# Prime avoidance lemma

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## 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.

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.

## 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 $a_i$ 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 $a_i$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. 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 $I_i$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 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 $j \ne i$.