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Dual numbers and their reduced quotient have the same prime set

Example

Let $k$ be a field and let $A = k [ε] / (ε^{2})$ . Then $Spec (A)$ consists of the single prime ideal $(\overline{ε})$ , and the reduced quotient $A / (\overline{ε}) ≅ k$ has the same prime spectrum.

Facts & Assumptions

Given: A field $k$ and the dual-number ring $A = k [ε] / (ε^{2})$ .

[L1]

Prime ideals of a quotient ring correspond 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]

Passing to the reduced quotient does not change the prime spectrum (Passing to the reduced quotient does not change the prime spectrum).

Verification

technique · direct

1.1L1givenalgebra

By [L1], prime ideals of $A$ correspond to prime ideals of $k [ε]$ containing $(ε^{2})$ . Any such prime contains $ε$ because $ε^{2}$ lies in it. The ideal $(ε)$ itself is prime since $k [ε] / (ε) ≅ k$ is a field. Therefore $Spec (A) = {(\overline{ε})}$ .

2.1L2step 1.1

The element $\overline{ε} \in A$ is nilpotent, so the reduced quotient of $A$ is exactly $A / (\overline{ε}) ≅ k$ . By [L2], this quotient has the same prime spectrum as $A$ , which is the singleton from step 1.1.

3.1step 1.1step 2.1∎

Hence the dual numbers and their reduced quotient have the same prime set.

Depends on

Used by

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Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, A Primer of Commutative Algebra, v4.03, §14 The spectrum of a ring (standard reference, not scraped)
The Stacks Project, Section 10.17: The spectrum of a ring (standard reference, not scraped)