Commutative Algebra

Let $R$ be a ring. For a subset $X\subset R$ define the radical of $X$ with respect to $R$ by

$$\sqrt{X}=\{x\in R\mid x^n\in X,n\ge 1\}$$

Clearly $\sqrt{{\cup }_{\alpha }{X}_{\alpha }}={\cup }_{\alpha }\sqrt{X_{\alpha }}$ for any family of subsets $X_{\alpha }$ of $R$.

Proposition: The set of zero-divisors of $R$ is equal to its radical ${\cup }_{x\ne 0}\sqrt{\mathrm{Ann}x}$

Proof: Let $D$ be the set of zero-divisors of $R$ so

$$D={\cup }_{x\ne 0}\mathrm{Ann}(x)$$

We have $D\subset \sqrt{D}$. Suppose $y\in \sqrt{D}$. Then $y^k\in D$ for some $k\ge 1$, hence $y^{k}x=0$ for some nonzero $x\in R$. If $k=1$ then $y\in D$, otherwise $y(y^{k-1}x)=0$. Then if $y^{k-1}x\ne 0$ then $y\in D$, otherwise if $y^{k-1}x=0$ we have $y\in D$ by induction.

Thus

$$D=\sqrt{D}=\sqrt{{\cup }_{x\ne 0}\mathrm{Ann}(x)}={\cup }_{x\ne 0}\sqrt{\mathrm{Ann}(x)}$$

Let $I◃R$. Then

$$\begin{array}{ccc}\sqrt{I}& =& \{x\in R\mid x^n\in I\}\\ & =& \{x\in R\mid I+x^n=I\}\\ & =& \{x\in R\mid (I+x{)}^n=I\}\\ & =& {\varphi }^{-1}(N_{R/I})◃R\end{array}$$

where $\varphi :R\to R/I$ is the natural map and $N_{R/I}$ is the nilradical of $A/I$.

The following are easily verified for any ideals $I,J$ in $R$:

1. $\sqrt{I}\supset I$

2. $\sqrt{\sqrt{I}}=\sqrt{I}$

3. $\sqrt{IJ}=\sqrt{I\cap J}=\sqrt{I}\cap \sqrt{J}$

4. $\sqrt{I}=R\iff I=R$

5. $\sqrt{I+J}=\sqrt{\sqrt{I}+\sqrt{J}}$

6. If $P◃R$ is prime then for all $n\ge 1$ we have $\sqrt{P^n}=P$.

Theorem: The radical of an ideal is the intersection of the prime ideals containing it.

Proof: Let $I◃R$. Then $\sqrt{I}={\varphi }^{-1}(N_{R/I})$. Recall that $N_{A/I}$ is the intersection of all prime ideals of $R/I$. The result follows since the prime ideals of $R/I$ are precisely those of the form $P/I$ for a prime ideal $P$ of $R$ containing $I$.

Example: Let $R=\mathbb{Z}$, $I=m\mathbb{Z}$. Then if $m=p_1^{a_1}...p_k^{a_k}$ we have $\sqrt{I}=p_1...p_k\mathbb{Z}$.

Proposition: Let $I,J◃R$. If $\sqrt{I},\sqrt{J}$ are coprime then $I,J$ are coprime.

Proof: $\sqrt{I+J}=\sqrt{\sqrt{I}+\sqrt{J}}=\sqrt{R}=R$, hence $I+J=R$.