# module homomorphism, ring, subset

• Nov 18th 2008, 08:11 PM
Erdos32212
module homomorphism, ring, subset
Background: Let $\displaystyle R$ be a ring, $\displaystyle S \subseteq R$ a multiplicatively closed subset, and let $\displaystyle M$, $\displaystyle N$ be $\displaystyle R-$modules. Let $\displaystyle \phi : M \rightarrow N$ be an $\displaystyle R-$module homomorphism, and let $\displaystyle \phi_s : S^{-1}M \rightarrow S^{-1}N$ denote the $\displaystyle S^{-1}R-$module homomorphism, and let $\displaystyle \phi_s(\frac{m}{s})=\frac{\phi(m)}{s}.$ You may assume without proof that $\displaystyle \phi_s$ is a well-defined $\displaystyle S^{-1}R-$module homomorphism.

Prove:
(a) Prove that if $\displaystyle \phi$ is injective, then $\displaystyle \phi_s$ is injective.
(b) Prove that if $\displaystyle \phi$ is surjective, then $\displaystyle \phi_s$ is surjective.
(c) Prove that if $\displaystyle M$ is a finitely generated free $\displaystyle R-$module with basis $\displaystyle \{ b_1,\ldots, b_m \}$ then $\displaystyle S^{-1}M$ is a finitely generated free $\displaystyle S^{-1}R-$module with basis $\displaystyle \{ \frac{b_1}{1},\ldots, \frac{b_m}{1} \}.$
(d) Now suppose $\displaystyle R$ is an integral domain and that $\displaystyle m$, $\displaystyle n$ are positive integers. Prove that if $\displaystyle R^m \cong R^n$, then $\displaystyle m=n$. (Hint: consider $\displaystyle S=R-\{0\}$.)
• Nov 18th 2008, 10:00 PM
NonCommAlg
Quote:

Originally Posted by Erdos32212
Background: Let $\displaystyle R$ be a ring, $\displaystyle S \subseteq R$ a multiplicatively closed subset, and let $\displaystyle M$, $\displaystyle N$ be $\displaystyle R-$modules. Let $\displaystyle \phi : M \rightarrow N$ be an $\displaystyle R-$module homomorphism, and let $\displaystyle \phi_s : S^{-1}M \rightarrow S^{-1}N$ denote the $\displaystyle S^{-1}R-$module homomorphism, and let $\displaystyle \phi_s(\frac{m}{s})=\frac{\phi(m)}{s}.$ You may assume without proof that $\displaystyle \phi_s$ is a well-defined $\displaystyle S^{-1}R-$module homomorphism.

Prove:
(a) Prove that if $\displaystyle \phi$ is injective, then $\displaystyle \phi_s$ is injective.
(b) Prove that if $\displaystyle \phi$ is surjective, then $\displaystyle \phi_s$ is surjective.
(c) Prove that if $\displaystyle M$ is a finitely generated free $\displaystyle R-$module with basis $\displaystyle \{ b_1,\ldots, b_m \}$ then $\displaystyle S^{-1}M$ is a finitely generated free $\displaystyle S^{-1}R-$module with basis $\displaystyle \{ \frac{b_1}{1},\ldots, \frac{b_m}{1} \}.$
(d) Now suppose $\displaystyle R$ is an integral domain and that $\displaystyle m$, $\displaystyle n$ are positive integers. Prove that if $\displaystyle R^m \cong R^n$, then $\displaystyle m=n$. (Hint: consider $\displaystyle S=R-\{0\}$.)

(a): if $\displaystyle \phi_S(m/s)=0,$ then $\displaystyle \frac{\phi(m)}{s}=0.$ thus there exists $\displaystyle t \in S$ such that $\displaystyle t \phi(m)=0.$ hence: $\displaystyle \phi(tm)=t\phi(m)=0.$ but $\displaystyle \phi$ is injective. thus $\displaystyle tm=0,$ i.e. $\displaystyle \frac{m}{s}=0.$

(b) if $\displaystyle x=\frac{n}{s} \in S^{-1}N,$ then since $\displaystyle \phi$ is surjective, there exists $\displaystyle m \in M$ such that $\displaystyle \phi(m)=n.$ thus: $\displaystyle x=\frac{\phi(m)}{s}=\phi_S(m/s).$ hence $\displaystyle \phi_S$ is surjective.

(c) let $\displaystyle x=\frac{v}{s} \in S^{-1}M.$ then $\displaystyle v=\sum_{i=1}^m r_ib_i, \ r_i \in R.$ thus: $\displaystyle x=\sum_{i=1}^m \frac{r_i}{s}\frac{b_i}{1},$ which proves that $\displaystyle I=\{\frac{b_i}{1} : i=1,2, \cdots m \}$ generates $\displaystyle S^{-1}M$ as $\displaystyle S^{-1}R$ module, because $\displaystyle \frac{r_i}{s} \in S^{-1}R.$

to prove that the set $\displaystyle I$ is linearly independent, we assume that $\displaystyle \sum_{i=1}^m \frac{r_i}{s_i}\frac{b_i}{1}=0,$ which after clearing denominators gives us: $\displaystyle \sum_{i=1}^n t_ir_ib_i=0,$ where $\displaystyle t_i \in S.$ since $\displaystyle \{b_i: \ i=1,2, \cdots, m\}$

is assumed to be a basis for M, it's a set of linearly independent elements of R. thus $\displaystyle t_i r_i = 0, \ i = 1, 2, \cdots , m.$ hence $\displaystyle \frac{r_i}{s_i}=0, \ i=1,2, \cdots , m.$

(d) let $\displaystyle S=R- \{0 \},$ which is obviously multiplicatively closed because R is a domain. now $\displaystyle Q(R)=S^{-1}R$ is the field of fractions of $\displaystyle R.$ on the other hand we have:

$\displaystyle \bigoplus_{i=1}^m Q(R)=\bigoplus_{i=1}^m S^{-1}R \simeq S^{-1}R^m \simeq S^{-1}R^n \simeq \bigoplus_{i=1}^n S^{-1}R=\bigoplus_{i=1}^n Q(R). \ \ \ \ (1)$

now since $\displaystyle Q(R)$ is a field, the LHS and RHS of (1) are two $\displaystyle Q(R)-$vector spaces. since they are isomorphic, their dimensions must be equal. (recall from linear algebra!) thus $\displaystyle m=n.$