----- > [!proposition] Proposition. ([[every ideal in a Noetherian ring contains a power of its radical]]) > Let $I$ be an [[ideal]] of a [[Noetherian ring]] $R$. Then there exists $n \geq 1$ such that $(\sqrt{ I })^{n} \subset I$, where $\sqrt{ I }$ denotes the [[radical of an ideal|radical]] of $I$. ^proposition > [!proposition] Corollary. > The [[nilradical of a ring|nilradical]] $\text{Nil }R=\sqrt{ (0) }$ is a [[nilpotent ideal]], as $\text{Nil }^{\ell}R \subset (0)$ for some $\ell \geq 1$. ^proposition > [!proof]- Proof. ([[every ideal in a Noetherian ring contains a power of its radical]]) > $\sqrt{ I }$ is [[ideal generated by a subset|finitely generated]] by hypothesis; write $\sqrt{ I }=\langle r_{1},\dots,r_{m} \rangle$, $m\geq0$, with $r_{1}^{e_{1}} \in I, \dots, r_{m}^{e_{m}} \in I$ for $e_{1},\dots,e_{m} \geq 0$. Note that, in general, $(\sqrt{ I })^{n}$ is generated by elements of the form $r_{1}^{\ell_{1}} \cdots r_{m}^{\ell_{m}}$ for $\ell_{1}+\dots+\ell_{m}=n$. > *Claim: putting $n=e_{1} + \dots + e_{m}$ proves the property.* > Indeed, it is enough to show that each generator of $(\sqrt{ I })^{e_{1}+\dots+e_{m}}$ lies in $I$. So let $r_{1}^{\ell_{1}} \cdots r_{m}^{\ell_{m}}$ such that $\ell_{1}+\dots+\ell_{m}=e_{1}+\dots+e_{m}$. It can't be the case that $\ell_{k}<e_{k}$ for all $k \in [m]$, hence for some $k$ we have $\ell_{k} \geq e_{k}$ and therefore $r_{k}^{\ell_{k}} \in I$. Then since $I$ is an ideal, the whole of $r_{1}^{\ell_{1}} \cdots r_{m}^{\ell_{m}}$is contained in $I$. ----- #### ---- #### 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 > ```