Commutative Algebra/Radicals, strong Nakayama

Radicals of ideals

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 $q\le R/I$ 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.$◻$

Proof:

Intersection of ideals is an 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 $I\subseteq m\Rightarrow j(I)\subseteq m$.$◻$

The radicals of the zero ideal

Note that by definition

$𝒩=\{r\in R|r^n=0\}$,

the set of nilpotent elements.

Proof:

Theorem 13.2.$◻$

We have $𝒥=j(\{0\})$.

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

Proof:

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

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

Radicals and localisation

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 $us{r}^n\in I$, whence $(usr{)}^n\in I$ and $usr\in r(I)$. Thus, $r/s=usr/u{s}^2\in S^{-1}r(I)$.

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 $r/s\in r(S^{-1}I)$.$◻$

Exercises

1. Prove that whenever $R$ is a reduced ring, then the canonical homomorphism $R\to \prod _{\genfrac{}{}{0ex}{}{p\le R}{p\text{ prime}}}R/p$ is injective.