Definition and basic properties of the Jacobson radical of an ideal

Let R be a commutative ring and let I \subseteq R be a proper ideal. Let \mathcal{M}_I denote the set of maximal ideals in R containing I; \mathcal{M}_I is nonempty by Proposition 11 in the text. Define \mathsf{Jac}(I) = \bigcap \mathcal{M}_I. By convention, \mathsf{Jac}(R) = R. Also, \mathsf{Jac}(0) is called the Jacobson radical of R.

  1. Prove that \mathsf{Jac}(I) is an ideal of R containing I.
  2. Prove that \mathsf{rad}(I) \subseteq \mathsf{Jac}(I).
  3. Let n > 1 be an integer and let R = \mathbb{Z}. Describe \mathsf{Jac}(n\mathbb{Z}) in terms of the prime factorization of n.

We begin with a lemma.

Lemma: If a,b \in \mathbb{Z} are relatively prime, then (a) + (b) = \mathbb{Z}. Proof: we can write xa + yb = 1 by Bezout’s identity for some integers x and y, so that 1 \in (a)+(b). \square

  1. By Proposition 11 in the text, \mathcal{M}_I is not empty. Thus by this previous exercise, \mathsf{Jac}(I) is an ideal of R. Since I \subseteq M for all M \in \mathcal{M}_I, we have I \subseteq \mathsf{Jac}(I).
  2. Let x \in \mathsf{rad}(I), and let M \in \mathcal{M}_I. Now x^m \in I \subseteq M for some m \geq 1; let m be minimal with this property. If m \geq 2, then xx^{m-1} \in M; since M is prime, we have x \in M or x^{m-1} \in M, a contradiction. Thus m = 1, and we have x \in M. Thus \mathsf{rad}(I) \subseteq M for all M \in \mathcal{M}_I, and hence \mathsf{rad}(I) \subseteq \mathsf{Jac}(I).
  3. Write the prime factorization of n as n = \prod p_i^{k_i}. Recall that the maximal ideals of \mathbb{Z} are precisely those of the form (p) for prime p. Now (n) \subseteq (p) precisely when p divides n; thus \mathcal{M}_{(n)} consists precisely of the ideals (p_i) for the primes p_i dividing n. Thus \mathsf{Jac}(n\mathbb{Z}) = \bigcap_{p_i | n} (p_i). Using the lemma (with induction) and this previous exercise, \mathsf{Jac}(n\mathbb{Z}) = (\prod p_i).
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