From 363f5fd35ba4afa4f1ccb9e094340147940e49fb Mon Sep 17 00:00:00 2001 From: user202729 <25191436+user202729@users.noreply.github.com> Date: Fri, 24 Nov 2023 09:56:47 +0700 Subject: [PATCH] Another bunch of changes --- patch-asy.sty | 2 +- references.bib | 2 +- tex/alg-NT/classgrp.tex | 5 ++- tex/alg-NT/dedekind.tex | 40 +++++++++++++++++-- tex/alg-NT/norm-trace.tex | 52 ++++++++++++++++++++++++- tex/alg-NT/ramification.tex | 39 ++++++++++++++++--- tex/alg-geom/affine-var.tex | 57 +++++++++++++++++++++------ tex/alg-geom/bezout.tex | 71 ++++++++++++++++++++++++++++++---- tex/alg-geom/localization.tex | 22 ++++++++--- tex/alg-geom/mor-scheme.tex | 22 ++++++++--- tex/alg-geom/proj-var.tex | 7 ++++ tex/alg-geom/sheaves.tex | 45 +++++++++++++++++---- tex/alg-geom/spec-examples.tex | 9 +++-- tex/alg-geom/spec-sheaf.tex | 17 +++++--- tex/alg-geom/spec-zariski.tex | 29 +++++++++++--- tex/alg-geom/zariski.tex | 49 ++++++++++++++++++++++- tex/preamble.tex | 10 ++--- 17 files changed, 409 insertions(+), 69 deletions(-) diff --git a/patch-asy.sty b/patch-asy.sty index 2727b71b..cd3f1a18 100644 --- a/patch-asy.sty +++ b/patch-asy.sty @@ -16,7 +16,7 @@ %% %% Licence: GPL2+ %% -\ProvidesPackage{asymptote} +\ProvidesPackage{patch-asy} [2012/08/25 v1.27 Asymptote style file for LaTeX] \def\Asymptote{{\tt Asymptote}} \InputIfFileExists{\jobname.pre}{}{} diff --git a/references.bib b/references.bib index b6461a0a..128fa88e 100644 --- a/references.bib +++ b/references.bib @@ -146,7 +146,7 @@ @unpublished{ref:gorin editor={Tony Zhang}, institution={MIT}, year={2018}, - url={https://www.mit.edu/~txz/links.html}, + url={https://web.archive.org/web/20190617235844/http://web.mit.edu/txz/www/links.html}, } // Notes used in passing diff --git a/tex/alg-NT/classgrp.tex b/tex/alg-NT/classgrp.tex index da65c98a..f70c4e48 100644 --- a/tex/alg-NT/classgrp.tex +++ b/tex/alg-NT/classgrp.tex @@ -358,7 +358,7 @@ \section{The signature of a number field} \[ \frac{\Norm(\ka)}{2^{r_2}} \sqrt{\left\lvert \Delta_K \right\rvert}. \] \end{lemma} \begin{proof}[Sketch of Proof] - Let \[ d = \Norm(\ka) \defeq [\OO_K : \ka]. \] + Let \[ d = \Norm(\ka) \defeq |\OO_K / \ka|. \] Then in the lattice $L_\ka$, we somehow only take $\frac 1d$th of the points which appear in the lattice $L$, which is why the area increases by a factor of $\Norm(\ka)$. To make this all precise I would need to do a lot more with lattices and geometry @@ -398,7 +398,8 @@ \section{Minkowski's theorem} Dissect the plane into $2 \times 2$ squares \[ [2a-1, 2a+1] \times [2b-1, 2b+1] \] and overlay all these squares on top of each other. - By the Pigeonhole Principle, we find there exist two points $p \neq q \in S$ which map to the same point. + By the Pigeonhole Principle, we find there exist two points $p \neq q \in S$ + which is mapped to the same point. Since $S$ is symmetric, $-q \in S$. Then $\half (p-q) \in S$ (convexity) and is a nonzero lattice point. I'll briefly sketch part (b): the idea is to consider $(1+\eps) S$ for $\eps > 0$ diff --git a/tex/alg-NT/dedekind.tex b/tex/alg-NT/dedekind.tex index c1444f27..6513fc5a 100644 --- a/tex/alg-NT/dedekind.tex +++ b/tex/alg-NT/dedekind.tex @@ -84,7 +84,7 @@ \section{Motivation} \end{remark} \section{Ideal arithmetic} -\prototype{$(x)(y) = (xy)$. In any case, think in terms of generators.} +\prototype{$(x)(y) = (xy)$, and $(x)+(y)=(\gcd(x, y))$. In any case, think in terms of generators.} First, I have to tell you how to add and multiply two ideals $\ka$ and $\kb$. \begin{definition} Given two ideals $\ka$ and $\kb$ of a ring $R$, we define @@ -125,6 +125,17 @@ \section{Ideal arithmetic} So ``scaling'' and ``multiplying by principal ideals'' are the same thing. This is important, since we'll be using the two notions interchangeably. \end{remark} +\begin{remark} + The addition of two ideals does not correspond to the addition of elements --- for example, $(4)+(6)=(4, 6)=(2)$, but $4+6=10$. + + This is the best we can hope for --- addition of elements does not make sense for ideals --- for example, $1+1=2$ and $1+(-1)=0$, but as ideals, $(1)=(-1)$. + + In fact, addition of ideal is the straightforward generalization of $\gcd$ of elements --- as you can check in the example above, $\gcd(4, 6)=2$. + + Nevertheless, I hope you agree that $\ka+\kb$ is a natural notation, compared to something like $(\ka, \kb)$. + + Because factorization involves \emph{multiplying}, instead of adding, the ideals together, we will not need to use the notation $\ka+\kb$ any time soon. +\end{remark} Finally, since we want to do factorization we better have some notion of divisibility. So we define: @@ -258,6 +269,27 @@ \section{Unique factorization works} \begin{moral} Unique factorization works perfectly in Dedekind domains! \end{moral} + +\begin{remark} + [Comparison between Dedekind domain and UFD] + If we temporarily forget about the Noetherian and integrally closed condition, we have: + \begin{itemize} + \ii An integral domain admits unique factorization of elements if the prime elements and the irreducible elements are the same. + \ii An integral domain admits unique factorization of ideals if the prime ideals and the maximal ideals are the same. + \end{itemize} + Notice the similarity --- in either case, the Noetherian condition is ``merely'' to ensure that, if you keep extracting prime factors, you will terminate in a finite time. +\end{remark} + +\begin{example} + [What went wrong if $\mathcal A$ is not integrally closed?] + Consider $\mathcal A = 2\ZZ$, which is an ideal of $\ZZ$. + Clearly, every nonzero prime ideal is maximal. + + Nevertheless, in $\mathcal A$, + $(2 \cdot 3 \cdot 5)=(60)$ is not a prime ideal (so of course it isn't a maximal ideal), + but we cannot break it down into, for example, $(2 \cdot 3) \cdot (5)$. +\end{example} + \begin{theorem}[Prime factorization works] Let $\ka$ be a nonzero proper ideal of a Dedekind domain $\mathcal A$. Then $\ka$ can be written as a finite product of nonzero prime ideals $\kp_i$, say @@ -442,7 +474,7 @@ \section{The factoring algorithm} \begin{theorem}[Factoring algorithm / Dedekind-Kummer theorem] \label{thm:factor_alg} Let $K$ be a number field. - Let $\theta \in \OO_K$ with $[\OO_K : \ZZ[\theta]] = j < \infty$, + Let $\theta \in \OO_K$ with $|\OO_K / \ZZ[\theta]| = j < \infty$, and let $p$ be a prime not dividing $j$. Then $(p) = p \OO_K$ is factored as follows: \begin{quote} @@ -489,7 +521,7 @@ \section{The factoring algorithm} % It turns out that we can still apply the % factoring algorithm to ``almost all primes'' as follows. % Suppose $K$ is a number field and $\alpha \in \OO_K$ such that -% $[\OO_K : \ZZ[\alpha]] = j < \infty$. +% $|\OO_K / \ZZ[\alpha]| = j < \infty$. % Then as long as $p \nmid j$, we can apply the above algorithm % with $f$ the minimal polynomial of $\alpha$. % The formulation we presented above was the special case where $j=1$. @@ -568,7 +600,7 @@ \section{The ideal norm} \begin{definition} The \vocab{ideal norm} (or absolute norm) of a nonzero ideal $\ka \subseteq \OO_K$ is defined as - $[\OO_K : \ka]$ and denoted $\Norm(\ka)$. + $|\OO_K / \ka|$ and denoted $\Norm(\ka)$. \end{definition} \begin{example}[Ideal norm of $(5)$ in the Gaussian integers] Let $K = \QQ(i)$, $\OO_K = \ZZ[i]$. diff --git a/tex/alg-NT/norm-trace.tex b/tex/alg-NT/norm-trace.tex index a2dcc19d..bee4c8b7 100644 --- a/tex/alg-NT/norm-trace.tex +++ b/tex/alg-NT/norm-trace.tex @@ -78,6 +78,18 @@ \section{Norms and traces} Nicely, the formula $a^2-2b^2$ and $2a$ also works when $b=0$. \end{example} +\begin{example}[{Norm of $a+b\sqrt[3] 2+c \sqrt[3] 4$}] + Let $\alpha = a+b\sqrt[3] 2+c\sqrt[3] 4 \in \QQ(\sqrt3) = K$. + As above, if $b \neq 0$ or $c \neq 0$, then $\alpha$ and $K$ have the + same degree $3$. + The conjugates of $\alpha$ are $a+b\sqrt[3] 2 \omega+c\sqrt[3] 4 \omega^2$ + and $a+b\sqrt[3] 2 \omega^2+c\sqrt[3] 4 \omega$, + and we can compute $\Norm_{K/\QQ}(\alpha) = a^3 + 2b^3 + 4c^3 - 6abc$ and + $\Tr_{K/\QQ}(\alpha) = 3a$. + + Note that in this case the conjugates of $\alpha$ does not + lie in the field $K$! +\end{example} Of importance is: \begin{proposition}[Norms and traces are rational integers] If $\alpha$ is an algebraic integer, its norm and trace @@ -91,6 +103,20 @@ \section{Norms and traces} why is the norm multiplicative? To do this, I have to give a new definition of norm and trace. +\begin{remark} + Another way to automatically add the ``corrective factor'' is to use the + embeddings of $K$ into $\CC$. + + As we will see later, in \Cref{thm:n_embeddings}, there are exactly + $d=\deg K$ embeddings of $K$ into $\CC$, say $\sigma_1, \dots, \sigma_d$. + Then, + $\Tr_{K/\QQ}(\alpha) = \sum_{i=1}^d \sigma_i(\alpha)$ and + $\Norm_{K/\QQ}(\alpha) = \prod_{i=1}^d \sigma_i(\alpha)$. + + There is exactly $d$ embeddings, regardless of the number of Galois + conjugates of $\alpha$. +\end{remark} + \begin{theorem}[Morally correct definition of norm and trace] Let $K$ be a number field of degree $n$, and let $\alpha \in K$. Let $\mu_\alpha \colon K \to K$ denote the map \[ x \mapsto \alpha x \] @@ -403,12 +429,36 @@ \section{On monogenic extensions} \ii the cyclotomic field in \Cref{prob:ring_int_cyclotomic}. \end{itemize} Unfortunately, it is not true in general: -the first counterexample is $\QQ(\alpha)$ for $\alpha$ a root of $X^3-X^2-2X-8$. +the first counterexample is $K=\QQ(\alpha)$ for $\alpha$ a root of $X^3-X^2-2X-8$. We call an extension with this nice property \vocab{monogenic}. As we'll later see, monogenic extensions have a really nice factoring algorithm, \Cref{thm:factor_alg}. +\begin{remark} + [What went wrong with $\OO_K$?] + As we have just mentioned above, as an abelian group, $\OO_K \cong \ZZ^3$, + so it's generated by finitely many elements. + + In fact, $\{ 1, \alpha, \beta \}$ is a basis of $\OO_K$, + where $\beta=\frac{\alpha+\alpha^2}{2}$. + The group generated by $\{ 1, \alpha, \alpha^2 \}$ has index 2 in $\OO_K$ -- + that is, $|\OO_K/\langle 1, \alpha, \alpha^2\rangle|=2$, and we misses + $\beta$. + + If we try to pick $\{ 1, \beta, \beta^2 \}$ as a basis instead, again we + get $|\OO_K/\langle 1, \beta, \beta^2\rangle|=2$, and we misses $\alpha$. + If you explicitly compute it out, you can get + $\beta^2 = \frac{3\alpha^2 + 7\alpha}{2} + 6 = 3 \beta+2 \alpha+6$. + + While this is not a proof that the extension is not monogenic, hopefully it + gives you a feeling of the structure of $\OO_K$. + + % TODO is it true that, there exist some \alpha and integers c_2, + % \dots, c_{n-1} such that \OO_K is generated by {1, \alpha, \alpha^2/c_2, + % \dots, \alpha^{n-1}/c_{n-1}}? +\end{remark} + \section{\problemhead} \begin{sproblem} % trivial diff --git a/tex/alg-NT/ramification.tex b/tex/alg-NT/ramification.tex index a0060ac4..92992942 100644 --- a/tex/alg-NT/ramification.tex +++ b/tex/alg-NT/ramification.tex @@ -163,6 +163,7 @@ \section{The magic of Galois extensions} But miraculously, it turns out there is only one orbit! \begin{theorem} [Galois group acts transitively] + \label{thm:galois_group_transitive} Let $K/\QQ$ be Galois with $G = \Gal(K/\QQ)$. Let $\{\kp_i\}$ be the set of distinct prime ideals in the factorization of $p \cdot \OO_K$ (in $\OO_K$). @@ -174,10 +175,34 @@ \section{The magic of Galois extensions} \begin{moral} All of the $\{\kp_i\}$ are Galois conjugates of each other. \end{moral} + +Before