Proof: Let $A=\{a_1,...,a_n\}$ generate $N$, and let $B=\{b_1,...,b_m\}$ be such that $\overline{B}$ generates $M/N$. We claim $A \cup B$ generates $M$. Note that for an arbitrary element $m∈M$ we have $\overline{m}=\overline{r_1b_1+...+r_mb_m}$ for some $r_i∈R$ so that $m=r_1b_1+...+r_mb_m+n$ for some $n∈N$, so that $m=r_1b_1+...+r_mb_m+s_1a_1+...+s_na_n$ for some $s_i∈R$, and now $M$ is generated by $A \cup B$.$~\square$
Tuesday, June 4, 2013
Quotients and Finite Generation of Modules (10.3.7)
Dummit and Foote Abstract Algebra, section 10.3, exercise 7:
MathJax TeX Test Page
If $N$ is a submodule of the $R$-module $M$, prove that if $M/N$ and $N$ are finitely generated, then so is $M$.
Proof: Let $A=\{a_1,...,a_n\}$ generate $N$, and let $B=\{b_1,...,b_m\}$ be such that $\overline{B}$ generates $M/N$. We claim $A \cup B$ generates $M$. Note that for an arbitrary element $m∈M$ we have $\overline{m}=\overline{r_1b_1+...+r_mb_m}$ for some $r_i∈R$ so that $m=r_1b_1+...+r_mb_m+n$ for some $n∈N$, so that $m=r_1b_1+...+r_mb_m+s_1a_1+...+s_na_n$ for some $s_i∈R$, and now $M$ is generated by $A \cup B$.$~\square$
Proof: Let $A=\{a_1,...,a_n\}$ generate $N$, and let $B=\{b_1,...,b_m\}$ be such that $\overline{B}$ generates $M/N$. We claim $A \cup B$ generates $M$. Note that for an arbitrary element $m∈M$ we have $\overline{m}=\overline{r_1b_1+...+r_mb_m}$ for some $r_i∈R$ so that $m=r_1b_1+...+r_mb_m+n$ for some $n∈N$, so that $m=r_1b_1+...+r_mb_m+s_1a_1+...+s_na_n$ for some $s_i∈R$, and now $M$ is generated by $A \cup B$.$~\square$
Subscribe to:
Post Comments (Atom)
No comments:
Post a Comment