# Thursday, March 31 :::{.remark} Last time: $A\in \Dedekind$ and $M\in \amod^\fg$ implies $A\cong \bigoplus A/I_i \oplus \bigoplus_{\mcl_i\in \Pic(A)} \mcl_i$. ::: :::{.exercise title="?"} Show that this implies the classification of finitely-generated modules over a PID: $M \cong \bigoplus A/p_i^{n_i} \oplus A^n$. ::: :::{.solution} Sketch: - PIDs are Dedekind, so ETS $\Pic(A) = \ts{A}$. - Use the SES $K\units \mapsvia{x\mapsto \sum v_p(x) p} \Div A\to \Pic A \to 0$, ETS the given map is surjective. - Reduce: ETS that for $g\in p\in \spec A$ with $p$ nonzero, there exists an $x\in K\units$ such that $v_p(x) = 1$ and $v_{p'}(x) = 0$ for all $p'\neq p$. - If $p = \gens{x}$, this works: $v_p(x) = 1$ since this is the maximal $k$ such that $x\in p^k$, but $x\in p^2\implies p=p^2$, contradicting Nakayama. $\contradiction$ - If $v_{p'}(x) > 0$, then $x\in p'$, $p \subseteq p'$ contradicting maximality of $p$. $\contradiction$ ::: :::{.remark} What about rings of dimension $d \geq 2$? E.g. $\dim \kxn = n$, regarded as functions on $\AA^n\slice k$, noting that we haven't quite defined dimension yet. ::: :::{.conjecture title="Serre-Swan, proved by Quillen"} Finitely-generated locally free modules over $\kxn$ are free. ::: ## Toward dimension: filtered/graded rings :::{.remark} It is easy to slightly modify this statement to make such a classification impossible! ::: :::{.question} How many homogeneous polynomials of degree $d$ in $n$ variables are there? Counting the dimension as a $k\dash$vector space is yields ${n+d\choose d} \sim d^n$. ::: :::{.remark} \envlist Recall: - An increasing filtration $\ts{M_i}$ of $M\in \amod$ satisfies $M_i \subseteq M_{i+1}$ and $\union_i M_i = M$ and there exists some $i_0$ with $M_{i_k} = 0$ for all $i_k\leq i_0$. - A module morphism $M \mapsvia{f} N$ respects filtrations iff $f(M_i) \subseteq N_i$. - For $N\leq M$ there is an induced filtrations $N_i \da M_i \intersect N$. - For $M \surjectsvia{f} Q$ a *surjective* morphism, there is an induced filtration $Q_i \da f(M_i)$. - A **graded abelian group** is a group $M \cong \bigoplus _{i\in \ZZ} M_i$ with $M_i = 0$ for $i\ll 0$. Typical example: $\kxn = \bigoplus _d \kxn_d$ is graded by homogeneous degree $d$ parts. - Passing between filtrations and gradings: given a grading $\oplus_i M_i$, set $\Fil^i M = \bigoplus _{k\leq i} M_i$. Given a filtration, take the associated graded $\bigoplus_i M_i/M_{i-1}$. - Taking the filtration is innocuous and doesn't change the module, but taking the associated graded does. Example: $M = \ZZ$, take the filtration $\gens{0} \subseteq \gens{2} \subseteq \ZZ$ whose associated graded is $\gens{2} \bigoplus \ZZ/2\ZZ \cong \ZZ \oplus C_2$, which now has torsion. - A morphism of filtered modules induces a well-defined map on the associated graded modules. ::: :::{.exercise title="?"} Show that if $f: M\to N$ is a morphism of filtered modules, then $\gr f$ is injective/surjective $\implies f$ is injective/surjective respectively. Show that the converse is not necessarily true unless $f$ is injective/surjective and the filtrations are induced. ::: :::{.solution} For injectivity, if $\gr f$ is injective choose $i$ minimally so that $M_i$ intersects $\ker f$, and contradict $i > -\infty$. By minimality, $x \not\in M_{i-1}$ so $\bar x\neq 0\in \gr^i(M) = M_{i} / M_{i-1}$, but $\gr(f)(\bar x) = 0$. For surjectivity, ETS $f: M_i\to N_i$ is surjective for all $i$. Induct on $i$, using that $M_i = N_i = 0$ for $i \ll 0$. Given $x\in N_i$, take $\bar x\in N_i/N_{i-1}$ and lift to $y\in M_i$ such that $\gr(f)(\bar y) = \bar x$; take $x-f(y)\in N_{i-1}$. ::: :::{.remark} Definitions: - A **graded ring** is a ring $A = \bigoplus_i A_i$ with $A_i A_j \to A_{i+j}$, a filtered ring has a filtration $\ts{A_i}$ with $A_i A_j \to A_{ i+j }$. - A **graded module $M$ over a graded ring $A$** satisfies $A_i M_j \subseteq M_{i+j}$ - A **filtered module over a filtered ring** satisfies $A_i M_j \subseteq M_{i+j}$. ::: :::{.exercise title="?"} Show that if $A$ is filtered then $\gr A$ is naturally graded. ::: :::{.solution} Show \[ {A_i \over A_{i-1}}\cdot {A_j \over A_{j-1}} \to {A_{i+j} \over A_{i+j-1} } .\] ::: :::{.exercise} Show that if $M$ is a filtered module over $A$ a filtered ring, if $\gr(M) \in \mods{\gr(A)}^\fg$ then $M\in \mods{A}^\fg$. ::: :::{.solution} Pick homogeneous generators $\ts{\bar x_i}$ for $\gr M$ and show there is a surjection \[ f: A^r &\to M \\ e_i &\to x_i .\] Reduce to showing $\gr f$ is surjective. ::: :::{.exercise title="?"} Show that if $A$ is a filtered ring and $\gr A$ is Noetherian then $A$ is Noetherian. ::: :::{.solution} Use that $I\in \Id(A)$ is a submodule with an induced filtration and $\gr I \subseteq \gr A$ is finitely-generated to show that $I \subseteq A$ is finitely-generated. ::: :::{.definition title="Good filtrations"} If $A \in \CRing$ is filtered and $M\in\amod$ is filtered, then $\Fil M$ is a **good filtration** if $\gr(M) \in \mods{\gr(A)}^\fg$. ::: :::{.example title="?"} Letting $M\in \amod$ for $A=\kxn$, if $\ts{\Fil^i M}$ is a good filtration then $P(i) \da \dim_k \Fil^i M$ will be a polynomial for $i \gg 0$ and one can define $\dim M = \deg P$. To be justified: - A good filtration exists, - $P$ is asymptotically polynomial, - $P$ is independent of good filtration chosen. In the free rank 1 case: $\dim \kxn_d = {d+n\choose d} \sim d^{n}$, so $\dim \kxn_{\leq d} = \sum_{k\leq n} d^k \sim d^n$. :::