Prime avoidance lemma

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This article is about the statement of a simple but indispensable lemma in commutative algebra
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Let R be a commutative unital ring. Let I_1, I_2, \ldots, I_n and J be ideals of R, such that J \subset \bigcup_j I_j. Then, if R contains an infinite field or if at most two of the I_js are not prime, then J is contained in one of the I_js.

Graded version

If R is graded, and J is generated by homogeneous elements of positive degree, then it suffices to assume that the homogeneous elements of J are contained in \bigcup_j I_j. However, we need to add the further assumption that all the I_js are prime.


The prime avoidance lemma is useful for establishing dichotomies; in particular, if J is an ideal which is not cintained in any of the I_js, then J has an element which is contained in none of the I_js.


If the ring contains an infinite field

In this case, the proof boils down to two observations:

  • Any ideal of the ring is also a vector space over the infinite field
  • A vector space over an infinite field cannot be expressed as a union of finitely many proper subspaces

If at most two of the ideals are not prime

The proof in this case proceeds by induction. The crucial ingredients to the proof are:

  1. If two of the three elements a,b,a+b belong to an ideal, so does the third (the fact that ideals are additive subgroups)
  2. If, for a product a_1a_2\ldots a_r, any a_i belongs to an ideal, so does the product
  3. If a product a_1a_2 \ldots a_r belongs to a prime ideal, then one of the a_is also belongs to that prime ideal

We now describe the proof by induction. The case n=1 requires no proof; the case n=2 follows from observation 1 (in other words, we only need to use that both are additive subgroups). Namely, suppose J is not a subset of either I_1 or I_2. Pick x_1 \in J \setminus I_2 and x_2 \in J \setminus I_1. Clearly, x_1 \in I_1, x_2 \in I_2. Then x_1 + x_2 is in J, hence it must be inside I_1 or I_2. This contradicts observation 1. Further information: Union of two subgroups is not a subgroup

For n > 2, we use induction. Suppose, without loss of generality, that I_n is a prime ideal. Also assume without loss of generality that J is not contained in the union of any proper subcollection of I_1, I_2, \ldots, I_n (otherwise, induction applies). Thus, we can pick x_i \in J \setminus \bigcup_{j \ne i} I_j for each i. Clearly x_i \in I_i.

We now consider the element:

x_1x_2 \ldots x_{n-1} + x_n

This is in J, hence it must be in one of the I_is. We consider two cases:

  • The sum is in I_n: Then, by observation 1, x_1x_2 \ldots x_{n-1} \in I_n. By observation 3, x_j \in I_n for some 1 \le j \le n-1, a contradiction.
  • The sum is in I_j for some 1 \le j \le n-1: Observation 2 tells us that x_1x_2\ldots x_{n-1} \in I_j, so observation 1 yields x_n \in I_j, a contradiction

In the graded case

In this case, the proof is the same, except that we can now get started on the proof only after raising x_1x_2\ldots x_{n-1} and x_n to positive powers so that the new terms have equal degrees, and can be added. In this case, we need all the I_js to be prime to ensure that even after taking powers, the elements x_i still avoid the ideals I_j, for j \ne i.