# Thursday, April 28 ## Dimension theory :::{.remark} Given an arbitrary grid, can you tile it with $2\times 1$ dominoes? What dominoes corresponding to Young's diagrams for partitions $\lambda = (2, 1)$? A principled way of approaching such problems: consider labeling the grid with monomials: \begin{tikzcd} \vdots \\ {y^2} & {xy^2} & {x^2y^2} \\ y & xy & {x^2y} \\ 1 & x & {x^2} & \cdots \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsMTEsWzAsMywiMSJdLFsxLDMsIngiXSxbMCwyLCJ5Il0sWzAsMSwieV4yIl0sWzEsMiwieHkiXSxbMiwzLCJ4XjIiXSxbMSwxLCJ4eV4yIl0sWzIsMiwieF4yeSJdLFsyLDEsInheMnleMiJdLFswLDAsIlxcdmRvdHMiXSxbMywzLCJcXGNkb3RzIl1d) Now labeling the $2\times 1$ tile with $1,x$ and the $1\times 2$ tiles with $1,y$. Note that there is a polynomial $f(x, y) = 1 + x + y + x^2 + xy + y^2+\cdots$ associated to the grid; if there admits a tiling then $f\in \gens{1+x, 1+y} \subseteq k[x,y]$. One can then check that $\bar f\in k[x,y]/\gens{1+x,1+y} = k$ satisfies $\bar f(-1,-1) = 2$, so $\bar f\neq 0$. Note that $\bar f = 0$ is a necessary but not sufficient condition. ::: :::{.remark} Last few topics: toward the Cohen structure theorem. Setup: $A\in \Noeth\Loc\Ring$ and let $\mfm\in \mspec A$, e.g. - $A = k\fps{x_1,\cdots, x_n}$, - $A = (\kxn/I)_{\mfp}$ - $A= \ZZpadic\fps{x_1,\cdots, x_n}/I$. We don't have a good dimension theory for these, since they aren't finitely generated algebras. Some approaches: - Hilbert dimension, - Krull dimension, - Generator dimension. ::: :::{.definition title="Hilbert dimension"} Setup: - Let $M\in\amod^\fg$ and choose an $\mfm\dash$good filtration $\Fil_i M$ on $M$, so $\mfm M_i \subseteq M_{i+1}$ for $i\gg 0$. Equivalently, $\Rees M$ is finitely-generated over $\Rees A$. - Let $\Phi(i) = \len(M/\Fil_i M)$, which is eventually polynomial. - Define $\dim_A M = \deg \Phi$. This defines a dimension for any finitely generated module. ::: :::{.remark} There is a naturally good filtration: $\Fil_i M \da \mfm^i M$, then \[ \Phi(i) = \len(M/\Fil_i M) = \sum_{0\leq j\leq i-1} \dim_{A/\mfm} \gr_j M \] where $\gr_j M = \mfm^j M/ \mfm^{j+1} M$. ::: :::{.lemma title="?"} \envlist - Given a SES $A\to B\to C$, $\hilbdim B = \max(\hilbdim A, \hilbdim C)$. - For $\Phi: M\injects M$, $\hilbdim M / m\phi \leq \hilbdim M - 1$ ::: :::{.example title="?"} Check $\hilbdim k\fps{x_1,\cdots, x_n} = n$ using $\mfm = \gens{x_1,\cdots, x_n}$; then $\Phi(i) = \len k\fps{x_1, \cdots, x_n}/\mfm^i = \sum_{0\leq j\leq i-1}{n+j\choose j}$ by counting monomials. ::: :::{.exercise title="?"} Show $\hilbdim \ZZpadic \fps{x_1,\cdots, x_n} = n+1$ using $\mfm = \gens{p, x_1,\cdots, x_n}$. ::: :::{.exercise title="?"} Find $\krulldim k\fps{x_1,\cdots, x_n}$ by finding a maximal chain of prime ideals. ::: :::{.definition title="Generator dimension"} Define $\gendim A$ to be the minimal $d$ such that there exist $x_1,\cdots, x_d\in \mfm = \sqrt{\gens{x_1,\cdots, x_d}}$. ::: :::{.theorem title="?"} If $A\in\kalg^\fg$ then $\dim A = \max_{m\in \mspec A} \dim A_m$. ::: :::{.exercise title="Dimension $\neq$ minimal number of generators"} Show that for $A = k\fps{x_1, x_2}/\gens{x_2^2}$ that $\gens{x_1,x_2}$ is a minimal set of generators for the maximal ideal but $\gendim A = 1$ since $m =\sqrt{x_i}$. ::: :::{.exercise title="?"} Show if $A\in \Loc\Noeth\CRing$ with $\mfm \in \mspec A$, then \[ \dim A \leq\dim_{A/\mfm} \mfm/\mfm^2 .