proving this, let us consider the easier problem +of factorization into elements. +\begin{quote} + Suppose $\OO_K$ is an UFD, and $p$ factors as $u p_1 p_2 \cdots p_n$ in + $\OO_K$, where $p_i$ are irreducibles and $u$ is an unit. + Show that the $p_i$ are all conjugates of each other, up to + multiplication by an unit. +\end{quote} +\begin{ques} + Try to prove it before reading it below. + (Hint: Galois theory. Alternatively, take the norm of $p_1$.) +\end{ques} \begin{proof} - Fairly slick. - Suppose for contradiction that no $\sigma \in G$ sends $\kp_1$ to $\kp_2$, say. - By the Chinese remainder theorem, we can find an $x \in \OO_K$ such that + Let $q=\NK(p_1)$ be the product of all conjugates + of $p_1$, then $q \in \QQ$. Thus $p \mid q$, so each $p_i$ is a factor of + $q$, and we're done by unique factorization. +\end{proof} + +Unfortunately, the product of all conjugates of an ideal $\kp_1$ is not +necessarily of the form $p \cdot \OO_K$ (for example, $K=\QQ[i]$ and $(1+i)$ +has no other conjugates). So in the proof, we pick $x$ which is an +``representative'' of $\kp_1$. + +\begin{proof} + [Proof of \Cref{thm:galois_group_transitive}] + Because $\kp_i$ are distinct primes, by the Chinese remainder theorem, + we can find an $x \in \OO_K$ such that \begin{align*} x &\equiv 0 \pmod{\kp_1} \\ x &\equiv 1 \pmod{\kp_i} \text{ for $i \ge 2$} @@ -190,9 +215,13 @@ \section{The magic of Galois extensions} Since $\NK(x)$ is an integer and divisible by $\kp_1$, we should have that $\NK(x)$ is divisible by $p$. Thus it should be divisible by $\kp_2$ as well. - But by the way we selected $x$, we have $x \notin \sigma\inv\kp_2$ for every $\sigma \in G$! - So $\sigma(x) \notin \kp_2$ for any $\sigma$, which is a contradiction. + Thus, for some $\sigma \in \Gal(K/\QQ)$, $\sigma(x)$ is divisible by + $\kp_2$, equivalently, $x$ is divisible by $\sigma\inv\kp_2$. + But by the way we selected $x$, we have within the factors of $p$, $x$ is + divisible by only $\kp_1$! + So $\sigma\inv\kp_2 = \kp_1$, and we're done. \end{proof} + \begin{theorem}[Inertial degree and ramification indices are all equal] Assume $K/\QQ$ is Galois. Then for any rational prime $p$ we have diff --git a/tex/alg-geom/affine-var.tex b/tex/alg-geom/affine-var.tex index a1eecf44..35da8c1a 100644 --- a/tex/alg-geom/affine-var.tex +++ b/tex/alg-geom/affine-var.tex @@ -84,7 +84,7 @@ \section{Naming affine varieties via ideals} Convince yourself that $\VV(I) = \{(3,4)\}$. \end{ques} So rather than writing $\VV(x-3, y-4)$ it makes sense to -think about this as $\VV\left( I \right)$, where $I = (x-3,y-4)$ is the \emph{ideal} +think about this as $\VV( I )$, where $I = (x-3,y-4)$ is the \emph{ideal} generated by the two polynomials $x-3$ and $y-4$. This is an improvement because \begin{ques} @@ -289,6 +289,43 @@ \section{Pictures of varieties in $\Aff^1$} the intersection $I \cap J$ is radical. \end{remark} +As another easy result concerning the relation between the ideal and variety, we +have: +\begin{proposition} + [$\VV(-)$ is inclusion reversing] + If $I \subseteq J$ then $\VV(I) \supseteq \VV(J)$. + Thus $\VV(-)$ is \emph{inclusion-reversing}. +\end{proposition} +\begin{ques} + Verify this. +\end{ques} +Thus, bigger ideals correspond to smaller varieties. + +These results will be used a lot throughout the chapter, so it would be useful +for you to be comfortable with the inclusion-reversing nature of $\VV$. +\begin{exercise} + Some quick exercises to help you be more familiar with the concepts. + \begin{enumerate} + \item Let $I=(y-x^2)$ and $J=(x+1, y+2)$. What is $\VV(I)$ and $\VV(J)$? + \item What is the ideal $K$ such that $\VV(K)$ is the union of the + parabola $y=x^2$ and the point $(-1, -2)$? + \item Let $L=(y-1)$. What is $\VV(L)$? + \item The intersection $\VV(I) \cap \VV(L)$ consist of two points $(1, + 1)$ and $(-1, 1)$. What's the ideal corresponding to it, in terms of + $I$ and $L$? + \item What is $\VV(I \cap L)$? What about $\VV(I L)$? + \end{enumerate} +\end{exercise} + +\begin{ques} +Note that the intersection of infinitely many ideals is still an ideal, but the +union of infinitely many affine varieties may not be an affine variety. + +Consider $I_k = (x-k)$ in $\CC[x]$, and take the infinite intersection +$I = \bigcap_{k \in \NN} I_k$. What is $\VV(I)$ and $\bigcup_{k \in \NN} +\VV(I_k)$? +\end{ques} + \section{Prime ideals correspond to irreducible affine varieties} \prototype{$(xy)$ corresponds to the union of two lines in $\Aff^2$.} @@ -418,16 +455,8 @@ \section{Pictures in $\Aff^2$ and $\Aff^3$} \section{Maximal ideals} \prototype{All maximal ideals are $(x_1-a_1, \dots, x_n-a_n)$.} -We begin by noting: -\begin{proposition} - [$\VV(-)$ is inclusion reversing] - If $I \subseteq J$ then $\VV(I) \supseteq \VV(J)$. - Thus $\VV(-)$ is \emph{inclusion-reversing}. -\end{proposition} -\begin{ques} - Verify this. -\end{ques} -Thus, bigger ideals correspond to smaller varieties. +Recall that bigger ideals correspond to smaller varieties. + As the above pictures might have indicated, the smallest varieties are \emph{single points}. Moreover, as you might guess from the name, @@ -526,6 +555,12 @@ \section{Motivating schemes with non-radical ideals} \section\problemhead \todo{some actual computation here would be good} +\begin{problem} + Show that $I \subseteq \sqrt I$ and $\sqrt{I \cap J} = + I J \subseteq I \cap J$, for two ideals + $I$ and $J$. +\end{problem} + \begin{problem} Show that a \emph{real} affine variety $V \subseteq \Aff_\RR^n$ can always be written in the form $\VV(f)$. diff --git a/tex/alg-geom/bezout.tex b/tex/alg-geom/bezout.tex index 1cfcfbd7..9a728498 100644 --- a/tex/alg-geom/bezout.tex +++ b/tex/alg-geom/bezout.tex @@ -15,7 +15,7 @@ \section{Non-radical ideals} For a first example, suppose we intersect $y=x^2$ with the line $y=1$; or more accurately, in projective coordinates of $\CP^2$, the parabola $zy=x^2$ and $y=z$. -The intersection of the ideals is +The ideal of the intersection is \[ (zy-x^2, y-z) = (x^2-z^2, y-z) \subseteq \CC[x,y,z]. \] So this corresponds to having two points; this gives two intersection points: $(1:1:1)$ and $(-1:1:1)$. @@ -77,9 +77,10 @@ \section{Hilbert functions of finitely many points} \end{definition} \begin{definition} If $V$ is a projective variety, we set $h_V = h_{\II(V)}$, - where $I$ is the \emph{radical} ideal satisfying $V = \Vp(I)$. - If $V = \varnothing$, we choose $I = (1)$. + where $I=\II(V)$ is the \emph{radical} ideal satisfying + $V = \Vp(I)$ as defined above. \end{definition} +In this case, $\CC[x_0, \dots, x_n]/I$ is just $\CC[V]$. \begin{example}[Examples of Hilbert functions in zero dimensions] \label{ex:hilbert_zero} For concreteness, let us use $\CP^2$. @@ -87,7 +88,7 @@ \section{Hilbert functions of finitely many points} \ii If $V$ is the single point $(0:0:1)$, with ideal $\II(V) = (x,y)$, then - \[ \CC[x,y,z] / (x,y) \cong \CC[z] + \[ \CC[V] = \CC[x,y,z] / (x,y) \cong \CC[z] \cong \CC \oplus z\CC \oplus z^2\CC \oplus z^3\CC \dots \] which has dimension $1$ in all degrees. Consequently, we have \[ h_I(d) \equiv 1. \] @@ -155,6 +156,7 @@ \section{Hilbert functions of finitely many points} \begin{proposition} [Hilbert functions of $I \cap J$ and $I+J$] + \label{prop:hilbert_function_intersection_union} Let $I$ and $J$ be homogeneous ideals in $\CC[x_0, \dots, x_n]$. Then \[ h_{I \cap J} + h_{I+J} = h_I + h_J. \] \end{proposition} @@ -276,6 +278,7 @@ \section{Hilbert polynomials} In fact, this behavior of ``eventually polynomial'' always works. \begin{theorem} [Hilbert polynomial] + \label{thm:hilbert_polynomial} Let $I \subseteq \CC[x_0, \dots, x_n]$ be a homogeneous ideal, not necessarily radical. Then \begin{enumerate}[(a)] @@ -303,7 +306,7 @@ \section{Hilbert polynomials} 0 \ar[r] & \left[ S/I \right]^{d-1} \ar[r, hook] & \left[ S / I \right]^d \ar[r, surjective head] - & \left[ S / (I+(f)) \right]^d \ar[r] & 0 \\ + & \left[ S / (I+(x_0)) \right]^d \ar[r] & 0 \\ & f \ar[r, mapsto] & f \cdot x_0 \\ && f \ar[r, mapsto] & f. \end{tikzcd} @@ -331,16 +334,56 @@ \section{B\'ezout's theorem} $m! \chi_I$ the \vocab{degree} of $I$, which is an integer, denoted $\deg I$. - Of course for projective varieties $V$ we let $h_V = h_{\II(V)}$. + Of course for projective varieties $V$ we let $h_V = h_{\II(V)}$, and $\deg + V = \deg \II(V)$. \end{definition} +\begin{remark} + Note that the degree of an ideal $\deg I$ is not the same as $\deg h_I$! +\end{remark} + +Let us show some properties of the degrees, which will allow us to compute the +degree of any projective variety from its irreducible components. +\begin{proposition}[Properties of degrees] + For two varieties $V$ and $W$, we have the following: + \begin{itemize} + \ii If $V$ and $W$ are disjoint and have the same dimension, then $\deg (V \cup W)=\deg + V+\deg W$. + \ii If $\dim V<\dim W$, then $\deg (V \cup W)=\deg W$. + \end{itemize} +\end{proposition} +So, +\begin{moral} + The degree is additive over components, + and it measures the ``degree'' of the highest-dimensional component. +\end{moral} + +% TODO maybe add https://en.wikipedia.org/wiki/Degree_of_an_algebraic_variety to +% make it easy to imagine the degree? No proof though so...? +% It's quite interesting that the geometric property is related to a purely +% algebraic concept (defined by Hilbert polynomial) though + +\begin{proof} + Follows from the properties of Hilbert polynomial in + \Cref{thm:hilbert_polynomial} and + \Cref{prop:hilbert_function_intersection_union}, + and that the leading coefficient + only depends on the largest-degree summand. +\end{proof} + \begin{example}[Examples of degrees] \listhack \begin{enumerate}[(a)] \ii If $V$ is a finite set of $n \ge 1$ points, it has degree $n$. \ii If $I$ corresponds to a double point, it has degree $2$. \ii $\CP^n$ has degree $1$. - \ii The parabola has degree $2$. + \ii Any line or plane, being ``isomorphic'' to $\CP^1$ and $\CP^2$ + respectively, has degree $1$. + % TODO check this, what if the line is embedded in \CP^3? + \ii The parabola has degree $2$. (Note that, as an algebraic variety, the parabola is isomorphic + to a line!) + % TODO is this still true if the parabola is embedded in \CP^3? + \ii The union of the parabola and a point has degree $2$. \end{enumerate} \end{example} @@ -358,6 +401,17 @@ \section{B\'ezout's theorem} Then \[ \deg\left( I + (f) \right) = k \deg I. \] \end{theorem} +Geometrically, +\begin{moral} + If $V$ is any projective variety, $\VV(f)$ is a hyperplane of degree $k$, then their + intersection $V \cap \VV(f)$ has degree $k \deg V$ --- unless + some irreducible component of $V$ is contained inside $\VV(f)$. +\end{moral} +This is what we mentioned at the beginning of the chapter. + +Because the ideal $I$ may not be radical, the geometric interpretation statement is not the most general +possible --- the problem will be rectified later with the generalization to schemes. + \begin{proof} Let $S = \CC[x_0, \dots, x_n]$ again. This time the exact sequence is @@ -461,6 +515,9 @@ \section{Applications} so they must lie on this line. \end{proof} +We'd like to remark that the Pascal's theorem is just a special case of the Cayley-Bacharach theorem, +which can be used to prove that the addition operation on an elliptic curve is associative. +Interested readers may want to try proving the Cayley-Bacharach