Definition 13.1:

Let ${\displaystyle R}$ be a ring, ${\displaystyle I\leq R}$ an ideal. The radical of ${\displaystyle I}$, denoted ${\displaystyle r(I)}$, is

${\displaystyle r(I):=\{r\in R|\exists n\in \mathbb {N} :r^{n}\in I\}}$.

A radical ideal is an ideal ${\displaystyle I}$ such that ${\displaystyle r(I)=I}$.

Theorem 13.2:

Let ${\displaystyle R}$ be a ring, ${\displaystyle I\leq R}$ an ideal. Then

${\displaystyle r(I)=\bigcap _{p{\text{ prime}} \atop I\subseteq p}p}$.

Proof:

For ${\displaystyle \subseteq }$, note that ${\displaystyle s^{n}\in I\subseteq p\Rightarrow s\in p}$. For ${\displaystyle \supseteq }$, assume for no ${\displaystyle n}$ ${\displaystyle s^{n}\in r(I)}$. Form the quotient ring ${\displaystyle R/I}$. By theorem 12.3, pick a prime ideal ${\displaystyle q\leq R/I}$ disjoint from the multiplicatively closed set ${\displaystyle \{s^{n}+I|n\in \mathbb {N} _{0}\}}$. Form the ideal ${\displaystyle q:=\pi ^{-1}(p)}$. ${\displaystyle p}$ is a prime ideal which contains ${\displaystyle I}$ and does not intersect ${\displaystyle \{s^{n}|n\in \mathbb {N} _{0}\}}$. Hence ${\displaystyle s}$ is not in the right hand side.${\displaystyle \Box }$

Corollary 13.3:

The radical of an ideal is an ideal.

Proof:

Intersection of ideals is an ideal.${\displaystyle \Box }$

Definition 13.4:

Let ${\displaystyle R}$ be a ring, ${\displaystyle I\leq R}$. The Jacobson radical of ${\displaystyle I}$ is defined as thus:

${\displaystyle j(I):=\bigcap _{m{\text{ maximal}} \atop I\subseteq m}m}$.

Theorem 13.5:

Let ${\displaystyle R}$ be a ring, ${\displaystyle I\leq R}$. ${\displaystyle j(I)}$ is a radical ideal.

Proof:

Clearly, ${\displaystyle j(I)\subseteq r(j(I))}$. Further ${\displaystyle r(j(I))\subseteq j(j(I))=j(I)}$ from theorem 13.2; the last equality from ${\displaystyle I\subseteq m\Rightarrow j(I)\subseteq m}$.${\displaystyle \Box }$

## The radicals of the zero ideal

Definition 13.6:

Let ${\displaystyle R}$ be a ring. ${\displaystyle \{0\}}$ is an ideal. The nilradical of ${\displaystyle R}$, written ${\displaystyle {\mathcal {N}}}$, is defined as

${\displaystyle {\mathcal {N}}:=r(\{0\})}$.

Note that by definition

${\displaystyle {\mathcal {N}}=\{r\in R|r^{n}=0\}}$,

the set of nilpotent elements.

Theorem 13.7:

${\displaystyle {\mathcal {N}}=\bigcap _{p\leq R \atop p{\text{ prime}}}p}$.

Proof:

Theorem 13.2.${\displaystyle \Box }$

Definition 13.8:

Let ${\displaystyle R}$ be a ring. ${\displaystyle \{0\}}$ is an ideal. The Jacobson radical of ${\displaystyle R}$, written ${\displaystyle {\mathcal {J}}}$, is defined as

${\displaystyle {\mathcal {J}}:=\bigcap _{m\leq R \atop m{\text{ maximal}}}m}$.

We have ${\displaystyle {\mathcal {J}}=j(\{0\})}$.

If ${\displaystyle R}$ is a ring, ${\displaystyle R^{\times }}$ is the set of units of ${\displaystyle R}$.

Theorem 13.9:

Let ${\displaystyle R}$ be a ring, ${\displaystyle {\mathcal {J}}}$ its Jacobson radical. Then

${\displaystyle r\in {\mathcal {J}}\Leftrightarrow \forall s\in R:1-rs\in R^{\times }}$.

Proof:

${\displaystyle \Rightarrow }$: Let ${\displaystyle r\in {\mathcal {J}}}$, ${\displaystyle s\in R}$. Assume ${\displaystyle 1-rs\notin R^{\times }}$. Form the ideal ${\displaystyle \langle 1-rs\rangle }$; by theorem 12.8 there exists ${\displaystyle m\leq R}$ maximal with ${\displaystyle \langle 1-rs\rangle \subseteq m}$, hence ${\displaystyle 1-rs\in m}$. If ${\displaystyle r\in {\mathcal {J}}\Rightarrow r\in m}$, then ${\displaystyle 1\in m}$, contradiction.

${\displaystyle \Leftarrow }$: Assume ${\displaystyle 1-rs\in R^{\times }}$ for all ${\displaystyle s}$ and ${\displaystyle r\notin {\mathcal {J}}}$. Then there is a maximal ideal ${\displaystyle m}$ not containing ${\displaystyle r}$. Hence ${\displaystyle m+\langle r\rangle =R}$ and ${\displaystyle 1=t+rs}$ for a ${\displaystyle t\in m}$ and an ${\displaystyle s\in R}$. Hence ${\displaystyle t=1-rs}$ is not a unit.${\displaystyle \Box }$

Definition 13.10:

If ${\displaystyle R}$ is a ring, ${\displaystyle S\subseteq R}$ is a multiplicative subset, ${\displaystyle I\leq R}$ an ideal, set

${\displaystyle S^{-1}I:=\{i/s|i\in I,s\in S\}\subseteq S^{-1}R}$,

the localisation of the ideal ${\displaystyle I}$ with respect to ${\displaystyle S}$.

Theorem 13.11:

Let ${\displaystyle R}$ be a ring, ${\displaystyle I\leq R}$ an ideal, ${\displaystyle S\subseteq R}$ a multiplicatively closed subset. Then

${\displaystyle r(S^{-1}I)=S^{-1}r(I)}$.

Proof:

Let ${\displaystyle r/s\in r(S^{-1}I)}$, that is, ${\displaystyle r^{n}/s^{n}\in S^{-1}(I)}$. Then ${\displaystyle r^{n}/t^{n}=i/s}$, ${\displaystyle i\in I}$, ${\displaystyle s\in S}$. There exists ${\displaystyle u\in S}$ such that ${\displaystyle u(r^{n}s-t^{n}i)=0}$. Thus ${\displaystyle usr^{n}\in I}$, whence ${\displaystyle (usr)^{n}\in I}$ and ${\displaystyle usr\in r(I)}$. Thus, ${\displaystyle r/s=usr{\big /}us^{2}\in S^{-1}r(I)}$.

Let ${\displaystyle r/s\in S^{-1}r(I)}$. We may assume ${\displaystyle r\in r(I)}$. Choose ${\displaystyle n}$ such that ${\displaystyle r^{n}\in I}$. Then ${\displaystyle (r/s)^{n}\in S^{-1}I}$, whence ${\displaystyle r/s\in r(S^{-1}I)}$.${\displaystyle \Box }$

## Exercises

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