What does it mean for two ideals to coincide locally on a Zariski open subset?

Question

Let $R$ be a Noetherian ring and $I,J$ $R$-ideals. Write $I^{\left<J\right>}=\bigcup\limits_{n\geq1}(I:J^n)$, where $(I:J^n)=\{r\in R\mid rJ^n\subset I\}$. Show $I^{\left<J\right>}$ is the unique largest $R$-ideal that coincides with $I$ locally on the open set $\operatorname{Spec}(R)\setminus V(J)$ where $\operatorname{Spec}(R)$ is the set of all prime ideals of $R$ with the Zariski topology and $V(J)$ is the set of all prime ideals containing $J$.

My question is what it means for two ideals to coincide locally on a Zariski open subset?

Answer 1

Let $U = \operatorname{Spec} R \setminus V(J)$. I’m guessing that means that for every $\mathfrak p ∈ U$, the localizations of these ideals at $\mathfrak p$ coincide, that is to say $I_{\mathfrak p} = I^{⟨J⟩}_{\mathfrak p}$. At least that’s what is true, if I have checked correctly.

$I^{⟨J⟩}$ does locally coincide with $I$ in this way:

• Obviously, $I ⊆ I^{⟨J⟩}$, so $I_{\mathfrak p} ⊆ I^{⟨J⟩}_{\mathfrak p}$ for all $\mathfrak p ∈ U$.
• Let $\mathfrak p ∈ U$. Then $\mathfrak p \not ⊇ J$, so there is some $c ∈ R\setminus \mathfrak p$ with $c ∈ J$. Hence for all $r ∈ I^{⟨J⟩}$ there is some $n ∈ ℕ$ with $rc^n ∈ I$ and so for any $s ∈ R\setminus \mathfrak p$, you have $r/s = rc^n/sc^n ∈ I_\mathfrak p$, so $I^{⟨J⟩}_{\mathfrak p} ⊆ I_{\mathfrak p}$.

I also tried checking that it’s the largest ideal, but I haven’t succeeded. But that wasn’t the question anyway, now was it?

Answer 2

$I^{\left<J\right>}$ is the unique largest $R$-ideal that coincides with $I$ locally on the open set $\mathrm{Spec}(R)\setminus V(J)$.

Proof. Since $R$ is Noetherian $I$ has a primary decomposition, $I=\bigcap\limits_{i=1}^{m}q_{i}$ with $q_i$ being $p_i$-primary. Then $$I^{\left<J\right>}= \bigcup\limits_{n≥1}^{}(\bigcap\limits_{i=1}^{m}q_{i}:J^n)= \bigcup\limits_{n≥1}^{}\bigcap\limits_{i=1}^{m}(q_{i}:J^n)= \bigcap\limits_{i=1}^{m}\bigcup\limits_{n≥1}^{}(q_{i}:J^n)= \bigcap\limits_{i=1}^{m}{q_i}^{\left<J\right>}=\bigcap\limits_{J\not\subset p_i}^{}q_i.$$ (The second equality from Exercise 1.12, the last equality from generalized Lemma 4.4 in Introduction to Commutative Algebra by M.F. Atiyah, I.G. Macdonald)

Let $K$ be an $R$-ideal that coincides with $I$ locally on the open set $\mathrm{Spec}(R)\setminus V(J)$. Take a primary decomposition of $K=\bigcap\limits_{i=1}^{l}s_{i}$. For every $p_j\in\{p_1,...,p_m\}$ with $J\not\subset p_j$, $K_{p_j}=(\bigcap\limits_{i=1}^{l}s_{i})_{p_j}=I_{p_j}=(\bigcap\limits_{J\not\subset p_i}^{}q_i)_{p_j}$. Then since the localization preserves intersections and the primary ideals in a localization are in one-to-one correspondence with the primary ideals of $R$ which don't meet the multiplicative closed subset (Proposition 3.11, 4.8 in the book above), for $q_t\subseteq p_j$, $\bigcap q_t$ is the intersection of some $s_i$. Then $$\{q_i\mid q_i\text{ is } p_i\text{-primary and } J\not\subset p_i \text{ for } i=1,2,...,m\}\subseteq\{s_1,...,s_l\}.$$ So $K\subseteq I^{\left<J\right>}$ and $I^{\left<J\right>}$ is the unique largest such ideal.