ExampleConstruction: Literature-sourcedVerification: Literature-sourcedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27

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Computing sqrt((x^2,xy)) from its containing primes

Example

Assume the Axiom of Choice.

In the polynomial ring $k [x, y]$ , let $I = (x^{2}, x y)$ . Then $\sqrt{I} = (x)$ , and the prime ideals containing $I$ are exactly the prime ideals containing $x$ .

Facts & Assumptions

Given: A field $k$ , the ideal $I = (x^{2}, x y) \subseteq k [x, y]$ , and the Axiom of Choice.

[L1]

Assuming the Axiom of Choice, the radical of an ideal is the intersection of the prime ideals containing it (The radical of an ideal is the intersection of the prime ideals containing it).

[L2]

A prime ideal contains a factor whenever it contains a product (Prime ideals and maximal ideals in a commutative ring).

Verification

technique · direct

1.1

If $p$ is a prime ideal containing $I$ , then $x^{2} \in p$ , so [L2] gives $x \in p$ . Conversely, any prime ideal containing $x$ also contains $x^{2}$ and $x y$ , hence contains $I$ . Thus the primes over $I$ are exactly the primes containing $x$ .

L2givenalgebra

2.1

The ideal $(x)$ is itself prime because $k [x, y] / (x) ≅ k [y]$ is an integral domain. Therefore [L1] and step 1.1 give $\sqrt{I} = ⋂_{\begin{array}{c} p \in Spec (k [x, y]) \\ I \subseteq p \end{array}} p = ⋂_{\begin{array}{c} p \in Spec (k [x, y]) \\ x \in p \end{array}} p = (x) .$

L1step 1.1algebra

3.1

Hence $\sqrt{(x^{2}, x y)} = (x)$ , as claimed.

step 2.1∎

Depends on

Used by

A radical of a product and intersection computation Example

Dependency tree · two levels

7 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

A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., §13 and §17 (standard reference, not scraped)
J. S. Milne, A Primer of Commutative Algebra, v4.03, §14 The spectrum of a ring (standard reference, not scraped)