Another bunch of changes

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}{}{}

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@@ -146,7 +146,7 @@ @unpublished{ref:gorin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

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 $\eps > 0$

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 $K = \QQ(i)$, $\OO_K = \ZZ[i]$.

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

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

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 $\VV(f)$.