----- $R$ is a ([[commutative ring|commutative]]) [[ring]], and $J(R)$ its [[Jacobson radical]]. > [!proposition] Proposition. ([[Nakayama's Lemma]]) > Let $M$ be a [[submodule generated by a subset|finitely generated]] $R$-[[module]], and $\mathfrak{a} \subset J(R)$ an [[ideal]] of $R$ such that $\mathfrak{a}M=M.$ > > Then $M=0$. ^proposition > [!proposition] Corollary. ($J(R)M$ is small when $M$ is finitely generated) > Let $M$ be a [[submodule generated by a subset|finitely generated]] $R$-[[module]], $N \subset M$ a [[submodule]], $\mathfrak{a} \subset J(R)$ an [[ideal]] of $R$ such that $\mathfrak{a}M+N=M$. Then $N=M$. > > Indeed, we have $\mathfrak{a}(M / N)=(\underbrace{\mathfrak{a}M+N}_{=M}) / N=M / N,$ > and so $M / N=0$ by Nakayama's Lemma. ^proposition > [!note] Remark. > The more larger the [[Jacobson radical]], the more useful Nakayama's Lemma is. It is most useful for a [[local ring]] $(R,\mathfrak{m})$, since then $J(R)=\mathfrak{m}$ is very large. ^note > [!proof]+ Proof. ([[Nakayama's Lemma]]) > Using the corollary in [[Cayley-Hamilton Theorem]], fix $a \in\mathfrak{a}$ such that $am=m$ for all $m \in M$, i.e., $(1-a)m=0$ for all $m \in M$. Because $a \in J(R)$, $1-a$[^1] is a [[unit]]. Thus we may divide by it, finding that $m=0$ for all $m \in M$. ----- #### [^1]: If it helps: $1-a=1-xy$ where $x=a$ and $y=1$. ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```