# Localization and Fractional Ideals (Tuesday, August 30) ## Continuing localization for rings :::{.remark} Recall - $S \leq (R, \cdot)$ a submonoid, $\iota: R\to S\inv R$ is universal for maps $f: R\to R'$ with $f(R) \subseteq (R')\units$. - $f\in \CRing(A, B)$ induces \[ f_* &\in \Set(\Id A, \Id B) \\ I &\mapsto IB = \gens{f(I)}_B \] and \[ f^* &\in \Set(\Id B, \Id A) \\ J &\mapsto f\inv(J) .\] - For $I\normal R$, $I \intersect S = \emptyset \implies \iota_*(I) \subset S\inv R$ is proper. - For $J\normal S\inv R$, $\iota_* \iota^* J = J$, so $\iota^*$ is an injection. Thus we can view $\Id(S\inv R) \subseteq \Id(R)$. - If $p\in \spec R$ and $p \intersect S = \emptyset$ then $\iota_*(p) \in \spec S\inv R$ and $\iota^* \iota_* p = p$. Thus there is a bijection \[ \spec S\inv R \mapstofrom \ts{p\in \spec R \st p \intersect S = \emptyset} .\] - For $p\in \spec R$, define $S\da R\sm p$ and $R_p \da S\inv R = (R\sm p)\inv R$. Thus there is a bijection \[ \spec R_p \mapstofrom \ts{q\in \spec R \st q \subseteq p} .\] - What this does: truncates the lattice $\Id(R)$ and keeps the underlattice (downward transitive closure?) of $p$: ![](figures/2022-08-30_11-22-42.png) - This is dual to quotienting, where $\spec R/I$ is the overlattice (upward transitive closure?) of $I$. - $R_p\in \Loc\CRing$ with $\mspec R_p = \ts{\mfm_p \da pR_p}$. - *Semilocal* ring are those with finitely many maximal ideals. - We defined **semilocalization**: given $\ts{p_i}$ with no pairwise containments, define $R_{p_1,\cdots, p_r} = \qty{\Intersect _{i\leq r} R\sm p_i} \inv R$. - $\spec R_{p_1,\cdots, p_r} = \ts{q\in \spec R \st q \intersect \qty{\Intersect _{i\leq r} R\sm p_i} } = \ts{q\in \spec R \st q \subseteq \Union_{i\leq r} p_i}$. - By **prime avoidance**, equivalently $q \subseteq p_i$ for some $i$, and so $\spec R_{p_1,\cdots, p_r} = \ts{p_1,\cdots, p_R}$. - Pushforward and localization commute, see \autocite[CA 7.7]{CA}: \[ {S\inv R\over IS\inv R}\cong {S\inv R \over I} .\] - For $R$ a Dedekind domain (Noetherian, $\krulldim = R$, and integral closure condition), let $\ts{p_i}_{i\leq v} \in \mspec R$. The quotient factors through the semilocalization: \begin{tikzcd} & {R_{p_1,\cdots, p_v}} \\ \\ R && {R/\prod p_i^{e_i}} \arrow[from=3-1, to=3-3] \arrow[from=3-1, to=1-2] \arrow[dashed, from=1-2, to=3-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwyLCJSIl0sWzEsMCwiUl97cF8xLFxcY2RvdHMsIHBfdn0iXSxbMiwyLCJSL1xccHJvZCBwX2lee2VfaX0iXSxbMCwyXSxbMCwxXSxbMSwyLCIiLDIseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=) ::: ## Localization for modules :::{.remark} Notes: - Every residue ring of a Dedekind domain is also a residue ring of a PID. - Projectives might not be free for Dedekind domains -- take any non-principal ideal. - On classification of modules over a Dedekind domain: see later remark. - Coming up later: lattices over Dedekind domains. - Local properties: properties true for $M$ iff true for $M_p$ for all $p$. Examples: being zero, flatness, finitely-generated, etc. - $f\in \rmod(M, N)$ is injective/surjective/bijection iff $f_p \in \mods{R_p}(M_p, N_p)$ is injective/surjective/bijective for all $p$. ::: :::{.warnings} A map being an *isomorphism* is a local property, but abstractly *being isomorphic* is not! For example, $\Pic(R)$ measures rank 1 projective modules, which are all locally isomorphic. ::: :::{.exercise title="Localization for modules"} To define localization for modules, given $S \subseteq R$ and $M\in \rmod$, we want $S\inv M \in \mods{S\inv R}$ and $\iota\in\rmod{M, S\inv M}$. One construct this as an equivalence relation on $S\times M$, or define $S\inv M \da M\tensor_R S\inv R$ with $M\to S\inv M$ by $x\mapsto x\tensor 1$. Show that these are equivalent. > Hint: use a universal property, where $S\acts S\inv M$ by *automorphisms* instead of just endomorphisms. ::: :::{.remark} On the classification of modules over a Dedekind domain: if $M\in \rmod^{\fg, \tors}$, so $\Ann_R(M) \neq 0$. These admit a classification, proved by reducing to the case when $R$ is a PID. Writing $M = \gens{x_1,\cdots, x_n}_R$, an exercise shows $\ann_R(M) = \Intersect _{i\leq n} \Ann_R(x_i) \contains \prod \Ann_R(x_i) \supsetneq \gens{0}$. Decomposing $\Ann_R(M) = \prod p_i^{e_i}$, $M$ is a faithful $(R/ \prod p_i^{e_i})\dash$module, and thus a finitely generated torsion module over $R_{p_1,\cdots, p_r}$ -- this is a PID! We can write $P\da \prod p_i^{e_i}$, we have $R/P = R_{p_1,\cdots, p_r}/ P R_{p_1,\cdots, p_r}$, and thus $M \cong \bigoplus _{i=1}^s R/p_i^{b_i}$ is a sum of cyclic modules. ::: :::{.theorem title="Super important!"