# module homomorphism, ring, subset

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

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

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

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

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

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

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

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

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

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

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

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