Prime avoidance lemma

This article may be expanded with text translated from the corresponding article in French. (October 2018) Click [show] for important translation instructions. View a machine-translated version of the French article. Machine translation like Deepl or Google Translate is a useful starting point for translations, but translators must revise errors as necessary and confirm that the translation is accurate, rather than simply copy-pasting machine-translated text into the English Wikipedia. Do not translate text that appears unreliable or low-quality. If possible, verify the text with references provided in the foreign-language article. You must provide copyright attribution in the edit summary accompanying your translation by providing an interlanguage link to the source of your translation. A model attribution edit summary (using German): Content in this edit is translated from the existing German Wikipedia article at [[:de:Exact name of German article]]; see its history for attribution. You should also add the template {{Translated|fr|Lemme d'évitement des idéaux premiers}} to the talk page. For more guidance, see Wikipedia:Translation.

In algebra, the prime avoidance lemma says that if an ideal I in a commutative ring R is contained in a union of finitely many prime ideals P_i's, then it is contained in P_i for some i.

There are many variations of the lemma (cf. Hochster); for example, if the ring R contains an infinite field or a finite field of sufficiently large cardinality, then the statement follows from a fact in linear algebra that a vector space over an infinite field or a finite field of large cardinality is not a finite union of its proper vector subspaces.^[1]

Statement and proof

The following statement and argument are perhaps the most standard.

Statement: Let E be a subset of R that is an additive subgroup of R and is multiplicatively closed. Let I1,I2,…,In,n≥1{displaystyle I_{1},I_{2},dots ,I_{n},ngeq 1} be ideals such that Ii{displaystyle I_{i}} are prime ideals for i≥3{displaystyle igeq 3}. If E is not contained in any of Ii{displaystyle I_{i}}'s, then E is not contained in the union ∪Ii{displaystyle cup I_{i}}.

Proof by induction on n: The idea is to find an element that is in E and not in any of Ii{displaystyle I_{i}}'s. The basic case n = 1 is trivial. Next suppose n ≥ 2. For each i choose

zi∈E−∪j≠iIj{displaystyle z_{i}in E-cup _{jneq i}I_{j}}

where the set on the right is nonempty by inductive hypothesis. We can assume zi∈Ii{displaystyle z_{i}in I_{i}} for all i; otherwise, some zi{displaystyle z_{i}} avoids all the Ii{displaystyle I_{i}}'s and we are done. Put

z=z1…zn−1+zn{displaystyle z=z_{1}dots z_{n-1}+z_{n}}.

Then z is in E but not in any of Ii{displaystyle I_{i}}'s. Indeed, if z is in Ii{displaystyle I_{i}} for some i≤n−1{displaystyle ileq n-1}, then zn{displaystyle z_{n}} is in Ii{displaystyle I_{i}}, a contradiction. Suppose z is in In{displaystyle I_{n}}. Then z1…zn−1{displaystyle z_{1}dots z_{n-1}} is in In{displaystyle I_{n}}. If n is 2, we are done. If n > 2, then, since In{displaystyle I_{n}} is a prime ideal, some zi,i<n{displaystyle z_{i},i<n} is in In{displaystyle I_{n}}, a contradiction.

Notes

1. 
   ^ Proof of the fact: suppose the vector space is a finite union of proper subspaces. Consider a finite product of linear functionals, each of which vanishes on a proper subspace that appears in the union; then it is a nonzero polynomial vanishing identically, a contradiction.
   

References

• 
  Mel Hochster, Dimension theory and systems of parameters, a supplementary note

This algebra-related article is a stub. You can help Wikipedia by expanding it. v t e

This page is only for reference, If you need detailed information, please check here