# Radicals

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

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

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 $$I\triangleleft R$$. Then

$\array { \sqrt{I} &=& \{x\in R| x^n \in I\} \\ &=& \{ x\in R| I + x^n= I\} \\ &=& \{ x\in R| (I+x)^n=I \} \\ &=& \phi^{-1}(N_{R/I}) \triangleleft R }$

where $$\phi : R \rightarrow R/I$$ 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. $$\sqrt{I J} = \sqrt{I \cap J} = \sqrt{I} \cap \sqrt{J}$$

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

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

6. If $$P\triangleleft R$$ 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 $$I \triangleleft R$$. Then $$\sqrt{I} = \phi^{-1}(N_{R/I})$$. 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 $$I,J\triangleleft R$$. 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 $$I+J = R$$.

Ben Lynn blynn@cs.stanford.edu 💡