\] ::: :::{.solution} Hint for using generator dimension: pick a basis $\ts{\bar x_i}_{1\leq i\leq n}$ of $\mfm/\mfm^2$. Lift to $\ts{x_i}$ and take $\mfm \da \gens{x_1,\cdots, x_n}$ and conclude by Nakayama. Hint for using Hilbert dimension: show $\hilbdim A = \hilbdim \oplus m^i/m^{i+1}$, since the LHS is $\deg(i\mapsto \len A/m_i)$ and the RHS is $\deg(i \mapsto \sum_{0\leq j\leq i-1} \dim_k m^j/m^{j+1})$. Then show there is an inequality $\hilbdim \bigoplus_i m^i/m^{i+1} \leq \dim_k m/m^2$ by showing there is an important multiplication map $\Symalg m/m^2 \surjects \oplus_i m^i/m^{i+1}$. Conclude using that the LHS is isomorphic to a power series rings $k\fps{\bar x_1,\cdots, \bar x_n}$ which is dimension $n$. ::: :::{.remark} $\oplus m^i/m^{i+1}$ is the tangent cone and $\Symalg m/m^2$ are functions on the tangent space, and the tangent cone is a subvariety of the tangent space. ::: :::{.remark} Let $M = k\fps{x,y}/\gens{x,y}$, $m = \gens{x,y}$, and $m/m^2 = \gens{\bar x, \bar y}$. Then $\spec \Symalg m/m^2$ is a curve (?) and $\oplus_i m^i/m^{i+1} = k\fps{x,y}/\gens{xy}$. ::: :::{.exercise title="?"} For $k\fps{x,y}/\gens{y^2-x^3}$, $m/m^2 = \gens{\bar x, \bar y}$, $m^2/m^3 = \gens{\bar x^2, \bar x\bar y}$, so $\oplus_i k\fps{x,y}/\gens{y^2}$. This yields a thickened line along $y=0$: ![](figures/2022-04-28_13-36-33.png) ::: :::{.slogan} The dimension of a variety is at most the dimension of its tangent spaces. ::: ## Smoothness and regularity :::{.remark} Idea: regularity is "almost smooth". ::: :::{.definition title="Regularity"} A ring $(A, m)\in \Noeth\Loc\CRing$ is **regular** iff the natural multiplication map $\Symalg m/m^2 \surjects \oplus_i m^i/m^{i+1}$ is an isomorphism. For arbitrary $A\in \Noeth\CRing$ (not necessarily local), $A$ is *regular* iff $A_m$ is regular for all $m\in \mspec A$. ::: :::{.example title="?"} \envlist - $\kxn$ is regular. - $k\fps{x_1,\cdots, x_n}$ is regular. ::: :::{.exercise title="?"} Any field is regular local, but regularity is not preserved under extensions. Take $\FF_p\functionfield{t^{1\over p}} \contains \FF_p\functionfield{t}$ and show \[ A \da \FF_p\functionfield{t^{1\over p}} \tensor_{\FF_p\functionfield{t}} \FF_p\functionfield{t^{1\over p}} \cong \FF_p\functionfield{t^{1\over p}}\adjoin{s}/\gens{s^p} ,\] which is not reduced. Use that $\FF_p(t^{1\over p}) = \FF_p(t)[x]/\gens{x^p-t}$, so \[ A = \FF_p(t)[x,y]/\gens{x^p-t, y^p-t} \\ = \FF_p(t^{1\over p})[y] / \gens{y^p - (t^{1\over p})^p } \\ = \FF_p(t^{1\over p})[y]/\gens{y - t^{1\over p}}^p .\] ::: :::{.definition title="Smoothness"} If $A\in \kalg^\fg$ and $\bar k = \algcl k$, then $A$ is **smooth** iff $A\tensor_k \kbar$ is regular. ::: :::{.exercise title="?"} Show that a ring $A\in \Noeth\Loc\CRing$ is regular iff $\dim A = \dim _{A/m} m/m^2$. ::: :::{.solution} Hint: use that regularity implies equality implies that the map $\pi: \Symalg m/m^2\to \oplus m^i/m^{i+1}$ is an isomorphism. Assume this is not an equality, let $f\in \ker \pi$, then $\dim A = \dim \oplus m^i / m ^{i+1} \leq \dim \Symalg m/m^2/f < \dim \Symalg m/m^2 = \dim m/m^2$, a contradiction. ::: :::{.exercise title="?"} Show that regular local rings are domains. ::: :::{.solution} Pick nonzero zero divisors, $a,b$ with $ab=0$. Then $\bar a, \bar b \in m^i/m^{i+1}$ with $\bar a \bar b = 0$, a contradiction. :::