LEC ALG3 5

Radicals

Proposition 110.1(Proposition).

Let $R$ be a commutative ring. The set of all nilpotent elements is an ideal in $R$.

Proof.

This proof works verbatim with nilpotent polynomials replaced with nilpotent elements of $R$ to show that the set of nilpotent elements form an abelian group. The fact that they are closed under multiplication by $R$ is immediate.□

041806

Definition 110.2(Definition).

1. $\sqrt{I}:=\{r\in R|r^n\in I\text{ for some }n\}$.
2. $\sqrt{(0)}$, called the nilradical, is the ideal of nilpotent elements.

e9f650

If $R$ is commutative, by the same argument as Proposition 1, $\sqrt{I}$ is an ideal for all $I$. If $R$ is not commutative, $\sqrt{I}$ is no longer an ideal, but a weaker condition holds:

Proposition 110.3(Proposition).

$\sqrt{I}$ consists of whole cosets of $I$.

Proof.

Let$r\in \sqrt{I}$, that is, $r^n\in I$ for some $n$. Clearly, $(r+i{)}^n\in I$ for any $i\in I$, since all the terms in the binomial expansion that contain $i$ must be in $I$, and the only one that does not, $r^n$, is in $I$ by hypothesis.□

Remark 110.4(Remark).

If $I$ is an ideal in $R$ and $\pi :R\to R/I$ is the projection map, then $(\pi (\sqrt{I}))=\sqrt{(\overline{0})}$.

Proposition 110.5(Proposition).

1. $\sqrt{\sqrt{I}}=\sqrt{I}$.
2. $\sqrt{I\cap J}=\sqrt{I}\cap \sqrt{J}$.
3. If $I$ is prime, then $\sqrt{I}=I$.

Lemma 110.6(Lemma).

If $P\subset R$ is a prime ideal, then $\sqrt{P^n}=P$ for all $n\in \mathbb{Z}$.

Proof.

Let$e\in \sqrt{P^n}$. Then, $e^k\in P^n\subseteq P$ for some $k\in \mathbb{Z}$. Since $P$ is prime, it follows that $e\in P$. Conversely, if $e\in P$, $e^n\in P^n$.□

6f6d3a

Proposition 110.7(Proposition).

If $n=p_1^{a_1}\dots p_k^{a_k}$, then $\sqrt{(\overline{0})}=(\overline{p_1{p}_2\dots p_n})$ in $\mathbb{Z}/n\mathbb{Z}$.

Proof.

$$\begin{array}{rl}\sqrt{(\overline{0})}& =\sqrt{(\overline{p_1^{a_1}\dots p_n^{a_n}})}\\ & \stackrel{!}{=}\sqrt{(\overline{p_1^{a_1}})\cap \cdots \cap (\overline{p_n^{a_n}})}\\ & =\sqrt{(\overline{p_1^{a_1}})}\cap \cdots \cap \sqrt{(\overline{p_n^{a_n}})},\end{array}$$

where the marked equality is true because $p_1^{a_1},\dots ,p_n^{a_n}$ are coprime. By Lemma 6, $\sqrt{(\overline{p^a})}=(\overline{p})$. Thus, we have

$$\begin{array}{rl}\sqrt{(\overline{0})}& =(\overline{p_1})\cap (\overline{p_2})\cap \cdots \cap (\overline{p_n})=(\overline{p_1{p}_2\dots p_n}).\end{array}$$ □

Example 110.8(Example).

Let $R=k[x,y]$ where $k$ is a field. Find $\sqrt{(x^2{y}^2)}$.

Clearly, $(x^2{y}^2)=(x^2)\cap (y^2)$. Also, $(x^2)=(x{)}^2$. Further, $(x)$ is a prime ideal (should be obvious, but Clare insists on showing $k[x,y]/(x)\cong k(y)$). Thus, using Lemma 6, $\sqrt{(x^2{y}^2)}=\sqrt{(x{)}^2}\cap \sqrt{(y{)}^2}=(x)\cap (y)=(xy)$.

Proposition 110.9(Proposition).

Every radical ideal in a noetherian ring is a finite intersection of prime ideals.

Proof.

First, we will show that if$I$ is a radical ideal and $ab\in I$, then $I=\sqrt{I+(a)}\cap \sqrt{I+(b)}$. $I\subseteq \sqrt{I+(a)}\cap \sqrt{I+(b)}$ is clear. To show the reverse inclusion, suppose $c\in \sqrt{I+(a)}\cap \sqrt{I+(b)}$. Then, $c^n=i+ar$ and $c^m=i^{\prime }+b{r}^{\prime }$ for some integers $n,m$ and $i,i^{\prime }\in I$, $r,r^{\prime }\in R$. We now have

$$\begin{array}{r}{c}^{n+m}=i{i}^{\prime }+ib{r}^{\prime }+i^{\prime }ar+abr{r}^{\prime }\in I.\end{array}$$

$c^{n+m}\in I$ implies $c\in I$, since $I$ is a radical ideal.

Now, If $I$ is not a prime ideal, there exist $a,b$ such that $ab\in I$ but $a,b\notin I$. We can thus write $I$ as the intersection of two radicals $I_1$ and $I_2$ containing $(a)$ and $(b)$ respectively. If either $I_1$ or $I_2$ is not prime, we repeat the process. We are guaranteed to reach a prime ideal eventually, since otherwise we’d have an infinite ascending chain of ideals.□