Example sheet 4
... 3. Let M be a module over an integral domain R. An element m ∈ M is a torsion element if rm = 0 for some non-zero r ∈ R. (i) Show that the set T of all torsion elements in M is a submodule of M , and that the quotient M/T is torsion-free—that is, contains no non-zero torsion elements. (ii) Is the Z- ...
... 3. Let M be a module over an integral domain R. An element m ∈ M is a torsion element if rm = 0 for some non-zero r ∈ R. (i) Show that the set T of all torsion elements in M is a submodule of M , and that the quotient M/T is torsion-free—that is, contains no non-zero torsion elements. (ii) Is the Z- ...
Solutions - NIU Math
... 0, which gives us the identity αn = −bn−1 αn−1 − · · · − b1 α − b0 . Multiplying this identity by α and substituting for αn gives us an identity expressing αn+1 in terms of αn−1 , αn−2 , . . . , α, 1. We can find similar identities for αn+2 , . . . , α2n−2 . In multiplying two elements of Z[α], the ...
... 0, which gives us the identity αn = −bn−1 αn−1 − · · · − b1 α − b0 . Multiplying this identity by α and substituting for αn gives us an identity expressing αn+1 in terms of αn−1 , αn−2 , . . . , α, 1. We can find similar identities for αn+2 , . . . , α2n−2 . In multiplying two elements of Z[α], the ...
WeekFive - Steve Watson
... If was inconsistent then |_ and |_ for some and |= and |= which is impossible. We have a proof of completeness for a language with cardinality that proceeds by creating a structure which has a cardinality at least , because that is the number of new constant symbols that are ...
... If was inconsistent then |_ and |_ for some and |= and |= which is impossible. We have a proof of completeness for a language with cardinality that proceeds by creating a structure which has a cardinality at least , because that is the number of new constant symbols that are ...
Chapter 7- counting techniques
... formed from a collection of n objects where the first object appears k1 times, the second object appears k2 times, and so on, is: n! ...
... formed from a collection of n objects where the first object appears k1 times, the second object appears k2 times, and so on, is: n! ...
Chapter 2: Sets
... Definition 3. Two sets A and B are equal (written A = B) if they have exactly the same members or elements. Two finite sets A and B are said to be equivalent (written A ∼ = B) if they have the same number of elements: that is, n(A) = n(B) a) {a, b, c} and {c, b, a} are equal. b) the set of all names ...
... Definition 3. Two sets A and B are equal (written A = B) if they have exactly the same members or elements. Two finite sets A and B are said to be equivalent (written A ∼ = B) if they have the same number of elements: that is, n(A) = n(B) a) {a, b, c} and {c, b, a} are equal. b) the set of all names ...
(ID ÈÈ^i+i)f(c)viVi.
... This is substantially equivalent to a definition given by H. Pnifer.f If 5 is a ring, every ovoid ideal is a Dedekind ideal, but the converse is not necessarily true. Ideals will be denoted by small German letters. An ideal is called a principal ideal if it is identical with the set of multiples of ...
... This is substantially equivalent to a definition given by H. Pnifer.f If 5 is a ring, every ovoid ideal is a Dedekind ideal, but the converse is not necessarily true. Ideals will be denoted by small German letters. An ideal is called a principal ideal if it is identical with the set of multiples of ...
Spaces with regular $ G_\ delta $
... By Theorem 3, there is a development {Sn} for X that satisfies the 3-link property. Since X is locally connected, we may assume that, for each n, the members of 9m are connected. Let x denote a point of X and let U be an open set containing x. We will show that there is an integer n such that if g e ...
... By Theorem 3, there is a development {Sn} for X that satisfies the 3-link property. Since X is locally connected, we may assume that, for each n, the members of 9m are connected. Let x denote a point of X and let U be an open set containing x. We will show that there is an integer n such that if g e ...
