Let R be a commutative ring, and let F be a free R-module. (i) If m is a maximal ideal in R, pro

step 1 Hello there. Okay, so in this exercise we have three vector spaces U, V and W, okay, and we have

Let R be a commutative ring, and let F be a free R-module....

Let $R$ be a commutative ring, and let $F$ be a free $R$-module. (i) If $\mathrm{m}$ is a maximal ideal in $R$, prove that $(R / \mathrm{m}) \otimes_{R} F$ and $F / \mathrm{m} F$ are isomorphic as vector spaces over $R / \mathrm{m}$. (ii) Prove that $\operatorname{rank}(F)=\operatorname{dim}\left((R / \mathrm{m}) \otimes_{R} F\right)$. (iii) If $R$ is a domain with fraction field $Q$, prove that $\operatorname{rank}(F)=$ $\operatorname{dim}\left(Q \otimes_{R} F\right)$.

Let $R$ be a commutative ring, and let $F$ be a free $R$-module.
(i) If $\mathrm{m}$ is a maximal ideal in $R$, prove that $(R / \mathrm{m}) \otimes_{R} F$ and $F / \mathrm{m} F$ are isomorphic as vector spaces over $R / \mathrm{m}$.
(ii) Prove that $\operatorname{rank}(F)=\operatorname{dim}\left((R / \mathrm{m}) \otimes_{R} F\right)$.
(iii) If $R$ is a domain with fraction field $Q$, prove that $\operatorname{rank}(F)=$ $\operatorname{dim}\left(Q \otimes_{R} F\right)$.

Key Concepts

Free Module

A free module is an R-module that has a basis; that is, every element of the module can be written uniquely as a finite linear combination of some set of elements called the basis. This concept mirrors the idea of a basis in vector spaces and is fundamental for defining the 'rank' of the module, which is the number of elements in such a basis.

Tensor Product

The tensor product of modules over a ring is a construction that allows one to 'change scalars' or extend modules to a new ring. It takes two modules over a commutative ring and produces another module that encapsulates all bilinear maps from the product of the modules. Tensoring with a quotient ring or a localization, for example, transforms certain module structures into vector spaces.

Module Rank

The rank of a free module is the cardinality of its basis, and it serves as a key invariant. When one tensor a free module with a field (obtained by quotienting by a maximal ideal or by localizing to the fraction field), the resulting vector space's dimension equals the rank of the original module, thereby linking module theory with linear algebra.

Quotient Module

A quotient module is formed by taking a module and factoring out a submodule. For example, given an ideal m in a ring R, the module F/mF is the set of cosets of F modulo the submodule mF. This construction is essential in bridging module theory with vector space theory when the quotient R/m has the structure of a field.

Maximal Ideal and Field Structure

A maximal ideal m in a commutative ring R has the property that the quotient ring R/m forms a field. This fact is pivotal when tensoring a module with R/m, since it endows the tensor product with the structure of a vector space over a field, allowing for the application of linear algebraic methods to study the module.

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