----
> [!definition] Definition. ([[radical of an ideal]])
> The **radical** of an [[ideal]] $I$ of a [[commutative ring|commutative]] [[ring]] $R$ the ideal $\sqrt{ I }:=\{ r \in R :r^{n} \in I \text{ for some } n \in \mathbb{N} \}.$
>
> We say $I$ is a **radical ideal** if $\sqrt{ I }=I$.
^definition
> [!basicexample]
> - [[nilradical of a ring]]
> - $\sqrt{ (12) } \subset \mathbb{Z}$ is equal to $(6)$.
> - any [[maximal ideal|maximal]] $\implies$ [[prime ideal|prime]] $\implies$ [[radical of an ideal|radical ideal]]
^basic-example
> [!basicproperties]
> - Assume $I=\sqrt{ I }$ is radical. If $x,y \in R$ and $xy \in I$, then $I=\sqrt{ I + \langle x \rangle } \cap \sqrt{ I+\langle y \rangle }$.
>
> >[!proof]- Proof.
> >
> >
> >
> > $\subset.$ Let $a \in I$. Then $a+xy \in I$. So $a=i+xy$ for some $i \in I$. Since $i+xy \in I + \langle x \rangle \subset \sqrt{ I + \langle x \rangle }$ and likewise $i+xy \in I+\langle y \rangle \subset \sqrt{ I + \langle y \rangle }$, we have $a \in \sqrt{ I + \langle x \rangle } \cap \sqrt{ I+\langle y \rangle }$.
> >
> > $\supset$. Let $b \in \sqrt{ I + \langle x \rangle } \cap \sqrt{ I+\langle y \rangle }$: fix $n,m \geq 0$ such that $b^{n}=i_{x}+rx$ and $b^{m}=i_{y}+sy$ for some $i_{x}, i_{y} \in I$ and $r,s \in R$. Then $\begin{align}
> > b^{n+m}=b^{n}b^{m}&= (i_{x}+rx)^{n} \cdot (i_{y}+sy)^{m} \\
> > &= \left( \sum_{k=0}^{n} {n \choose k} i_x^{n-k} r^{k}x^{k} \right) \cdot \left( \sum_{\ell=0}^{m} {m \choose \ell} i_y^{m-\ell} s^{\ell}y^{\ell} \right) \\
> > &= \sum_{k=0}^{n} \sum_{\ell=0}^{m} {n \choose k}{m \choose \ell} \underbrace{i_x^{n-k} r^{k}x^{k} i_y^{m-\ell} s^{\ell}y^{\ell} }_{\in I} \\
> > &\in I,
> > \end{align}$
> > implying $b \in I$ since $I=\sqrt{ I }$ is radical. Note that we needed $xy \in I$ here to handle the term where $k=n$ and $\ell=m$.
> > (definitely there are quicker ways to convince oneself of this. e.g. $b^{n} b^{m} \in (I+\langle x \rangle)(I + \langle y \rangle)=I + \langle xy \rangle = I$.)
> - $\sqrt{ I }=\bigcap_{\mathfrak{p} \supset I \text{ prime }}^{}\mathfrak{p}$, per the corollary [[nilradical equals intersection of all prime ideals|here]].
>
> - $\sqrt{ I+J }=\sqrt{ \sqrt{ I } + \sqrt{ J } }$
>
> > [!proof]-
> > $\subset.$ Take $a \in \sqrt{ I+J }$, say, $a^{\ell}=i+j$ for some $\ell$, $i \in I, j \in J$. Since $i+j=i^{1}+j^{1}$, $a^{\ell}\in { \sqrt{ I } + \sqrt{ J } }$.
> >
> > $\supset.$ Take $r \in \sqrt{ \sqrt{ I } + \sqrt{ J } }$, so $\ex \ell$ s.t. $r^{\ell}=a+b$, where $a^{m} \in I$ $(\ex m)$ and $b^{n} \in J$ ($\ex n$). Want to find $e$ s.t. $r^{e} \in I+J$. *Claim*: $e=\ell(m+n)$ suffices. Indeed, $\begin{align}
> > (r ^{\ell(m+n)}) &= (r^{\ell})^{m+n} \\
> > &= (a+ b)^{m+n} \\
> > &= \sum_{k=0}^{m+n} {m + n \choose k} a ^{k} b^{m+n-k} \\
> > &= (\sum_{k=0}^{m} {m + n \choose k} a ^{k} \underbrace{ b^{\overbrace{m+n-k}^{ \geq n}} }_{ \in J } ) + (\sum_{k=m+1}^{m+n} {m + n \choose k} \underbrace{ a ^{\overbrace{k}^{ \geq m}} }_{ \in I } b^{m+n-k}).
> > \end{align}$
> > Since $I,J$ are [[ideal|ideals]], the each term of the first sum lives in $J$ and each term of the second sum lives in $I$; hence the total summation is over elements of $I+J$ and thus lives in $I+J$, as desired.
> >
>
>
> - $\sqrt{ I^{n} }=\sqrt{ I }$ for all $n \geq 1$.
>
> > [!proof]-
> > $\subset.$ Let $a \in \sqrt{ I^{n} }$, so $a^{m}$ ( $\ex m$) is a combination of elements of the form $i_{1}\cdots i_{n}$. Clearly each $i_{1} \cdots i_n \in I$, thus $a^{m} \in I$, so $a \in \sqrt{ I }$.
> >
> > $\supset.$ Let $a \in \sqrt{ I }$, so $a^{m} \in I$ for some $m$. Then $a^{mn} =(a^{m})^{n}\in I^{n}$, so $a \in \sqrt{ I^{n} }$.
>
> $\sqrt{ I } \neq \langle 1 \rangle$ whenever $I \neq \langle 1 \rangle$.
>
> > [!proof]-
> > - Contrapositively suppose $\sqrt{ I }=\langle 1 \rangle$. Then there exists $n$ such that $1^{n}\in I$, but $1^{n}=1$, hence $I=\langle 1 \rangle$.
>
>
^4e9b7c
----
####
----
#### 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
> ```