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 $(Q:f^\infty)=(1)$.
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