Problem Score 1 2 3 4 or 5 Total - Mathematics
... (a) Z[X]/(X 2 + 1) and Z ⊕ Z are isomorphic as Z-modules and as rings. (b) Q[X]/(X 2 − X − 1) and Q[X]/(X 2 − 1) are isomorphic as rings and Q-vector spaces. Solution: (a) Z[X]/(X 2 + 1) ' Z[ i ] and Z ⊕ Z are isomorphic as abelian groups (i.e. as Z-modules) in fact ϕ : Z[ i ] → Z ⊕ Z, ϕ(a + bi) = ( ...
... (a) Z[X]/(X 2 + 1) and Z ⊕ Z are isomorphic as Z-modules and as rings. (b) Q[X]/(X 2 − X − 1) and Q[X]/(X 2 − 1) are isomorphic as rings and Q-vector spaces. Solution: (a) Z[X]/(X 2 + 1) ' Z[ i ] and Z ⊕ Z are isomorphic as abelian groups (i.e. as Z-modules) in fact ϕ : Z[ i ] → Z ⊕ Z, ϕ(a + bi) = ( ...
µ-Compactness with Respect to a Hereditary Class Key Words
... 3. µ-compactness with respect to an ideal In this section we introduce and study the notion of compactness with respect to a GTS and a hereditary class in order to obtain many well known generalized forms of compactness in the literature. Definition 3.1. Let (X, µ) be a GTS and H be a hereditary cla ...
... 3. µ-compactness with respect to an ideal In this section we introduce and study the notion of compactness with respect to a GTS and a hereditary class in order to obtain many well known generalized forms of compactness in the literature. Definition 3.1. Let (X, µ) be a GTS and H be a hereditary cla ...
Profinite Heyting algebras
... ↓U is clopen for every clopen U ⊆ X Let Upτ (X) denote the Heyting algebra of clopen upsets of X, where U → V = X− ↓(U − V ). Theorem (Esakia 1974). For every Heyting algebra A, there exists an Esakia space (X, ≤) such that A is isomorphic to Upτ (X). ...
... ↓U is clopen for every clopen U ⊆ X Let Upτ (X) denote the Heyting algebra of clopen upsets of X, where U → V = X− ↓(U − V ). Theorem (Esakia 1974). For every Heyting algebra A, there exists an Esakia space (X, ≤) such that A is isomorphic to Upτ (X). ...
Subrings of the rational numbers
... THEOREM. Suppose that A is a subdomain of the rational numbers. Then there is a set of primes S such that A is isomorphic to the ring Z S generated by the integers and the inverses of all elements of Z. The ring ZS consists of all fractions of the form a/b where a is an integer and b is a monomial i ...
... THEOREM. Suppose that A is a subdomain of the rational numbers. Then there is a set of primes S such that A is isomorphic to the ring Z S generated by the integers and the inverses of all elements of Z. The ring ZS consists of all fractions of the form a/b where a is an integer and b is a monomial i ...
Birkhoff's representation theorem
This is about lattice theory. For other similarly named results, see Birkhoff's theorem (disambiguation).In mathematics, Birkhoff's representation theorem for distributive lattices states that the elements of any finite distributive lattice can be represented as finite sets, in such a way that the lattice operations correspond to unions and intersections of sets. The theorem can be interpreted as providing a one-to-one correspondence between distributive lattices and partial orders, between quasi-ordinal knowledge spaces and preorders, or between finite topological spaces and preorders. It is named after Garrett Birkhoff, who published a proof of it in 1937.The name “Birkhoff's representation theorem” has also been applied to two other results of Birkhoff, one from 1935 on the representation of Boolean algebras as families of sets closed under union, intersection, and complement (so-called fields of sets, closely related to the rings of sets used by Birkhoff to represent distributive lattices), and Birkhoff's HSP theorem representing algebras as products of irreducible algebras. Birkhoff's representation theorem has also been called the fundamental theorem for finite distributive lattices.