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.


\[ D = \sqrt{D} = \sqrt{\cup_{x\ne 0} \mathrm{Ann}(x)} = \cup_{x\ne 0} \sqrt{\mathrm{Ann}(x)} \]

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 💡