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.

  • $\begingroup$ Just for the record: this is Davis lemma, and can be found in Kaplansky and Matsumura. $\endgroup$
    – user26857
    Commented Aug 4, 2019 at 16:34

2 Answers 2


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 $i$.


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. $\square$

Two things that are notable about this proof:
(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.

  • $\begingroup$ 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. $\endgroup$ Commented Aug 2, 2019 at 17:58
  • $\begingroup$ @Geoffrey Trang See the community wiki I posted for more on this. You were right to ask for counterexamples, since point (1) in my answer was actually misleading, being a limitation of a particular standard proof and not a limitation of any truth. $\endgroup$ Commented Aug 5, 2019 at 16:40

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