Commutative Algebra/Radicals, strong Nakayama

Radicals of ideals
Proof:

For $$\subseteq$$, note that $$s^n \in I \subseteq p \Rightarrow s \in p$$. For $$\supseteq$$, assume for no $$n$$ $$s^n \in r(I)$$. Form the quotient ring $$R/I$$. By theorem 12.3, pick a prime ideal $$q \le R/I$$ disjoint from the multiplicatively closed set $$\{s^n + I|n \in \mathbb N_0\}$$. Form the ideal $$q := \pi^{-1}(p)$$. $$p$$ is a prime ideal which contains $$I$$ and does not intersect $$\{s^n|n \in \mathbb N_0\}$$. Hence $$s$$ is not in the right hand side.

Proof:

Intersection of ideals is an ideal.

Proof:

Clearly, $$j(I) \subseteq r(j(I))$$. Further $$r(j(I)) \subseteq j(j(I)) = j(I)$$ from theorem 13.2; the last equality from $$I \subseteq m \Rightarrow j(I) \subseteq m$$.

The radicals of the zero ideal
Note that by definition
 * $$\mathcal N = \{r \in R| r^n = 0\}$$,

the set of nilpotent elements.

Proof:

Theorem 13.2.

We have $$\mathcal J = j(\{0\})$$.

If $$R$$ is a ring, $$R^\times$$ is the set of units of $$R$$.

Proof:

$$\Rightarrow$$: Let $$r \in \mathcal J$$, $$s \in R$$. Assume $$1 - rs \notin R^\times$$. Form the ideal $$\langle 1 - rs \rangle$$; by theorem 12.8 there exists $$m \le R$$ maximal with $$\langle 1 - rs \rangle \subseteq m$$, hence $$1 - rs \in m$$. If $$r \in \mathcal J \Rightarrow r \in m$$, then $$1 \in m$$, contradiction.

$$\Leftarrow$$: Assume $$1 - rs \in R^\times$$ for all $$s$$ and $$r \notin \mathcal J$$. Then there is a maximal ideal $$m$$ not containing $$r$$. Hence $$m + \langle r \rangle = R$$ and $$1 = t + rs$$ for a $$t \in m$$ and an $$s \in R$$. Hence $$t = 1 - rs$$ is not a unit.

Radicals and localisation
Proof:

Let $$r/s \in r(S^{-1}I)$$, that is, $$r^n/s^n \in S^{-1}(I)$$. Then $$r^n/t^n = i/s$$, $$i \in I$$, $$s \in S$$. There exists $$u \in S$$ such that $$u(r^n s - t^n i) = 0$$. Thus $$u s r^n \in I$$, whence $$(usr)^n \in I$$ and $$usr \in r(I)$$. Thus, $$r/s = usr \big/ us^2 \in S^{-1} r(I)$$.

Let $$r/s \in S^{-1}r(I)$$. We may assume $$r \in r(I)$$. Choose $$n$$ such that $$r^n \in I$$. Then $$(r/s)^n \in S^{-1}I$$, whence $$r/s \in r(S^{-1}I)$$.

Exercises

 * 1) Prove that whenever $$R$$ is a reduced ring, then the canonical homomorphism $$R \to \prod_{p \le R \atop p \text{ prime}} R/p$$ is injective.