theorem using the same technique. \section\problemhead \begin{problem} diff --git a/tex/alg-geom/localization.tex b/tex/alg-geom/localization.tex index 7836c234..c56a7850 100644 --- a/tex/alg-geom/localization.tex +++ b/tex/alg-geom/localization.tex @@ -53,7 +53,7 @@ \section{Spoilers} if $R = \CC[V]$ is the coordinate ring, then the above will become abbreviated to just \begin{align*} - \OO_{V}(D(g)) &= R_{g} \\ + \OO_{V}(D(g)) &= R[g^{-1}] \\ \OO_{V, p} &= R_{\km} \quad \text{where } \{p\} = \VV(\km). \end{align*} The former will be pronounced @@ -68,7 +68,7 @@ \section{Spoilers} We will construct a ringed space called $X = \Spec A$, whose elements are \emph{prime ideals} of $A$ and is equipped with the Zariski topology and a sheaf $\OO_X$. -It will turn out that, in analogy to what we had before, +It will turn out that, the right way to define the sheaf $\OO_X$ is to use localization, \begin{align*} \OO_X(D(f)) &= A[f\inv] \\ \OO_{X,\kp} &= A_\kp @@ -78,6 +78,13 @@ \section{Spoilers} localizations will give us a way to more or less describe the sheaf $\OO_X$ completely. +In other words, +\begin{moral} + Localization is the purely algebraic way to \emph{define} the + ring of regular functions on a smaller open set from the ring of ``global'' + regular functions. +\end{moral} + \section{The definition} \begin{definition} A subset $S \subseteq A$ is a \vocab{multiplicative set} @@ -181,7 +188,8 @@ \section{Localization away from an element} if we localize $\Zc{60}$ away from the element $5$, we get $(\Zc{60})[1/5] \cong \Zc{12}$. You should try to think about why this is the case. - We will see a ``geometric'' reason later. + We will see a ``geometric'' reason later, + in \Cref{sec:example_spec_Zc60}. \end{enumerate} \end{example} @@ -221,6 +229,10 @@ \section{Localization away from an element} so $y = 0$ in $A$, and thus $y$ just goes away completely. From this we get a ring isomorphism \[ A[1/x] \cong \CC[x,1/x].\] + + Later, we will be able to use our geometric intuition to ``see'' this + at \Cref{sec:example_spec_kxy_div_xy}, once + we have defined the affine scheme. \end{example} \section{Localization at a prime ideal} @@ -403,12 +415,12 @@ \section{Prime ideals of localizations} and even $(0)$ give prime ideals of $\ZZ[1/6]$. But $(2)$ and $(3)$ no longer correspond to - prime ideals; in fact in $A_6$ we have $(2) = (3) = (1)$, + prime ideals; in fact in $\ZZ[1/6]$ we have $(2) = (3) = (1)$, the whole ring. \end{example} \begin{example} - [Prime ideals of $A_{(5)}$] + [Prime ideals of $\ZZ_{(5)}$] Suppose we localize $\ZZ$ at the prime $(5)$. As we saw, \[ \ZZ_{(5)} = \left\{ \frac{m}{n} \mid m,n \in \ZZ, diff --git a/tex/alg-geom/mor-scheme.tex b/tex/alg-geom/mor-scheme.tex index 686e0187..fb000586 100644 --- a/tex/alg-geom/mor-scheme.tex +++ b/tex/alg-geom/mor-scheme.tex @@ -53,6 +53,12 @@ \section{Morphisms of ringed spaces via sections} which may not be so well-behaved. So the solution is that we \emph{include} the data of $f^\sharp$ as part of the definition of a morphism. +\begin{remark} + As we will see in \Cref{ex:spec_C_morphism}, unlike the situation in algebraic varieties + where the morphism is uniquely determined by the map of topological space, + here $\pi^\sharp$ is not necessarily uniquely determined by the map $\pi$. + Thus, including the $\pi^\sharp$ is necessary. +\end{remark} \begin{definition} A \vocab{morphism of ringed spaces} $(X, \OO_X) \to (Y, \OO_Y)$ consists of a pair $(\pi, \pi^\sharp)$ @@ -246,6 +252,7 @@ \subsection{One-point schemes} \end{example} \begin{example} [$\Spec \CC$ horror story] + \label{ex:spec_C_morphism} There are multiple maps $X = \Spec \CC \to \Spec \CC = Y$, horribly enough! Indeed, these are spaces with one point, @@ -291,10 +298,15 @@ \subsection{Examples of constant maps} so this time we want to specify a map \[ \pi^\sharp_U \colon \OO_Y(U) \to \OO_X(X) = \RR[x] \] which satisfies restriction maps. - Note that for any $U$, the element $y$ must map to a unit in $\RR[x]$; + Note that for any $U$, the element $y$ must be mapped to a unit in $\RR[x]$; since $1/y$ is a section too for a subset of $U$ not containing $(y)$. + In more detail, let $W = U \cap D(y)$ so that $(y) \notin W$, then + \[ + \pi^\sharp_W(y) = \pi^\sharp_U(y) \quad \text{and} \quad + \pi^\sharp_W(y) \pi^\sharp_W(1/y) = 1. + \] Actually for any real number $c \neq 3$, - $y-c$ must be map to a unit in $\RR[x]$. + $y-c$ must be mapped to a unit in $\RR[x]$. This can only happen if $y \mapsto 3 \in \RR[x]$. As we have specified $\RR[y] \mapsto \RR[x]$ with $y \mapsto 3$, @@ -526,7 +538,7 @@ \section{The big theorem} in the Zariski topology. \end{exercise} \ii Now we want to also define maps on the stalks, - and so for ecah $\pi(\kp) = \kq$ we set + and so for each $\pi(\kp) = \kq$ we set \[ B_\kq \ni \frac fg \mapsto \frac{\psi(f)}{\psi(g)} \in A_\kp. \] This makes sense since $g \notin \kq \implies \psi(g) \notin \kp$. Also $f \in \kq \implies \psi(f) \in \kp$, @@ -587,7 +599,7 @@ \section{More examples of scheme morphisms} \end{tikzcd} commutes. Often, if $S = \Spec k$, we will refer - to schemes over $k$ or $k$-schemes for short. + to $X$ by schemes over $k$ or $k$-schemes for short. \end{definition} \begin{example} [{$\Spec k[\dots]$}] @@ -674,7 +686,7 @@ \section{A little bit on non-affine schemes} Show that in fact $U$ can be covered by two open sets which are both affine. \end{ques} -However, we show now that you really do need two open sets. +However, we show now that you really do need two distinguished open sets. \begin{proposition} [Famous example: punctured plane isn't affine] The punctured plane $U = (U, \OO_U)$, diff --git a/tex/alg-geom/proj-var.tex