} Let $R\in \CRing$ by either Noetherian or a domain and let $M\in \rmod^{\fg}$. Then $M$ is projective iff $M$ is *locally free*, i.e. for every $m\in\mspec R$, the localization $M_m$ is a free $R_m\dash$module. ::: :::{.slogan} The local version of being projective is just being free.[^loc_triv] [^loc_triv]: Similar to vector bundles being locally trivial. ::: :::{.remark} Every ideal of a Dedekind domain $R$ is projective, which thus become free after localizing. Geometrically, $\spec R$ is an integral affine curve -- dimension 1, nonsingular, Noetherian. It's also reasonable in NT to consider finite products of Dedekind domains. However, tensoring to the completion may not yield a Dedekind domain (which shows up in NT2). ::: :::{.corollary title="?"} If $R$ is a domain and $P\in \rmod^{\proj, \rank=1}$, \[ \Endo_R(P) = R .\] ::: :::{.proof title="?"} Take the structure map $f: R\to \endo_R(M)$ where $x\mapsto x\cdot$; it's ETS $f_m: R_m \to \endo_R(P) \tensor_R R_m \cong \Endo_{R_m}(P_m)$$ is an isomorphism. Since $P_m$ is a rank 1 projective $R_m\dash$module at $R_m$ is local, all such modules are isomorphic and thus $P_m\cong R_m$. Conclude since $\Endo_{R_m}(R_m) = R_m$. ::: ## Fractional Ideals :::{.definition title="Fractional ideals"} Let $R$ be a domain with $K \da \ff(R)$. A **fractional $R\dash$ideal** is a nonzero $R\dash$module of $K$ for which there exists an $a\in R\smz$ such that $aM \subseteq R$.[^how_to] A usual ideal is sometimes referred to as an **integral ideal**. Define $\Frac R$ to be the set of fractional ideals. [^how_to]: How to think of this: $M \subseteq {1\over a} R$, so this is a "bounded denominators" result. ::: :::{.exercise title="?"} Show that a. Every finitely-generated nonzero $R\dash$submodule of $K$ is a fractional ideal. This should be reasonable: think of finding common denominators. b. If $R$ is Noetherian then every fractional ideal if a finitely-generated $R\dash$module. Thus defining fractional ideals as finitely-generated submodules only coincides with the above definition in the Noetherian case. ::: :::{.remark} Operations on $\Frac(R)$: - $I+J \da \gens{I, J}_R = \ts{i+j}$, - $IJ \da \gens{\ts{ij \st i\in I, j\in J}} = \ts{i_1j_1 + \cdots i_n j_n}$, - $I \intersect J$, - $(I: J) \da \ts{x\in K \st xJ \subseteq I}$ (**by far the most important operation on ideals!**) Note that $(\Frac R, \cdot, R)$ is a commutative monoid -- how far is this from being a group? A fractional ideal $I$ is **invertible** iff it is invertible in this monoid, i.e. there is a $J$ such that $IJ = R$. Define $\Inv(R)$, the commutative group of invertible fractional ideals, and define $I^* \da (R:I)$. ::: :::{.exercise title="?"} If $I \in \Frac(R)$ and $J\in \Inv(R)$, \[ (I:J) = IJ\inv .\] ::: :::{.slogan} For Dedekind domains, to contain is to divide. ::: :::{.exercise title="?"} \envlist a. For $I\in \Frac(R)$, show that $I\inv \iff II^* = R$. b. Show that if $I \subseteq J$ are in $\Frac(R)$ and $J\in \Inv(R)$ then $I = J(I: J)$ so $J\divides I$. ::: :::{.remark} There is always an injection $\iota: I I^* \injects R$, so invertibility can be checked locally: $I\in \Inv(R) \iff \iota$ is an isomorphism $\iff \iota_p: I_m I_m^* \cong R_m$ for all $m$. Also note that if $I$ is principal, so $I = \gens{x} = Rx$ for some $x\in K\units$ implies $I\inv = \gens{x}\inv = \gens{x\inv}$. So a fractional $R\dash$ideal is invertible iff locally principal. :::