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# Prime avoidance lemma

View other indispensable lemmata

## 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 Ijs are not prime, then J is contained in one of the Ijs.

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 Ijs 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 Ijs, then J has an element which is contained in none of the Ijs.

## Proof

### 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 ai 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 ais 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 I1 or I2. 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 x1 + x2 is in J, hence it must be inside I1 or I2. 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 In 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 Iis. We consider two cases:

• The sum is in In: 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 Ij 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 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 xn to positive powers so that the new terms have equal degrees, and can be added. In this case, we need all the Ijs to be prime to ensure that even after taking powers, the elements xi still avoid the ideals Ij, for $j \ne i$.