b/tex/alg-geom/proj-var.tex index 72dd5021..d128c864 100644 --- a/tex/alg-geom/proj-var.tex +++ b/tex/alg-geom/proj-var.tex @@ -284,6 +284,13 @@ \section{As ringed spaces} is defined as the ring \[ \CC[V] = \CC[x_0, \dots, x_n] / I. \] \end{definition} + +\begin{remark} + Unlike the case of \Cref{remark:meaning_name_coordinate_ring}, an element + of $\CC[V]$ no longer correspond to a function from $V$ to $\CC$; + nevertheless, it is a function from $\VV(I) \subseteq \Aff^{n+1}$ to $\CC$. +\end{remark} + However, when we define a rational function we must impose a new requirement that the numerator and denominator are the same degree. \begin{definition} diff --git a/tex/alg-geom/sheaves.tex b/tex/alg-geom/sheaves.tex index b2df0ff5..172bab27 100644 --- a/tex/alg-geom/sheaves.tex +++ b/tex/alg-geom/sheaves.tex @@ -37,8 +37,24 @@ \section{Pre-sheaves} Recall that $\OO_V$ took \emph{open sets of $V$} to \emph{rings}, with the interpretation that $\OO_V(U)$ was a ``ring of functions''. +Recall from \Cref{sec:sheaf_regular_functions} that $\OO_V$, as a set, consist +of simply the algebraic functions. However, if we view $\OO_V$ purely as a set, +the structure of the functions is essentially thrown way. + +Let us see how the functions in $\OO_V$ are related to each other: +\begin{itemize} + \ii Each function in $\OO_V$ is defined on a open set $U \subseteq V$. + \ii If two functions are defined on the same open set, you can add and + multiply them together. In other words, $\OO_V(U)$ is a ring. + \ii Given a function $f \in \OO_V(U)$, we can restrict it to a smaller open + subset $W \subseteq U$. +\end{itemize} +These are the operations that we will impose on a pre-sheaf. + \subsection{Usual definition} So here is the official definition of a pre-sheaf. +We will only define a pre-sheaf of rings, however it's possible to define a +pre-sheaf of sets, pre-sheaf of abelian groups, etc. \begin{definition} For a topological space $X$ let $\Opens(X)$ denote the open sets of $X$. \end{definition} @@ -417,8 +433,10 @@ \section{Sheaves} but sadly I have to give the definition first for the examples to make sense. \begin{definition} - A \vocab{sheaf} $\mathscr F$ is a pre-sheaf obeying two additional axioms: - Suppose $U$ is covered by open sets $U_\alpha \subseteq U$. Then: + A \vocab{sheaf} $\mathscr F$ on a topological space $X$ + is a pre-sheaf obeying two additional axioms: + Suppose $U$ is an open set in $X$, and $U$ is covered by open sets + $U_\alpha \subseteq U$. Then: \begin{enumerate} \ii (Identity) If $s, t \in \mathscr F(U)$ are sections, and $s\restrict{U_\alpha} = t\restrict{U_\alpha}$ @@ -435,7 +453,8 @@ \section{Sheaves} \end{definition} \begin{remark} [For keepers of the empty set] - The above axioms imply $\SF(\varnothing) = 0$ (the zero ring). + The above axioms imply $\SF(\varnothing) = 0$ (the zero ring), when $\SF$ is + a sheaf of rings. This is not worth worrying about until you actually need it, so you can forget I said that. \end{remark} @@ -501,6 +520,16 @@ \section{Sheaves} This is because to verify a function is continuous, one only needs to look at small open neighborhoods at once. + \ii Let $X = \RR$, and define the presheaf of rings $\SF$ by: + \[ + \SF(U) = \{ f \colon U \to \RR: \text{there exists continuous $g + \colon \RR \to \RR$ such that $g\restrict{U} = f$} \}. + \] + Then $\SF$ is not a sheaf. Indeed, $s_1(x) = 0$ in $\SF((-1, 0))$ + and $s_2(x) = 1$ in $\SF((0, 1))$ agrees on the (empty) overlap, + but they cannot be glued together to an element in $\SF((-1, 0) \cup(0, + 1))$. + \ii For a complex variety $V$, $\OO_V$ is a sheaf, precisely because our definition was \emph{locally} quotients of polynomials. @@ -513,11 +542,13 @@ \section{Sheaves} while $s_2$ is the constant function $2$ on $U_2$, then we cannot glue these to a constant function on $U_1 \cup U_2$. - \ii On the other hand, \emph{locally constant} functions + \ii \label{ex:sheaf_of_locally_constant} On the other hand, + \emph{locally constant} functions do produce a sheaf. (A function is locally constant if for every point it is constant on some open neighborhood.) \end{enumerate} - In fact, the sheaf in (d) is what is called a \emph{sheafification} + In fact, the sheaf in \ref{ex:sheaf_of_locally_constant} is what is called + a \emph{sheafification} of the pre-sheaf constant functions, which we define momentarily. \end{example} @@ -553,7 +584,7 @@ \section{For sheaves, sections ``are'' sequences of germs} \end{example} -From the above example it's obvious that if we know each germ $[s]_p$ +From the above example it's obvious that if we know each germ $[s]_p$, this should let us reconstruct the entire section $s$. Let's check this from the sheaf axioms: \begin{exercise} @@ -784,7 +815,7 @@ \section{Sheafification (optional)} \end{problem} \begin{problem} - Let $\SF$ be a sheaf on a space $X$ and let $s \in \SF(X)$ + Let $\SF$ be a sheaf of rings on a space $X$ and let $s \in \SF(X)$ be a global section. Define the \vocab{support} of $s$ as \[ Z = \left\{ p \in X \mid [s]_p \neq 0 \in \SF_p \right\}. \] diff --git a/tex/alg-geom/spec-examples.tex b/tex/alg-geom/spec-examples.tex index 7f5a5441..61493726 100644 --- a/tex/alg-geom/spec-examples.tex +++ b/tex/alg-geom/spec-examples.tex @@ -333,6 +333,7 @@ \section{$\Spec k[x]/(x^3-5x^2)$, a double point and a single point} The derivative is meant formally here! \section{$\Spec \Zc{60}$, a scheme with three points} +\label{sec:example_spec_Zc60} We've being seeing geometric examples of ring products coming up, but actually the Chinese remainder theorem you are used to with integers is no different. @@ -413,12 +414,13 @@ \section{$\Spec k[x,y]$, the two-dimensional plane} \ii The stalk above $(x-1, y+2)$ is the set of rational functions $\frac{f(x,y)}{g(x,y)}$ such that $g(1,-2) \neq 0$. - \ii The stalk above the non-closed point $(y-x^2)$ + \ii The stalk above the non-closed point $(y-x^2)$ is the set of rational functions $\frac{f(x,y)}{g(x,y)}$ such that $g(t, t^2) \neq 0$. For example the function $\frac{xy}{x+y-2}$ is still fine; - despite the fact it vanishes at the point $(1,1)$ and $(-2,4)$ - on the parabola it is a function + despite the fact that the denominator vanishes at the point $(1,1)$ + and $(-2,4)$ + on the parabola, it is a function on a ``generic point'' (crudely, ``most points'') of the parabola. \ii The stalk above $(0)$ is the entire fraction field $k(x,y)$ of rational functions. @@ -556,6 +558,7 @@ \section{$\Spec \ZZ[i]$, the Gaussian integers (one-dimensional)} which is a degree two $\FF_3$-extension. \section{Long example: $\Spec k[x,y]/(xy)$, two axes} +\label{sec:example_spec_kxy_div_xy} This is going to be our first example of a non-irreducible scheme. \subsection{Picture} diff --git a/tex/alg-geom/spec-sheaf.tex b/tex/alg-geom/spec-sheaf.tex index 38b61b8d..4ea9bf77 100644 --- a/tex/alg-geom/spec-sheaf.tex +++ b/tex/alg-geom/spec-sheaf.tex @@ -5,6 +5,13 @@ \chapter{Affine schemes: the sheaf} defining the sheaf $\OO_X$ on it, making it into a ringed space. This is done quickly in the first section. +As before, our goal is: +\begin{moral} + The sheaf $\OO_X$ coincides with the sheaf of regular functions on + affine varieties, so that we can apply our geometric intuition to $\Spec A$ + when $A$ is an arbitrary ring. +\end{moral} + However, we will then spend the next several chapters trying to convince the reader to \emph{forget} the definition we gave, in practice. @@ -319,7 +326,7 @@ \subsection{Localizations give local rings} Then the localization $A_\kp$ has exactly one maximal ideal, given explicitly by \[ \kp A_\kp = - \left\{ \frac fg \mid f \in \kp \; g \notin \kp \right\}. \] + \left\{ \frac fg \mid f \in \kp, \; g \notin \kp \right\}. \] \end{theorem} The ideal $\kp A_\kp$ thus captures the idea of ``germs vanishing at $\kp$''.\footnote{The notation $\kp A_\kp$ really means @@ -467,7 +474,7 @@ \subsection{Computing values: a convenient square} can also be described as $\Frac(A/\kp)$. \end{theorem} So for example, if $A = \CC[x,y]$ and $\kp = (x,y)$, -then $A/\kp = \CC$ and $\Frac(A_\kp) = \Frac(\CC) = \CC$, as we expected. +then $A/\kp = \CC$ and $\Frac(A/\kp) = \Frac(\CC) = \CC$, as we expected. In practice, $\Frac(A/\kp)$ is probably the easier way to compute $\kappa(\kp)$ for any prime ideal $\kp$. @@ -576,7 +583,7 @@ \section{\problemhead} \begin{hint} $k[x,y] \times k[z,z\inv]$. \end{hint} - \begin{soln} + \begin{sol} Let $V = D(x) \cup D(y) \subset U$ denote the punctured plane, so its complement $D(z)$ looks like a punctured line. Then $V \cap D(z) = \varnothing$ and the following diagram @@ -595,7 +602,7 @@ \section{\problemhead} On the other hand $\OO_X(V) = k[x,y]$ as shown in \S4.4.1 of Vakil. So \[ \OO_X(U) = k[x,y] \times k[z,z\inv]. \] - \end{soln} + \end{sol} \end{problem} \begin{problem} @@ -615,6 +622,6 @@ \section{\problemhead} \begin{problem} Let $A$ be a ring, and $\km$ a maximal ideal. - Consider $\km$ as point of $\Spec A$ + Consider $\km$ as a point of $\Spec A$. Show that $\kappa(\km) \cong A/\km$. \end{problem} diff --git a/tex/alg-geom/spec-zariski.tex b/tex/alg-geom/spec-zariski.tex index fd7a2147..8c17490c 100644 --- a/tex/alg-geom/spec-zariski.tex +++ b/tex/alg-geom/spec-zariski.tex @@ -28,10 +28,21 @@ \section{Some more advertising} and generate the entire structure from just $A$ itself. The final result is called $\Spec A$, the \vocab{spectrum} of $A$. The affine varieties $\VV(I)$ we met earlier will just be -$\CC[x_1, \dots, x_n] / I$, but now we will be able to take +$\Spec \CC[\VV(I)] = \Spec \CC[x_1, \dots, x_n] / I$ +, but now we will be able to take \emph{any} ideal $I$, thus finally completing the table at the end of the ``affine variety'' chapter. +To emphasize the point: +\begin{moral} + For affine varieties $V$, the spectrum of the coordinate ring $\CC[V]$ is + $V$. +\end{moral} +Thus, we may also think of $\Spec$ as the opposite operation of taking the +ring of global sections, defined purely-algebraically in order to depend only on +the intrinsic properties of the affine variety itself (the ring $\OO_V$) and not +the embedding. + The construction of the affine scheme in this way will have three big generalizations: \begin{enumerate} @@ -125,6 +136,13 @@ \section{The set of points} \footnotesize Image from \cite{img:calvin_hobbes_fly}. \end{center} +\begin{remark} + [Why don't the prime non-maximal ideals correspond to the whole parabola?] + We have already seen a geometric reason in \Cref{sec:localize_prime_ideal} earlier: + localizing a ring at a prime non-maximal ideal gives the functions that may blow up somewhere in the + parabola, but not \emph{generically}. +\end{remark} + \begin{example} [More examples of spectrums] \listhack @@ -306,7 +324,7 @@ \section{The Zariski topology on the spectrum} \end{proposition} \begin{proof} Just note $(0)$ is a prime ideal, - and in every open set. + and is in every open set. \end{proof} You should compare this with our old classical result that $\CC[x_1, \dots, x_n]/I$ @@ -412,9 +430,10 @@ \section{On radicals} Conversely, suppose $x \notin \sqrt I$, meaning $1, x, x^2, x^3, \dots \notin I$. - Then, consider the localization $A[1/x]$. + Then, consider the localization $(A/I)[1/x]$, which is not the zero ring. Like any ring, it has some maximal ideal (Krull's theorem). - This