# $x+ I \subset p_1 \cup \cdots \cup p_n$ then for some $i$, $x + I \subset p_i.$

Let $$R$$ be a commutative ring with $$1$$. Let $$I$$ be an ideal of $$R$$ and $$x$$ be an element in the ring $$R.$$ If $$p_1, \ldots,p_n$$ be prime ideals in $$R$$ with $$x+ I \subset p_1 \cup \cdots \cup p_n$$ then for some $$i$$, $$x + I \subset p_i.$$

It is well known when $$x=0,$$ the prime avoidance theorem. Can someone give me some idea to prove this. Thanks.

• Just for the record: this is Davis lemma, and can be found in Kaplansky and Matsumura. Commented Aug 4, 2019 at 16:34

In response to Geoffrey Trang's comment on my previous answer, I will add a sharper answer as community wiki.

The following proof can be found in O.A.S. Karamzadeh's note, The Prime Avoidance Lemma Revisited, as Theorem B. Here I simplify slightly and assume the ring is commutative.

Let $$P_1, \ldots, P_n$$ be ideals of a ring $$R$$ at most one of which is not prime. Let $$I$$ be an additive subset of $$R$$ having the structure of an ideal with respect to each $$P_i$$, except perhaps one of the prime $$P_i$$. If $$T$$ is a subset of $$R$$ such that $$I + T \subseteq \bigcup P_i$$, then there exists a $$t \in T$$ such that $$(I, t) \subseteq P_i$$ for some $$i$$.

Proof:

We proceed by induction on $$n$$, with the case $$n=1$$ being trivial. We can assume that $$P_i \nsubseteq P_j$$ for any $$i,j$$. For the inductive step, we first order the $$P_i$$ so that $$P_1$$ is prime and $$I$$ has ideal structure with respect to the remaining $$P_i$$. The remaining $$P_i$$ will still have at most $$1$$ non-prime $$P_i$$ amongst them. We separate two cases: (1) $$I + T \subseteq \bigcup_{i=2}^n P_i$$, and applying the inductive hypothesis we are immediately done (2) $$I + T \nsubseteq \bigcup_{i=2}^n P_i$$, and thus there exist $$x \in I, t \in T$$ such that $$x + t \in P_1 \setminus \bigcup_{i=2}^n P_i$$. In this case we claim that $$(I, t) \subseteq P_1$$.

For convenience set $$J = \bigcap_{i=2}^n P_i$$. Note that $$J \nsubseteq P_1$$ by primeness of $$P_1$$ and the assumption that $$P_i \nsubseteq P_j$$ for all $$i,j$$. It thus suffices to show that $$I \cap J \subseteq P_1$$ because that would force $$I \subseteq P_1$$, and consequently force $$t \in P_1$$ (note that we needed $$I$$ to have ideal structure w.r.t. $$J$$ in order to say that $$IJ \subseteq I \cap J$$). Thus let $$y \in I \cap J$$. Since $$x + t$$ isn't in any $$P_i$$ for $$i \geq 2$$ and $$y \in P_i$$ for all $$i \geq 2$$, we deduce $$x + t + y \notin \bigcup_{i=2}^n P_i$$, and hence $$x + t + y \in P_1$$. Since $$x + t$$ is also in $$P_1$$, we deduce that $$y \in P_1$$. This shows that indeed $$I \cap J \subseteq P_1$$, and completes the proof. $$\square$$

Finally I'll give a couple of simple examples that demonstrate why the assumptions on the algebraic structure of $$I$$ and on the number of prime ideals are crucial in the above statement.

That $$I$$ needs to have ideal structure with respect to (all but a particular one of) the $$P_i$$

Consider the ring $$R = F_2[x]$$ and the coset $$x + F_2$$. This coset is formed from a bonafide subring $$F_2 \subset F_2[x]$$, and consists of the two non-unit elements $$x, x+1$$, so of course we can cover it by proper (prime) ideals of $$R$$ (take $$(x), (x+1))$$. However, the coset $$x + F_2$$ generates $$R$$ as an ideal, and so cannot be contained in any proper prime ideal.

