# Thursday, April 07 ## Hilbert dimension of finitely generated algebras over a field :::{.remark} Prove: - $f: M\injects M$ implies $\dim \coker f < \dim M$. - If $A = \kxn$ and $f\in A\smz$ then $\dim A/f < \dim A$. - If $A\in \mods{\kxn}^\fg$ and $f\in A$ is a nonzero divisor, then $\dim A/F < \dim A$. ::: :::{.definition title="Dimension"} If $A\in \kalg^\fg$, choosing a surjection $\kxn \surjects A$, define $\dim A \da \dim_{\kxn} A$. ::: :::{.exercise title="?"} Show this is well-defined. It follows from the following: ::: :::{.theorem title="?"} Let $M\in \mods{k[x_1,\cdots, x_n, y_1,\cdots, x_n]}$ with $M\in \mods{\kxn}^\fg$. Then \[ \dim_{\kxn} M = \dim_{k[x_1,\cdots, x_n, y_1,\cdots, y_n] } .\] ::: :::{.solution} Why the theorem implies the exercise: \begin{tikzcd} && {} \\ \textcolor{rgb,255:red,214;green,92;blue,92}{k[x_1,\cdots, x_n, y_1, \cdots, y_m]} && \textcolor{rgb,255:red,214;green,92;blue,92}{k[y_1,\cdots, y_m]} \\ \\ \textcolor{rgb,255:red,92;green,92;blue,214}{k[x_1,\cdots, x_n]} && A \\ \\ {} && {} \arrow["g", color={rgb,255:red,214;green,92;blue,92}, two heads, from=2-1, to=2-3] \arrow[color={rgb,255:red,214;green,92;blue,92}, two heads, from=2-3, to=4-3] \arrow[color={rgb,255:red,92;green,92;blue,214}, two heads, from=4-1, to=4-3] \arrow["{\exists: x_i\mapsto x_i, y_i\mapsto g\inv(y_i)}"', dashed, from=2-1, to=4-1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) This can be filled in to a commutative diagram, so \[ \dim_{\kxn} A = \dim_{k[x_1,\cdots, x_n, y_1,\cdots, y_m]} A = \dim_{k[y_1, \cdots, y_m]} A .\] ::: :::{.proof title="of theorem"} It's ETS this when $m=1$, so suppose $M\in \mods{ \kxn[y] }$ with $M\in \mods{\kxn}^\fg$. Take the surjection $\kxn^r\surjects M$ where $e_i\mapsto m_i$ for generators $m_i$, and let $M_i$ be the induced filtration. Similarly take $k[x_1,\cdots, x_n, y]^r \surjects M$ with $e_i\mapsto m_i$ and let $\tilde M_i$ be the induced filtration. Since $M_i \subseteq \tilde M_i$, $\dim_{\kxn} M \leq \dim_{\kxn[y]} M$. The claim is that one can choose $k$ such that $yM_0 \subseteq M_k$ and $\tilde M_i \subseteq M_{ik}$. Why: $\tilde M_i = \ts{m\in M \st m \text{ is in the image of } (f_1,\cdots, f_r) \text{ of degree at most } i}$. So write $f_j = \sum_{n} f_{j, n}(x_1,\cdots, x_n)y^n$, then $\tilde M_i \subseteq \spanof (M_i + yM_{i-1} + y^2 M_{i-2} + \cdots)$. Why this finishes the proof: $\dim_{\kxn} M = \dim_{\kxn[y]} M$, so - $\Phi(i) = \dim M_i$ - $\tilde \Phi(i) = \dim \tilde M_i$ - $\Phi(i) \leq \tilde \Phi(i)\leq \Phi(ik)$, which forces $\deg \Phi = \deg \tilde \Phi$. ::: :::{.definition title="Dimensions of modules"} If $A\in \kalg^\fg$ and $M\in \amod^\fg$, choose $\kxn \surjects A$ to write $\dim_A M \da \dim_{\kxn} M$. ::: :::{.remark} Upshot: we have a notion of dimension for $M\in \kalg^\fg$ when $k\in \Field$ which doesn't depend on choices. ::: :::{.exercise} Show the following properties: - $\dim \kxn = n$. - If $A,B\in \kalg^\fg$ and $A\to B$ with $B\in\amod^\fg$, then $\dim_A B = \dim B$ (this just involves adding variables