irreducible polynomials (possibly repeated). More precise ...
... A polynomial is irreducible if it is of positive degree and cannot be factored into polynomials of strictly smaller degree. So for instance every polynomial of degree one is irreducible. In fact, every polynomial can be uniquely factored into irreducible polynomials (possibly repeated). More precise ...
Solution
... Ia J ⊆ I by the definition of J, and it is generated by αβ by the definition of multiplication of principal ideals. (c) Let x ∈ I. Ia ⊇ I, so x ∈ Ia ; thus in particular, x = sα for some s. But since sα ∈ I this means that sIa ⊆ I, and s ∈ J, as desired. We have shown that I = Ia J = (αβ), so it is ...
... Ia J ⊆ I by the definition of J, and it is generated by αβ by the definition of multiplication of principal ideals. (c) Let x ∈ I. Ia ⊇ I, so x ∈ Ia ; thus in particular, x = sα for some s. But since sα ∈ I this means that sIa ⊆ I, and s ∈ J, as desired. We have shown that I = Ia J = (αβ), so it is ...