That at most one of the $$P_i$$ can generally be non-prime

Consider the ring $$R = F_2[x,y]/(x^2, y^2)$$. Consider the coset $$x + (y)$$. It consists of the elements $$x, x + yx, x+y, x + yx + y$$, and can be covered by the non-prime ideals $$(x), (x + y)$$. However, $$y$$ is not contained in either of these ideals, so of course $$(x,y)$$ isn't either.

We can view this as a corollary to the statement with $$x = 0$$. What you want is:

If $$x + I \subseteq p_1 \cup \cdots \cup p_n$$ then $$(x) + I \subseteq p_1 \cup \cdots \cup p_n$$

Proof: Suppose $$x + I \subseteq p_1 \cup \cdots \cup p_n$$. In particular $$x$$ must be contained in at least one of the $$p_i$$. Reorder the $$p_i$$ and pick $$k$$ such that $$x \in p_1 \cap \cdots \cap p_k$$ but $$x \notin p_{k+1} \cup \cdots \cup p_{n}$$. If $$I$$ were contained in $$p_1 \cup \cdots \cup p_k$$, then by prime avoidance we'd have $$I$$ in some $$p_j$$, and we'd be done (since then $$x, I \subseteq p_j$$. Similarly, if $$x$$ were contained in all of the $$p_i$$, then we would find that $$I \subseteq p_1 \cup \cdots \cup p_n$$ and conclude similarly.

In fact these are the only possible scenarios. In the remaining case, we would have that $$1 \leq k < n$$ and that $$I \nsubseteq p_1 \cup \cdots \cup p_k$$. Moreover we can safely assume that there are no comparability relations among the $$p_i$$, i.e. $$p_i \nsubseteq p_j$$ for all $$i,j$$ (otherwise just remove the redundant ideal). These assumptions would allow us to choose firstly an element $$a \in I \setminus (p_1 \cup \cdots \cup p_k)$$ and secondly an element $$b \in (p_{k+1} \cap \cdots \cap p_n) \setminus (p_{1} \cup \cdots \cup p_k)$$

In more detail, we can choose this $$b$$ because otherwise we'd have $$p_{k+1} \cap \cdots \cap p_n \subseteq p_{1} \cup \cdots \cup p_k$$, and since $$p_i \subseteq p_{k+1} \cap \cdots \cap p_n$$ for $$k+1 \leq i \leq n$$ (by primeness), another application of prime avoidance would then imply a relation $$p_i \subseteq p_j$$. Now by our choice of $$a,b$$, we have that $$ab \in (I \cap p_{k+1} \cap \cdots \cap p_n) \setminus (p_1 \cup \cdots \cup p_k)$$. Since $$x \in (p_1 \cap \cdots \cap p_k) \setminus (p_{k+1} \cup \cdots \cup p_n)$$, we then have $$x + ab \in x + I \setminus (p_1 \cup \cdots \cup p_n)$$, which is the desired contradiction. $$\square$$

(1) whereas the standard prime avoidance can be stated with the weakened assumption that at least $$n-2$$ of the $$p_i$$ are prime, our choice of $$b$$ did depend on the primeness of all the $$p_i$$.
(2) whereas the standard prime avoidance can be stated with $$I$$ generalized to a subrng of $$R$$ (not necessarily containing $$1$$), the construction of the absurd $$x + ab$$ in the proof depended on $$ab \in I$$, which in turn depended on $$I$$ having an ideal structure.
• Perhaps, you should give three specific counterexamples showing that the assumptions that $I$ is an ideal and that the $p_i$ are all prime cannot be weakened unlike in the prime avoidance lemma. The first counterexample should merely have $I$ as a subrng that is not an ideal. The second counterexample should still have $I$ as an ideal, but have all but one of the $p_i$s (WLOG $p_1$) prime. Finally, the third counterexample should again have $I$ as an ideal, but have all but two of the $p_i$s (WLOG $p_1$ and $p_2$) prime. Commented Aug 2, 2019 at 17:58