“Only by avoiding the beginning of things can we escape their end.”
―Cyril Connolly
Remark11.34.
We will now discuss an important lemma known as Prime Avoidance. This is an elementary fact, but it is very helpful. Prime Avoidance says that if an ideal \(I\) is not contained in any of the primes \(P_1, \ldots, P_n\text{,}\) then \(I\) cannot be contained in their union. Set-theoretically this is possible, of course; but if the ideals \(P_1, \ldots, P_n\) are all prime, then it is actually not possible for \(I \subseteq P_1 \cup \cdots \cup P_n\) unless \(I\) is contained in one of the \(P_i\text{.}\) In fact, for this to work we can even allow two of the \(P_i\) to be just any ideals, as long as the remaining \(P_i\) are prime.
Lemma11.35.Prime Avoidance.
Let \(R\) be a ring, \(I_1,\dots,I_n,J\) be ideals, and suppose that \(I_i\) is prime for \(i>2\text{.}\) If \(J\not\subseteq I_i\) for all \(i\text{,}\) then \(J\not\subseteq \bigcup_i I_i\text{.}\) Equivalently, if \(J\subseteq \bigcup_i I_i\text{,}\) then \(J\subseteq I_i\) for some \(i\text{.}\)
Moreover, if \(R\) is \(\N\)-graded, and all of the ideals are homogeneous, all \(I_i\) are prime, and \(J\not\subseteq I_i\) for all \(i\text{,}\) then there is a homogeneous element in \(J\) that is not in \(\bigcup_i I_i\text{.}\)
Proof.
We proceed by induction on \(n\text{.}\) If \(n=1\text{,}\) there is nothing to show. When \(n=2\text{,}\) we have two ideals \(I_1\) and \(I_2\) and an ideal \(J\) such that \(J \nsubseteq I_1\) and \(J \nsubseteq I_2\text{,}\) and we want to show that \(J \nsubseteq I_1 \cup I_2\text{.}\) By assumption there exist elements \(a_1, a_2 \in J\) with \(a_1 \notin I_1\) and \(a_2 \notin I_2\text{.}\) If \(a_2 \notin I_1\) or \(a_1 \notin I_2\text{,}\) then we have an element that is not in \(I_1 \cup I_2\text{,}\) and we are done. On the other hand, if \(a_2 \in I_1\) and \(a_1 \in I_2\text{,}\) then consider \(c = a_1 + a_2 \in J\text{.}\) Since \(a_1 \in I_1\) but \(a_2 \notin I_1\text{,}\) then \(c \notin I_1\text{.}\) Similarly, \(c \notin I_2\text{.}\) Therefore, \(c \notin I_1 \cup I_2\text{.}\)
Now suppose that the statement holds for some \(n-1 \geq 2\text{,}\) and consider ideals \(I_1, \ldots, I_{n}\) with \(I_i\) prime for all \(i \geq3\) such that \(J \notin I_i\) for any \(i\text{.}\) By induction hypothesis, for each \(i\) we have
\begin{equation*}
a_i \notin \bigcup_{j\neq i} I_j \textrm{ and } a_i \in J.
\end{equation*}
If some \(a_i\notin I_i\text{,}\) we are done, so let’s assume that \(a_i\in I_i\) for each \(i\text{.}\) Consider
\begin{equation*}
a := a_n + a_1 \cdots a_{n-1} \in J.
\end{equation*}
Notice that \(a_1\cdots a_{n-1}=a_i(a_1\cdots \widehat{a_i} \cdots a_{n-1})\in I_i\text{.}\) If \(a\in I_i\) for \(i<n\text{,}\) then we also have \(a_n\in I_i\text{,}\) a contradiction. If \(a\in I_n\text{,}\) then we also have \(a_1\cdots a_{n-1} = a - a_n \in I_n\text{,}\) since \(a_n\in I_n\text{.}\) Since \(n>2\text{,}\) our assumption is that \(I_n\) is prime, so one of \(a_1,\dots,a_{n-1}\in I_n\text{,}\) which is a contradiction. So \(a \notin I_i\) for all \(i\text{,}\) and thus \(a\) is the element we were searching for.
If all \(I_i\) are homogeneous and prime, then we proceed as above but replacing \(a_n\) and \(a_1,\dots,a_{n-1}\) with suitable powers so that \(a_n + a_1 \cdots a_{n-1}\) is homogeneous. For example, we could take
The primeness assumption guarantees that noncontainments in ideals is preserved.
Theorem11.36.Stronger Prime Avoidance.
Let \(R\) be a ring, \(P_1, \ldots, P_n\) prime ideals, \(x \in R\) and \(I\) be an ideal in \(R\text{.}\) If \((x) + I \not\subseteq P_i\) for each \(i\text{,}\) then there exists \(y \in J\) such that
\begin{equation*}
x + y \notin \bigcup_{i=1}^n P_i.
\end{equation*}
Proof.
We proceed by induction on \(n\text{.}\) When \(n=1\text{,}\) if every element of the form \(x+y\) with \(y \in R\) is in \(P= P_1\text{,}\) then multiplying by \(r \in R\) we conclude that every \(rx + y \in P\text{,}\) meaning \((x) + I \subseteq P\text{.}\)
Now suppose \(n>1\) and that we have shown the statement for \(n-1\) primes. If \(P_i \subseteq P_j\) for some \(i \neq j\text{,}\) then we might as well exclude \(P_i\) from our list of primes, and the statement follows by induction. So assume that all our primes \(P_i\) are incomparable.
If \(x \notin P_i\) for all \(i\text{,}\) we are done, since we can take \(x+0\) for the element we are searching for. So suppose \(x\) is in some \(P_i\text{,}\) which we assume without loss of generality to be \(P_n\text{.}\) Our induction hypothesis says that we can find \(y \in I\) such that \(x+y \notin P_1 \cup \cdots \cup P_{n-1}\text{.}\) If \(x+y \notin P_n\text{,}\) we are done, so suppose \(x+y \in P_n\text{.}\) Since we assumed \(x \in P_n\text{,}\) we must have \(I \not\subseteq P_n\text{,}\) or else we would have had \((x) + I \subseteq P_n\text{.}\) Now \(P_n\) is a prime ideal that does not contain \(P_1, \ldots, P_{n-1}\text{,}\) nor \(I\text{,}\) so
\begin{equation*}
P \not\supseteq I P_1 \cdots P_{n-1}.
\end{equation*}
Choose \(z \in I P_1 \cdots P_{n-1}\) not in \(P_n\text{.}\) Then \(x+y+z \notin P_n\text{,}\) since \(z \notin P_n\) but \(x+y \in P_n\text{.}\) Moreover, for all \(i<n\) we have \(x+y+z \notin P_i\text{,}\) since \(z\in P_i\) and \(x+y \notin P_i\text{.}\)
