Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $R$ be a commutative ring and $P_1,\ldots ,P_n$ be prime ideals of $R$. If $S=\bigcap_{i=1}^n (R\setminus P_i)$ then show that the ring of fractions $S^{-1}R$ has only finitely many maximal ideals.

The above result will also follow if we can show that the set $\lbrace P\in\mbox{ Spec }(R) : P\subseteq\bigcup_{i=1}^n P_i\rbrace$ has only finitely many maximal elements ( where the partial order is natural inclusion).

share|improve this question
2  
Google "prime avoidance". –  Bill Dubuque Jun 16 '12 at 19:06

2 Answers 2

up vote 4 down vote accepted

It is a standard result in the theory of localizations that there is a one-to-one, inclusion preserving correspondence between the prime ideals of $S^{-1}R$ and the prime ideals of $R$ that are disjoint from $S$.

Thus, the prime ideals of $S^{-1}R$ in the case where $S=R\setminus (\cup P_i)$ corresponds precisely the prime ideals of $R$ that are contained in $\cup P_i$. So we just need to determine what are the ideals of $R$ that are maximal with respect to being contained in $\cup P_i$.

Suppose that $J$ is an ideal of $R$ that is contained in $\cup P_i$. I claim that in fact it is contained in one of the $P_j$.

We proceed by induction on $n$. If $n=1$, there is nothing to prove. If $n=2$, let $J\subseteq P_1\cup P_2$. If $J$ is not contained in either $P_1$ nor $P_2$, let $a\in J\setminus P_1$ (and hence, necessarily, $a\in P_2$) and $b\in J\setminus P_2$ (and hence, necessarily, $b\in P_1$). Then $a+b\in J\subseteq P_1\cup P_2$. But $a\in P_2$, $b\notin P_2$, so $a+b\notin P_2$; and $a\notin P_1$, $b\in P_1$, so $a+b\notin P_1$. Thus, $a+b\notin P_1\cup P_2$, a contradiction.

Assume the result holds for fewer than $n$ primes, $n\gt 2$. If $J$ is contained in the union of any proper subcollection of $P_i$, then we can apply induction. Thus, we may assume that $J$ is not contained in the union of any $n-1$ of the ideals. So for each $j$ we can pick $a_j$ such that $$a_j \in J\setminus \bigcup_{\stackrel{i=1}{i\neq j}}^{n} P_i.$$ Note in particular that $a_j\in P_j$.

Now, consider $a_1\cdots a_{n-1}+a_n$. Since $a_i\in P_i$, then $$a_1\cdots a_{n-1}\in P_1\cap\cdots \cap P_{n-1}\subseteq P_1\cup\cdots\cup P_n.$$ But since $a_i\notin P_n$ for $i=1,\ldots,n-1$ and $P_n$ is prime, then $a_1\cdots a_{n-1}\notin P_n$. On the other hand, $a_n\in P_n\setminus(P_1\cup\cdots \cup P_{n-1})$.

But then $a_1\cdots a_{n-1}+a_n\notin P_n$, since $a_n\in P_n$ but $a_1\cdots a_{n-1}\notin P_n$; and $a_1\cdots a_{n-1}+a_n\notin P_1\cup\cdots \cup P_{n-1}$, since $a_1\cdots a_{n-1}\in P_1\cup\cdots \cup P_{n-1}$ but $a_1\cdots a_{n-1}\notin P_n$. This is a contradiction, since it plainly lies in $J$. This completes the proof of the claim.

Therefore, if $J$ is an ideal of $R$ that is maximal among those that are contained in $P_1\cup\cdots\cup P_n$, then it must be equal to one of the $P_i$ (since it is contained in at least one, and if the inclusion is proper, then that $P_i$ would contradict the maximality).

Therefore, the maximal ideals of $S^{-1}R$ are exactly the ideals that correspond to $P_1,\ldots,P_n$: if $M$ is a maximal ideal of $S^{-1}R$, then it is also a prime ideal, hence corresponds to a prime $Q$ ideal of $R$ that is contained in $P_1\cup\cdots\cup P_n$, and maximality of $M$ implies maximality of $Q$ in $R$ among ideals contained in $P_1\cup\cdots\cup P_n$, which implies that $Q$ is equal to one of the $P_i$, which means $M$ corresponds to one of the $P_i$. In particular, there are only finitely many such maximal ideals in $S^{-1}R$.

share|improve this answer

Hint $\ $ This is known as prime avoidance: ideal $\rm\:I\not\subset P_i\ prime\:\Rightarrow\:I\not\subset \cup\, P_i.\:$ Here is the inductive step (for $\rm\:n=3)$. By induction there is $\rm\:r_1 \in I\:$ but $\rm\:r_1\not\in\:P_2\cup P_3.\:$ If $\rm\:r_1\not\in P_1\:$ we are done. Similarly there are $\rm\:r_2,r_3\in I,\:$ but $\rm\: r_2 \not\in P_1\cup P_3,\:$ and $\rm\: r_3\not\in P_1\cup P_2.\:$ Again, all $\rm\: r_i \in P_i,\:$ else the proof is complete. Hence $\rm\:r_i\in P_j \iff i = j,\: $ therefore

$$\rm I\, \ni\, r = r_2 r_3 + r_1 r_3 + r_1 r_2\, \not\in\, P_1\cup P_2\cup P_3$$

since, e.g. $\rm\:mod\ P_1\!:\ r_1\equiv 0\:\Rightarrow\: r\equiv r_2 r_3\not\equiv 0\:$ by $\rm\:r_2,r_3\not\equiv 0.\:$ Ditto for the other $\rm\:P_i.$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.