to $\kxn$). - If $M\in \amod^\fg$ then $\dim_A M\leq \dim A$. Hint: $A^n\surjects M$. - If $A,B\in \kalg^\fg$ and $A\to B$ with $B\in\amod^\fg$, then $\dim B \leq \dim A$. - If $A,B\in \kalg^\fg$ with $A\injects B$ then $\dim A < \dim B$. Hint: consider \begin{tikzcd} \kxn && A \\ \\ {\kxn[y_1,\cdots, y_m]} && B \arrow[hook, from=1-3, to=3-3] \arrow[two heads, from=3-1, to=3-3] \arrow[two heads, from=1-1, to=1-3] \arrow[hook, from=1-1, to=3-1] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXGt4biJdLFsyLDAsIkEiXSxbMCwyLCJcXGt4blt5XzEsXFxjZG90cywgeV9tXSJdLFsyLDIsIkIiXSxbMSwzLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFsyLDMsIiIsMix7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6ImVwaSJ9fX1dLFswLDEsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6ImVwaSJ9fX1dLFswLDIsIiIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d) Take the induced filtrations $A_i, B_i$ induced by degree on $\kxn$, then $\iota(A_i) \subseteq B_i$. - If $I \normal A\in \kalg^\fg$ then $\dim A/I \leq \dim A$ (geometrically this is passing to a closed subspace). - If $A\in \kalg^\fg$ is a domain with $I\normal A$ then $\dim A/I < \dim A$. Hint: it's ETS $\dim A/f < \dim A$, so use the SES $0\to A \mapsvia{f} A \to A/f \to 0$. - If $A$ is a domain and $f\in A\smz$ then $\dim A/f = \dim A - 1$. ::: :::{.example title="?"} Using these properties: - $A = k[x,y]$ satisfies $\dim A = 2$, while $B=k[x,y]/y = k[x]$ satisfies $\dim B = 1$. - For $A = k[x,y]/xy$ and $B = A/y = k[x]$, both are dimension 1. Geometrically, $A$ is the union of the $x$ and $y$ axes, and $B$ sets $y=0$ which results in the $x$ axis. - If $f: A\to B$ yields $B\in \amod^\fg$, the induced map $f^*: \spec B\to \spec A$ has finite fibers. - If additional $f$ is an injection, the dimensions are equal. > Injection means dense image, and dense image + finite fibers preserves dimension. ::: ## Other notions of dimension: transcendence degree :::{.remark} Setup: let - $k \subseteq K$ be an inclusion of fields (with no finiteness conditions) - $\alpha \in K$ is **integral over $k$** iff $f(\alpha) = 0$ for some monic $f\in k[x]$. - $K$ is **algebraic over $k$** iff every $\alpha\in K$ is integral over $k$. - $K$ is **transcendental over $k$** if it is not algebraic. ::: :::{.example title="?"} Some examples: - $\QQ(\sqrt 5)$ is algebraic over $\QQ$. - $\QQ(t)$ is transcendental over $\QQ$. ::: :::{.definition title="Algebraically independent elements"} A subset $\ts{x_a} \subseteq K$ is **algebraically independent over $k$** iff there does not exist a nonzero polynomial $p\in k[t_1,\cdots, t_n]$ with $p(\cdots, x_a, \cdots) = 0$. Such a subset is **maximal** if it is not strictly contained in another such subset. ::: :::{.definition title="Transcendence degree"} If $K/k$ admits a maximal finite algebraically independent subset $S$, define $\trdeg_k K = \size S$. ::: :::{.claim} This is well-defined. To be justified next time! ::: :::{.definition title="?"} If $A$ is a finitely generated domain over $k$, then $\dim A = \trdeg_k \ff(A)$. :::