symmetric monoidal (∞,1)-category of spectra
(also nonabelian homological algebra)
A free module over some ring $R$ is freely generated on a set of basis elements.
Under the interpretation of modules as generalized vector bundles a free module corresponds to a trivial bundle.
Let $C$ be a monoidal category, and $Alg(C)$ the category of monoids in $C$; and for $A \in Alg(C)$ let $A$Mod$(C)$ be the category of $A$-modules in $C$.
There is the evident forgetful functor $U : A Mod(C) \to C$ that sends each module $(N,\rho)$ to its underlying object $N \in C$.
The left adjoint $C \to A Mod(C)$ is the corresponding free construction. The modules in the image of this functor are free modules.
Let $R$ be a ring. We discuss free modules over $R$.
For $R \in$ Ring a ring and $S \in$ Set, the free $R$-module on $S$ is isomorphic to the ${\vert S\vert}$-fold direct sum of $R$ with itself
Let $R$ be a commutative ring, and let $R\{X\}$ denote the free $R$-module on a set $X$.
The free $R$-module functor is strong monoidal with respect to the Cartesian monoidal structure on sets, and the tensor product of $R$-modules.
In other words, the free module construction turns set-theoretic products into tensor products. Thus, it preserves algebraic objects (such as monoid objects, Hopf monoid objects, etc.) and their homomorphisms. In particular, if $M$ is a monoid in the category of sets (and hence a bimonoid with the canonical comonoid structure) then $R\{M\}$ is a bimonoid object in $R \mathsf{Mod}$, which is precisely a $K$-bialgebra. A group $G$ in the category of sets is a Hopf monoid, and hence $R\{G\}$ is a Hopf algebra — this is precisely the group algebra of $G$.
Let $R$ be a commutative ring.
Assuming the axiom of choice, the following are equivalent
every submodule of a free $R$-module is itself free;
every ideal in $R$ is a free $R$-module;
$R$ is a principal ideal domain.
(See also Rotman, pages 650-651.) Condition 1. immediately implies condition 2., since ideals of $R$ are the same as submodules of $R$ seen as an $R$-module. Now assume condition 2. holds, and suppose $x \in R$ is any nonzero element. Let $\lambda_x$ denote multiplication by $x$ (as an $R$-module map). We have a sequence of surjective $R$-module maps
(where $\nabla$ is the codiagonal map); by the Yoneda lemma, the composite map $R \to R$ is of the form $\lambda_r$, where $r \in R$ is the value of the composite at $1 \in R$. Since $\lambda_r$ is surjective, we have $\lambda_r(s) = r s = 1$ for some $s$, so that $r$ is invertible. Hence $\lambda_r$ is invertible, and this implies $\lambda_x$ is monic. Therefore $R$ is a domain. From that, we infer that if $f$ and $g$ belong to a basis of an ideal $I$, then
whence $f$ and $g$ are not linearly independent, so $f = g$ and $I$ as an $R$-module is generated by a single element, i.e., $R$ is a principal ideal domain.
That condition 3. implies condition 1. is proved here.
Assuming the axiom of choice, over a ring $R$ which is a principal ideal domain, every module has a projective resolution of length 1.
See at projective resolution – Resolutions of length 1 for more.
Assuming the axiom of choice, if $R = k$ is a field then every $R$-module is free: it is $k$-vector space and by the basis theorem every such has a basis.
finitely generated module, finitely presented module?
free module $\Rightarrow$ projective module $\Rightarrow$ flat module $\Rightarrow$ torsion-free module
projective object, projective presentation, projective cover, projective resolution
injective object, injective presentation, injective envelope, injective resolution
flat object, flat resolution
Textbooks:
Last revised on January 5, 2021 at 05:38:23. See the history of this page for a list of all contributions to it.