means our usual bijection between prime ideals of $A[1/x]$ + This means our usual bijection between prime ideals of $(A/I)[1/x]$, + prime ideals of $A/I$ and prime ideals of $A$ gives some prime ideal $\kp$ of $A$ containing $I$ but not containing $x$. Thus $x \notin \bigcap_{\kp \supseteq I} \kp$, as desired. @@ -424,7 +443,7 @@ \section{On radicals} The longer direction of this proof is essentially saying that for any $x \in A$, there is a maximal ideal of $A$ not containing $x$. - The ``short'' proof is to use Krull's theorem on $A[1/x]$ as above, + The ``short'' proof is to use Krull's theorem on $(A/I)[1/x]$ as above, but one can also still prove it directly using Zorn's lemma (by copying the proof of the original Krull's theorem). \end{remark} diff --git a/tex/alg-geom/zariski.tex b/tex/alg-geom/zariski.tex index 76d6c366..1c6ce7c9 100644 --- a/tex/alg-geom/zariski.tex +++ b/tex/alg-geom/zariski.tex @@ -224,6 +224,25 @@ \section{The Zariski topology on affine varieties} Similarly, the intersection of two distinguished open sets is distinguished, just as the product (not intersection!) of two principal ideals is principal. +\begin{proposition} + [Properties of distinguished open set] + Recall that $\VV$ is inclusion-reversing, so being the complement of $\VV$, + we would expect $D$ to be ``inclusion-preserving''. Indeed: + \begin{itemize} + \item If $(f) \subseteq (g)$ (that is, $g \mid f$), then $D(f) \subseteq + D(g)$. + \item Recall that $(fg) \subseteq(f) \cap(g)$. For distinguished open + set, we have $D(fg) = D(f) \cap D(g)$. + \end{itemize} +\end{proposition} + +It is useful to be familiar with the behavior of $D$. +\begin{ques} + If $V=\AA^2$, then $D(x)$ is the plane minus the $y$-axis, and + $D(y)$ is the plane minus the $x$-axis. What is $D(xy)$? +\end{ques} + + % There used to be an exercise here about D(f) % being a distinguished base @@ -252,6 +271,15 @@ \section{Coordinate rings} (Notation explained next section.) \end{definition} +\begin{remark} + [Meaning of the name ``coordinate ring''] + \label{remark:meaning_name_coordinate_ring} + We call the functions $x$, $y$ and $z$ above as the \vocab{coordinate + functions}, as they maps each point in the variety $V$ to its coordinate. + So, the coordinate ring $\CC[V]$ is simply the ring generated by $\CC$ and + the coordinate functions. +\end{remark} + At first glance, we might think this is just $\CC[x_1, \dots, x_n]$. But on closer inspection we realize that \emph{on a given variety}, some of these functions are the same. @@ -277,10 +305,29 @@ \section{Coordinate rings} Thus properties of a variety $V$ correspond to properties of the ring $\CC[V]$. \section{The sheaf of regular functions} +\label{sec:sheaf_regular_functions} \prototype{Let $V = \Aff^1$, $U = V \setminus \{0\}$. Then $1/x \in \OO_V(U)$ is regular on $U$.} Let $V$ be an affine variety and let $\CC[V]$ be its coordinate ring. -We want to define a notion of $\OO_V(U)$ for any open set $U$: +As mentioned in the start of the chapter, we want to define a variety +based on its intrinsic properties only, which is done by studying the collection +of algebraic functions on it. + +In \cite{ref:vakil} ``Motivating example: The sheaf of differentiable +functions'' section, you can see a comparison of how a differentiable manifold +can be studied by studying the differentiable functions on it. + +Denote the set of all rational functions on $V$ by $\OO_V$ (as will be seen +later, this terminology is not quite accurate as we need to allow multiple +representations). We can view this as a set, however this does not capture the +full structure of the rational functions: +\begin{ques} + For any two elements $f$ and $g$ in $\CC[V]$, show that the set where + $\frac{f(x)}{g(x)}$ is well-defined is open in the Zariski topology. + (Hint: $g\pre(0)$ is closed.) +\end{ques} + +So, we want to define a notion of $\OO_V(U)$ for any open set $U$: the ``nice'' functions on any open subset. Obviously, any function in $\CC[V]$ will work as a function on $\OO_V(U)$. However, to capture more of the structure we want to diff --git a/tex/preamble.tex b/tex/preamble.tex index ed3b0a04..779c06aa 100644 --- a/tex/preamble.tex +++ b/tex/preamble.tex @@ -4,6 +4,7 @@ %% which is annoying... font issues \usepackage[T1]{fontenc} \usepackage{lmodern} +\usepackage[pdfusetitle]{hyperref} % These are evan.sty \usepackage{amsmath,amssymb,amsthm} @@ -40,7 +41,6 @@ pdfkeywords={napkin,math}, pdfsubject={web.evanchen.cc}, colorlinks, - pdfusetitle, } %%fakesection Commutative diagrams @@ -146,7 +146,6 @@ frametitlefont=\bfseries, innertopmargin=4pt, innerbottommargin=8pt, - nobreak=true, linecolor=RawSienna, backgroundcolor=Salmon!5, } @@ -209,7 +208,6 @@ mdframed={style=mdblackbox} ]{thmblackbox} -\theoremstyle{theorem} \declaretheorem[name=Question,sibling=theorem,style=thmblackbox]{ques} \declaretheorem[name=Exercise,sibling=theorem,style=thmblackbox]{exercise} \declaretheorem[name=Remark,sibling=theorem,style=thmgreenbox]{remark} @@ -290,10 +288,10 @@ \DeclareFieldFormat[inbook,thesis]{title}{\mkbibemph{#1}\addperiod} % italic title with period \DeclareFieldFormat[article]{title}{#1} % title of journal article is printed as normal text \DeclareFieldFormat[article]{volume}{\textbf{#1}\addcolon\space} -\renewcommand{\mkbibnamefirst}[1]{\textsc{#1}} -\renewcommand{\mkbibnamelast}[1]{\textsc{#1}} +\renewcommand{\mkbibnamegiven}[1]{\textsc{#1}} +\renewcommand{\mkbibnamefamily}[1]{\textsc{#1}} \renewcommand{\mkbibnameprefix}[1]{\textsc{#1}} -\renewcommand{\mkbibnameaffix}[1]{\textsc{#1}} +\renewcommand{\mkbibnamesuffix}[1]{\textsc{#1}} \renewcommand{\finentrypunct}{} %%fakesection Mini ToC