Splitting Polynomial And Saturations

Problem: Prove the following statements from lecture:

• A splitting polynomial for $I$ is an $f \in R$ such that if $(I : f^\infty)=(I : f^\ell)$, then both $(I:f^\infty)$ and $(I:f^\ell)$ are strictly larger ideals than $I$. Show there is a splitting polynomial for $I$ if and only if $I$ is not primary.
• If $Q$ is $P$-primary, then
• $f \notin P \implies (Q : f) = Q$, and $(Q: f^\infty)=Q$,
• $f \in P, f \notin Q \implies (Q:f)$ is $P$-primary and $(Q:f^\infty)=(1)$.

Solution:

• Suppose there is a splitting polynomial $f$ for $I$. Suppose $I$ were primary. Consider $g \in (I:f^\infty)$. Then $gf^\ell \in I$ for some $\ell$. Since $I$ is primary, one has that either $g \in I$ or $f^\ell \in I$. If [[f^\ell \in I$]], then$(I,f^\ell)=I$, contradicting that$(I,f^\ell)$is strictly larger than$I$. To avoid a contradiction, one must assume that one had$g \in I$for all$g \in (I:f^\infty)$. But then one has that$(I:f^\infty)\subset I$, which is also a contradiction. Thus,$I$cannot be primary. • Now suppose$I$is not primary. Consider some$fg \in I$. Since$I$is not primary, one must have that$g \notin I$and$f^N \not I \forall N$. One claims that$f$is a splitting polynomial for$I$. • Consider$(I:f^\infty)$. Recall that, by assumption$fg \in I$, meaning that$g \in (I \colon f^\infty)$. From above, one has that$g \notin I$, so$(I \colon f^\infty)$is strictly larger than$I$. • Consider$(I,f)$. From above, one has that$f \notin I$, so$(I,f)$must be strictly larger than$I$. • Suppose$Q$is$P$-primary. • Suppose$f \notin P$. Suppose$g \in (Q : f)$. Then$gf \in Q$, so since$Q$is primary,$g \in Q$or$f^N \in Q$. If$f^N \in Q$, then$f \in \sqrt{Q}=P$, contradicting that$f \notin P$. Therefore, one must have that$g \in Q$, so$(Q:f) \subset Q$. One then has that$Q = (Q:f)$. • Suppose$f \in P, f \notin Q$. • Suppose$gh \in (Q:f)$. Then$ghf \in Q$. Since$Q$is primary, one has that either$f \in Q$or$(gh)^N \in Q$. But$f \notin Q$, so one must have that$(gh)^N \in Q$. Since$Q$is primary, one has that either$g^M \in Q$or$h^M \in Q$for some sufficiently large$M$. In either case, one has that either$g^M \in (Q:f)$or$h^M \in (Q:f)$, which is exactly what is needed for$(Q:f)$to be primary. One now needs to show that$\sqrt{(Q:f)}=P$. Since$Q \subset (Q:f)$, one has that$P=\sqrt{Q}\subset \sqrt{(Q:f)}$. To show inclusion the other way, suppose$g \in \sqrt{(Q:f)}$. Then$g^kf \in Q$, so either$g^k \in Q$or$f^N \in Q$and either$g^{km} \in Q$or$f \in Q$. Since$f \notin Q$, one must have that$g^kM \in Q$, so$g \in \sqrt{Q}=P$, meaning$\sqrt{(Q:f)} \subset P$. Thus,$(Q:f)$is$P$-primary. • Now we consider$(Q:f^\infty)$. Since$f \in P=\sqrt{Q}=\sqrt{(Q:f)}$, one has that$f^k \in Q$. Then$1 \cdot f^k \in Q$, so one must have$1 \in (Q:f^\infty)$. Thus$(Q:f^\infty)=(1)\$.