We have a famous theorem for a vector space which states that "every set of generators of a vector space $V$ over a field $F$ contains a basis of $V$". Is this statement true for a free module (probably over a PID) instead of the vector space $V$?
2026-04-18 15:51:35.1776527495
On the basis of a free module over a PID
113 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
1
There are 1 best solutions below
Related Questions in ABSTRACT-ALGEBRA
- Feel lost in the scheme of the reducibility of polynomials over $\Bbb Z$ or $\Bbb Q$
- Integral Domain and Degree of Polynomials in $R[X]$
- Fixed points of automorphisms of $\mathbb{Q}(\zeta)$
- Group with order $pq$ has subgroups of order $p$ and $q$
- A commutative ring is prime if and only if it is a domain.
- Conjugacy class formula
- Find gcd and invertible elements of a ring.
- Extending a linear action to monomials of higher degree
- polynomial remainder theorem proof, is it legit?
- $(2,1+\sqrt{-5}) \not \cong \mathbb{Z}[\sqrt{-5}]$ as $\mathbb{Z}[\sqrt{-5}]$-module
Related Questions in MODULES
- Idea to make tensor product of two module a module structure
- $(2,1+\sqrt{-5}) \not \cong \mathbb{Z}[\sqrt{-5}]$ as $\mathbb{Z}[\sqrt{-5}]$-module
- Example of simple modules
- $R$ a domain subset of a field $K$. $I\trianglelefteq R$, show $I$ is a projective $R$-module
- $S_3$ action on the splitting field of $\mathbb{Q}[x]/(x^3 - x - 1)$
- idempotent in quiver theory
- Isomorphism of irreducible R-modules
- projective module which is a submodule of a finitely generated free module
- Exercise 15.10 in Cox's Book (first part)
- direct sum of injective hull of two modules is equal to the injective hull of direct sum of those modules
Trending Questions
- Induction on the number of equations
- How to convince a math teacher of this simple and obvious fact?
- Find $E[XY|Y+Z=1 ]$
- Refuting the Anti-Cantor Cranks
- What are imaginary numbers?
- Determine the adjoint of $\tilde Q(x)$ for $\tilde Q(x)u:=(Qu)(x)$ where $Q:U→L^2(Ω,ℝ^d$ is a Hilbert-Schmidt operator and $U$ is a Hilbert space
- Why does this innovative method of subtraction from a third grader always work?
- How do we know that the number $1$ is not equal to the number $-1$?
- What are the Implications of having VΩ as a model for a theory?
- Defining a Galois Field based on primitive element versus polynomial?
- Can't find the relationship between two columns of numbers. Please Help
- Is computer science a branch of mathematics?
- Is there a bijection of $\mathbb{R}^n$ with itself such that the forward map is connected but the inverse is not?
- Identification of a quadrilateral as a trapezoid, rectangle, or square
- Generator of inertia group in function field extension
Popular # Hahtags
second-order-logic
numerical-methods
puzzle
logic
probability
number-theory
winding-number
real-analysis
integration
calculus
complex-analysis
sequences-and-series
proof-writing
set-theory
functions
homotopy-theory
elementary-number-theory
ordinary-differential-equations
circles
derivatives
game-theory
definite-integrals
elementary-set-theory
limits
multivariable-calculus
geometry
algebraic-number-theory
proof-verification
partial-derivative
algebra-precalculus
Popular Questions
- What is the integral of 1/x?
- How many squares actually ARE in this picture? Is this a trick question with no right answer?
- Is a matrix multiplied with its transpose something special?
- What is the difference between independent and mutually exclusive events?
- Visually stunning math concepts which are easy to explain
- taylor series of $\ln(1+x)$?
- How to tell if a set of vectors spans a space?
- Calculus question taking derivative to find horizontal tangent line
- How to determine if a function is one-to-one?
- Determine if vectors are linearly independent
- What does it mean to have a determinant equal to zero?
- Is this Batman equation for real?
- How to find perpendicular vector to another vector?
- How to find mean and median from histogram
- How many sides does a circle have?
No. Consider the set $\{2,3\}$, which generates $\mathbb{Z}$ as a $\mathbb{Z}$-module; it does not contain a basis of $\mathbb{Z}$.