Prime avoidance lemma

Statement
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 $$I_j$$s are not prime, then $$J$$ is contained in one of the $$I_j$$s.

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 $$I_j$$s are prime.

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

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)
 * If, for a product $$a_1a_2\ldots a_r$$, any $$a_i$$ belongs to an ideal, so does the product
 * 1) If a product $$a_1a_2 \ldots a_r$$ belongs to a prime ideal, then one of the $$a_i$$s 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.

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 $$I_i$$s. 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 $$I_j$$s to be prime to ensure that even after taking powers, the elements $$x_i$$ still avoid the ideals $$I_j$$, for $$j \ne i$$.