# Commutative Algebra/Radicals, strong Nakayama
## Radicals of ideals

Definition 13.1:
Let $R$ be a ring, $I\leq R$ an ideal. The radical of $I$, denoted $r(I)$, is
    $r(I):=\{r\in R|\exists n\in \mathbb {N} :r^{n}\in I\}$.
A radical ideal is an ideal $I$ such that $r(I)=I$.
Theorem 13.2:
Let $R$ be a ring, $I\leq R$ an ideal. Then
    $r(I)=\bigcap _{p{\text{ prime}} \atop I\subseteq p}p$.
Proof:
For $\subseteq$, note that $s^{n}\in I\subseteq p\Rightarrow s\in p$. For $\supseteq$, assume for no $n$ $s^{n}\in r(I)$. Form the quotient ring $R/I$. By theorem 12.3, pick a prime ideal $q\leq R/I$ disjoint from the multiplicatively closed set $\{s^{n}+I|n\in \mathbb {N} _{0}\}$. Form the ideal $q:=\pi ^{-1}(p)$. $p$ is a prime ideal which contains $I$ and does not intersect $\{s^{n}|n\in \mathbb {N} _{0}\}$. Hence $s$ is not in the right hand side.$\Box$
Corollary 13.3:
The radical of an ideal is an ideal.
Proof:
Intersection of ideals is an ideal.$\Box$
Definition 13.4:
Let $R$ be a ring, $I\leq R$. The Jacobson radical of $I$ is defined as thus:
    $j(I):=\bigcap _{m{\text{ maximal}} \atop I\subseteq m}m$.
Theorem 13.5:
Let $R$ be a ring, $I\leq R$. $j(I)$ is a radical ideal.
Proof:
Clearly, $j(I)\subseteq r(j(I))$. Further $r(j(I))\subseteq j(j(I))=j(I)$ from theorem 13.2; the last equality from $I\subseteq m\Rightarrow j(I)\subseteq m$.$\Box$

## The radicals of the zero ideal

Definition 13.6:
Let $R$ be a ring. $\{0\}$ is an ideal. The nilradical of $R$, written ${\mathcal {N}}$, is defined as
    ${\mathcal {N}}:=r(\{0\})$.
Note that by definition
    ${\mathcal {N}}=\{r\in R|r^{n}=0\}$,
the set of nilpotent elements.
Theorem 13.7:
    ${\mathcal {N}}=\bigcap _{p\leq R \atop p{\text{ prime}}}p$.
Proof:
Theorem 13.2.$\Box$
Definition 13.8:
Let $R$ be a ring. $\{0\}$ is an ideal. The Jacobson radical of $R$, written ${\mathcal {J}}$, is defined as
    ${\mathcal {J}}:=\bigcap _{m\leq R \atop m{\text{ maximal}}}m$.
We have ${\mathcal {J}}=j(\{0\})$.
If $R$ is a ring, $R^{\times }$ is the set of units of $R$.
Theorem 13.9:
Let $R$ be a ring, ${\mathcal {J}}$ its Jacobson radical. Then
    $r\in {\mathcal {J}}\Leftrightarrow \forall s\in R:1-rs\in R^{\times }$.
Proof:
$\Rightarrow$: Let $r\in {\mathcal {J}}$, $s\in R$. Assume $1-rs\notin R^{\times }$. Form the ideal $\langle 1-rs\rangle$; by theorem 12.8 there exists $m\leq R$ maximal with $\langle 1-rs\rangle \subseteq m$, hence $1-rs\in m$. If $r\in {\mathcal {J}}\Rightarrow r\in m$, then $1\in m$, contradiction.
$\Leftarrow$: Assume $1-rs\in R^{\times }$ for all $s$ and $r\notin {\mathcal {J}}$. Then there is a maximal ideal $m$ not containing $r$. Hence $m+\langle r\rangle =R$ and $1=t+rs$ for a $t\in m$ and an $s\in R$. Hence $t=1-rs$ is not a unit.$\Box$

## Radicals and localisation

Definition 13.10:
If $R$ is a ring, $S\subseteq R$ is a multiplicative subset, $I\leq R$ an ideal, set
    $S^{-1}I:=\{i/s|i\in I,s\in S\}\subseteq S^{-1}R$,
the localisation of the ideal $I$ with respect to $S$.
Theorem 13.11:
Let $R$ be a ring, $I\leq R$ an ideal, $S\subseteq R$ a multiplicatively closed subset. Then
    $r(S^{-1}I)=S^{-1}r(I)$.
Proof:
Let $r/s\in r(S^{-1}I)$, that is, $r^{n}/s^{n}\in S^{-1}(I)$. Then $r^{n}/t^{n}=i/s$, $i\in I$, $s\in S$. There exists $u\in S$ such that $u(r^{n}s-t^{n}i)=0$. Thus $usr^{n}\in I$, whence $(usr)^{n}\in I$ and $usr\in r(I)$. Thus, $r/s=usr{\big /}us^{2}\in S^{-1}r(I)$.
Let $r/s\in S^{-1}r(I)$. We may assume $r\in r(I)$. Choose $n$ such that $r^{n}\in I$. Then $(r/s)^{n}\in S^{-1}I$, whence $r/s\in r(S^{-1}I)$.$\Box$

## Exercises

* Prove that whenever $R$ is a reduced ring, then the canonical homomorphism $R\to \prod _{p\leq R \atop p{\text{ prime}}}R/p$ is injective.