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Passing to the reduced quotient does not change the prime spectrum

Statement

Let $R$ be a commutative ring, let $N = Nil (R)$ , and let $π : R \to R / N$ be the quotient map. Then contraction along $π$ induces an inclusion-preserving bijection $Spec (R / N) \to Spec (R)$ . If $J ⊴ R / N$ and $I = π^{- 1} (J)$ , this bijection identifies $V (J)$ with $V (I)$ .

Facts & Assumptions

Given: A commutative ring $R$ , its nilradical $N = Nil (R)$ , and the quotient map $π : R \to R / N$ .

[L1]

Prime ideals of a quotient correspond exactly to prime ideals of the original ring containing the kernel (Prime ideals of a quotient ring are exactly the prime ideals containing the ideal).

[L2]

Every element of $N$ is nilpotent (The nilradical and reduced rings).

Proof

technique · direct

1.1L2givenalgebra

Let $p \in Spec (R)$ . If $x \in N$ , then $x^{m} = 0$ for some $m \geq 1$ , so $x^{m} \in p$ . Because $p$ is prime, this forces $x \in p$ . Thus every prime ideal of $R$ contains $N$ .

2.1L1step 1.1

Applying [L1] to the quotient map $π$ and using step 1.1, one gets an inclusion-preserving bijection from $Spec (R / N)$ onto all of $Spec (R)$ . Moreover, for an ideal $J ⊴ R / N$ with pullback $I$ , a prime $q$ of $R / N$ contains $J$ exactly when its contraction $π^{- 1} (q)$ contains $I$ . So the same bijection carries $V (J)$ onto $V (I)$ .

3.1step 2.1∎

Therefore passing from $R$ to its reduced quotient does not change the prime spectrum or the vanishing sets attached to quotient ideals.

Passing to the reduced quotient does not change the prime spectrum

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources