M README.md → README.md

@@ -1,6 +1,8 @@
 # math/cryptography
 
-Usually while studying papers and books I take handwritten notes, this section contains some of them re-written to LaTeX. The notes are not complete, don’t include all the steps neither the explanations neither most of the proofs. I use these notes to revisit the concepts after some time of reading the paper.
+> Note: this repo is human-written, no AI generated content.
+
+Usually while studying papers and books I take handwritten notes, this section contains some of them re-written to LaTeX. I use these notes to revisit the concepts after some time of reading the paper.
 
 ## mathematics
 
@@ -16,6 +18,7 @@
 - [Notes on the DFT & FFT](fft-notes.pdf)
 - [Notes on NTT](notes_ntt.pdf)
 - [Notes on Reed-Solomon codes](notes_reed-solomon.pdf)
+- [Notes on Lattices](lattices.pdf)
 
 ## cryptography
 

M commutative-algebra-notes.pdf → commutative-algebra-notes.pdf

M commutative-algebra-notes.tex → commutative-algebra-notes.tex

@@ -8,6 +8,7 @@ \usepackage{enumerate}
 \usepackage{hyperref}
 \usepackage{amssymb}
 \usepackage{tikz} % diagram
+\usepackage{changepage} % for adjustwidth
 
 \begin{filecontents}[overwrite]{commutative-algebra-notes.bib}
 @misc{am,
@@ -1815,7 +1816,8 @@
   If $k$ algebraically closed, then $K=k$.
 \end{lemma}
 \begin{proof}
-  $K$ a fingen $k$-algebra and a field $\Longleftrightarrow$ $K=k[X_1, \ldots, X_n]$ a field.\\
+  $K$ a fingen $k$-algebra $\Longleftrightarrow$ $\exists~ x_1,\ldots,x_n\in K$ such that
+  $$K=k[x_1,\ldots,x_n].$$
   Want to show that $K$ is algebraic over $k$.
 
   Since $K$ is a fingen $k$-algebra, by Noether normalization \ref{noether-normalization},\\
@@ -1826,7 +1828,7 @@ \item $K$ is integral over the polynomial ring $R=k[y_1, \ldots, y_r] \subset K$.
   \end{itemize}
 
   By \ref{R.4.9}:\\
-  if $A \subset B$ integral extension of integral domain,\\
+  if $A \subset B$ integral extension of integral domains,\\
   then $A$ is a field $\Longleftrightarrow~ B$ is a field.
 
   which applied to our case:
@@ -1837,11 +1839,11 @@ \vspace{0.3cm}
   Since $y_i \in K$ are algebraically independent\\
   $\Longrightarrow~ R=k[y_1, \ldots, y_r]$ is a polynomial ring in $r$ indeterminates.
 
-  Then, if $r \geq 1$, we have at least one variable ($k[y_1]$);\\
-  in $k[y_1]$, the element $y_1$ is not invertible, since $1/y_1$ is a rational function, not a polynomial.\\
-  (alternatively: the only invertible elements in a polynomial ring over a field are the non-zero constants of the field $k$).
+  Then, if $r \geq 1$, we have at least one variable $y_1$ (with $k[y_1]$);\\
+  but the only invertible elements in a polynomial ring over a field are the non-zero constants of the field $k$, so $y_1$ is not invertible in $R$.
+  (alternatively: the element $y_1$ is not invertible in the polynomial ring, since $1/y_1$ is a rational function, not a polynomial).
 
-  Therefore, $R=k[y_1, \ldots, y_r]$ can be a field iff $r=0$, and we know that is a field, therefore $r=0$ and $R=k$.
+  Therefore, $R=k[y_1, \ldots, y_r]$ can be a field iff $r=0$, and we know that it is a field, therefore $r=0$ and $R=k$.
 
   By Noether normalization, at the previous step we knew that $K$ is integral over $R$ ($=k$).
 
@@ -1849,6 +1851,8 @@ \vspace{0.3cm}
   By \ref{integral-implies-finite}, a fingen algebra that is integral over its base field is a finite vector space over that field.
 
   Thus $[K:k]<\infty$, and $K$ is algebraic over $k$.
+
+  If $k$ is algebraically closed, every element $\alpha\in K$ is algebraic over $k$, and its minimal polynomial over $k$ must have degree one. Therefore $\alpha\in k$, and thus $K=k$.
 \end{proof}
 
 \vspace{0.5cm}
@@ -1860,7 +1864,9 @@ \begin{itemize}
     \item $K$ fingen as a $k$-algebra
     \item $K$ a field
   \end{itemize}
-  Then $K$ is algebraic over $k$, so that $k \subset K$ is a finite field extension; that is $[K:k] < \infty$.
+  Then $K$ is finite-dimensional over $k$, and hence algebraic over $k$; that is $[K:k] < \infty$.
+
+  If $k$ is algebraically closed, then $K=k$.
 \end{thm}
 \begin{proof}
   proven at previous proof.
@@ -1869,116 +1875,186 @@
 
 
 \vspace{0.8cm}
-Now, let $m\subset k[X_1, \ldots, X_n]$ a maximal ideal. Let $K=k[X_1, \ldots, X_n]$.
+Now, assume that $k$ is algebraically closed, and let $m\subset k[X_1, \ldots, X_n]$ be a maximal ideal. Let $A=k[X_1, \ldots, X_n]$.
 
-Then $\frac{k[X_1, \ldots, X_n]}{m} = \frac{K}{m}$ is a field (by basic ideal quotient theory \ref{maximal-quotient-field}).
+Then $\frac{k[X_1, \ldots, X_n]}{m} = \frac{A}{m}$ is a field (by basic ideal quotient theory \ref{maximal-quotient-field}).
 
-If it's a fingen $k$-algebra $\Longrightarrow$ by Zariski's lemma, $\frac{K}{m}$ is algebraic over $k$.
+Moreover, $\frac{A}{m}$ is a fingen $k$-algebra, generated by the classes
+$$X_1+m,\ldots,X_n+m.$$
 
-And since $k$ is algebraically closed $\Longrightarrow~ \frac{K}{m}=k$.
+Therefore, by Zariski's lemma, $\frac{A}{m}$ is algebraic over $k$.
+
+And since $k$ is algebraically closed $\Longrightarrow~ \frac{A}{m}\cong k$.
+
+Let
+$$\pi:A\longrightarrow \frac{A}{m}$$
+be the quotient map. By definition, $\pi$ is the surjective homomorphism that sends each polynomial to its coset modulo $m$, and the kernel of this map is exactly $ker(\pi)=m$.
+
+By Zariski's lemma, $\frac{A}{m}$ is a finite field extension of $k$, and since $k$ is algebraically closed, by Zariski's lemma, this extension must be $k$ itself; denote this relation by
+$$\theta:\frac{A}{m}\stackrel{iso}{\longrightarrow} k$$
+the $k$-algebra isomorphism.
 
-By definition of $K$ being a $k$-algebra,
-$$\exists~ \psi: k \longrightarrow \frac{K}{m} ~~~\Longrightarrow~ k \stackrel{iso}{\cong} \frac{K}{m}$$
+Define the composite map
+$$\phi=\theta\circ\pi:A\longrightarrow k.$$
+which is a $k$-algebra homomorphism.
 
-Thus, $\forall x_i \in k,~ \psi(x_i)=a_i \in k$.
-And by definition of a quotient ring,
+Thus, $\forall~ i\in[n]$, define
+$$a_i=\phi(X_i)\in k.$$
 
-$\psi(x_i)$ is the coset
+Since $\phi$ is a $k$-algebra homomorphism, it must map any scalar $c \in k$ to itself, ie. $\phi(c) = c$. Therefore
+$$\phi(X_i - a_i) = \phi(X_i) - \phi(a_i) = a_i - a_i = 0$$
+
+Since $\phi = \theta \circ \pi$, and $\theta$ is an isomorphism (by definition injective), the only way for $\phi(f)$ to be $0$ is if $\pi(f)=0$:
 \begin{align*}
-  \psi(x_i)&=x_i+m\\
-           &\Longrightarrow a_i = x_i +m\\
-           &x_i - a_i \in m
+  &\phi(X_i-a_i)=\theta(\pi(X_i - a_i)) = 0\\
+  &\Longrightarrow \pi(X_i-a_i)=0\\
 \end{align*}
 
-Since $x_i - a_i \in m ~~\Longrightarrow~~ J=(X_1 - a_1, \ldots, X_n -a_n) \subseteq m$,
+Since $ker(\pi)=m$, any element that maps to $0$ under $\pi$ must belong to $m$, thus
+$$\pi(X_i - a_i) =0 ~~\Longrightarrow~ X_i-a_i\in m$$
+
+Alternative way: by definition of a quotient ring, and elements of $\frac{A}{m}$ are cosets of the form $f+m$. $\pi(x_i)$ sends $X_i$ to the coset $X_i+m$, ie. $\pi(x_i)=X_i+m$. Thus, if $a_i$ is the image of $X_i$, then
+$$X_i \equiv a_i \pmod m,$$
+which by the definition of congruence modulo ideal, this means $X_i -a_i \in m$.
+
+\vspace{0.5cm}
+
+Since $X_i - a_i \in m ~~\Longrightarrow~~ J=(X_1 - a_1, \ldots, X_n -a_n) \subseteq m$,
 
 since $J$ is the kernel of the evaluation map at point $(a_1, \ldots, a_n)$, then $J$ is a maximal ideal.
 
-(Corollary \ref{5.2}) Thus $J=m$, ie. $m=(X_1 - a_1, \ldots, X_n -a_n)$, with $a_i \in k$.
 
-\vspace{0.4cm}
-We had $k \stackrel{iso}{\cong} \frac{K}{m} = \frac{k[X_1, \ldots, X_n]}{m}$.
+\begin{adjustwidth}{4em}{0pt}
+  (Corollary \ref{5.2}) Thus $J=m$, ie. $m=(X_1 - a_1, \ldots, X_n -a_n)$, with $a_i \in k$.
 
-We have the quotient map
-\begin{align*}
-  \phi: k[X_1, \ldots, X_n] &\longrightarrow \frac{k[X_1, \ldots, X_n]}{m} \cong k\\
-  x_i &\mapsto a_i \in k
-\end{align*}
-which is the natural evaluation map $f(X_1, \ldots, X_n) \longrightarrow f(a_1, \ldots, a_n)$.
+  \vspace{0.4cm}
+  We had $k \stackrel{iso}{\cong} \frac{A}{m} = \frac{k[X_1, \ldots, X_n]}{m}$.
+
+  We have the map
+  \begin{align*}
+    \phi: k[X_1, \ldots, X_n] &\longrightarrow \frac{k[X_1, \ldots, X_n]}{m} \cong k\\
+    X_i &\mapsto a_i \in k
+  \end{align*}
+  which is the natural evaluation map $f(X_1, \ldots, X_n) \longmapsto f(a_1, \ldots, a_n)$.
 
-Since $\phi$ is a $k$-algebra homomorphism, then
-$$\forall~ f \in K,~ \phi(f(X_1, \ldots, X_n)) = f(\phi(x_1), \ldots, \phi(x_n)) = f(a_1, \ldots, a_n)$$
+  Since $\phi$ is a $k$-algebra homomorphism, then
+  $$\forall~ f \in A,~ \phi(f(X_1, \ldots, X_n)) = f(\phi(X_1), \ldots, \phi(X_n)) = f(a_1, \ldots, a_n).$$
 
-Thus, $\exists$ a one-to-one correspondence:
-\begin{align*}
-  \text{points in}~ k^n ~~~ &\longleftrightarrow~~~ m-Spec~ A, ~\text{maximal ideals in}~ k[X_1, \ldots, X_n]\\
-  (a_1, \ldots, a_n) ~~~&\longleftrightarrow~~~ (X_1 - a_1, \ldots, X_n - a_n)
-\end{align*}
+  Thus, $\exists$ a one-to-one correspondence:
+  \begin{align*}
+    \text{points in}~ k^n ~~~ &\longleftrightarrow~~~ m-Spec~A, ~\text{maximal ideals in}~ k[X_1, \ldots, X_n]\\
+    (a_1, \ldots, a_n) ~~~&\longleftrightarrow~~~ (X_1 - a_1, \ldots, X_n - a_n)
+  \end{align*}
+\end{adjustwidth}
 
 \vspace{0.5cm}
-Observe: $x_i - a_i \in ker(\psi) =m~ \forall~ i \in [n]$.
+Now let $J\subsetneq A$ be an arbitrary proper ideal. Since every proper ideal is contained in a maximal ideal, $\exists$ a maximal ideal $m$ such that
+$$J\subseteq m$$
 
-We also have seen that $m=(X_1-a_1, \ldots, X_n-a_n)$.
+We have seen that
+$$m=(X_1-a_1, \ldots, X_n-a_n)$$
+for some $P=(a_1,\ldots,a_n)\in k^n$.
 
-So for $P=(a_1, \ldots, a_n) \in k^n$, it is a zero for every polynomial in $m$, ie.
-$$\forall~ f \in m ~\Longrightarrow~ f(P)=0$$
+So $P$ is a zero for every polynomial in $m$, ie.
+$$\forall~ f \in m ~\Longrightarrow~ f(P)=0.$$
 
 Since $J \subseteq m$, $P$ is also a zero for every polynomial in $J$; ie. every element of $J$ vanishes at $P$.
 
 $\Longrightarrow~ P \in V(J)$, and thus $V(J) \neq \emptyset$. This is what the Nullstellensatz first part states:
-\begin{thm}{5.6}[Nullstellensatz] \label{nullstellensatz}
+\begin{thm}{5.6}[Nullstellensatz] \label{nullstellensatz-again}
+  Let $k$ be an algebraically closed field and let $J\subseteq k[X_1,\ldots,X_n]$ be an ideal.
   \begin{enumerate}[a.]
     \item if $J \subsetneq k[X_1, \ldots, X_n]$ then $V(J) \neq \emptyset$
-    \item $I(V(J)) = rad J$, in other words, for $f \in k[X_1, \ldots, X_n]$,
-      $$f(P)=0 ~\forall~ P \in V ~~\Longleftrightarrow~ f^n \in J ~\text{for some $n$.}$$
+    \item $I(V(J)) = rad(J)~~(=\sqrt{J})$, in other words, for $f \in k[X_1, \ldots, X_n]$,
+      $$f(P)=0 ~\forall~ P \in V(J) ~~\Longleftrightarrow~~ f^N \in J ~\text{for some $N\geq 1$.}$$
   \end{enumerate}
 \end{thm}
 \begin{proof}
   \begin{enumerate}[a.]
   \item already proven in the previous steps.
     
-  \item $I(V(J)) = rad J$\\
+  \item $I(V(J)) = \sqrt{J}$\\
   ie. vanishing ideal of a variety = radical of the ideal defining the variety
 
-  where $rad~J = \{ f \in R | f^n \in J ~\text{for some}~ n>0 \}$.
+  where
+  $$\sqrt{J} = \{ f \in A ~|~ f^N \in J ~\text{for some}~ N>0 \}.$$
+
+  First, want to show that
+  $$\sqrt{J}\subseteq I(V(J)).$$
+
+  Let $f\in\sqrt{J}$. Then $f^N\in J$ for some $N\geq 1$.
+
+  Thus, for every $P\in V(J)$,
+  $$f(P)^N=f^N(P)=0,$$
+  and therefore $f(P)=0$.
+
+  Hence $f\in I(V(J))$, so
+  $$\sqrt{J}\subseteq I(V(J)).$$
 
-  Want to show that if a polynomial vanishes at all points where $g_1, \ldots, g_m$ vanish, then $f \in rad(g_1, \ldots, g_m)$.
+  Now, want to show that if a polynomial vanishes at all points where $J$ vanishes, then $f \in \sqrt{J}$.
 
-  Consider the ring $k[X_1, \ldots, X_n, Y]$ and the ideal $J'$ generated by $\{ g_1, \ldots, g_m, 1-Y f \}$
+  Since $k[X_1,\ldots,X_n]$ is Noetherian, write
+  $$J=(g_1,\ldots,g_m).$$
 
-  Suppose there is a point $(a_1, \ldots, a_n, a_{n+1})$ that is a zero of $J'$. ie.
-  $$\exists~ \underbrace{(a_1, \ldots, a_n, a_{n+1})}_{a} \in V(J')$$
+  Consider the ring $k[X_1, \ldots, X_n, Y]$ and the ideal $J'$ generated by $\{ g_1, \ldots, g_m, 1-Y f \}$.
+
+  Suppose there is a point $(a_1, \ldots, a_n,b)$ that is a zero of $J'$. ie.
+  $$\exists~ (a_1, \ldots, a_n,b) \in V(J').$$
+
+  Since
+  $$g_i(a_1,\ldots,a_n)=0,$$
+  our hypothesis says
+  $$f(a_1,\ldots,a_n)=0.$$
 
-  Since $g_i(a)=0$, our hypothesis says $f(a)=0$. However, the last generator $(1-Yf)$ requires
-  $$1 - a_{n+1} f(a) = 0 ~~\Longrightarrow~ \text{implies}~ 1 - a_{n+1} \cdot 0 = 0 ~\Longrightarrow~ 1-0=0$$
+  However, the last generator $(1-Yf)$ requires
+  $$1 - b f(a_1,\ldots,a_n) = 0 ~~\Longrightarrow~ 1 - b \cdot 0 = 0 ~\Longrightarrow~ 1=0,$$
   a contradiction.
 
   Therefore, $V(J') = \emptyset$.
 
   \vspace{0.4cm}
 
-  By the Weak Nullstellensatz (\ref{weak-nullstellensatz-again}),\\
-  \hspace*{2em} if $V(J')=\emptyset$ then $J'=(1)$, so $1 \in J'=(1)$.
+  By part a., applied to $k[X_1,\ldots,X_n,Y]$,\\
+  \hspace*{2em} if $V(J')=\emptyset$ then $J'=k[X_1,\ldots,X_n,Y]$, so $1 \in J'$.
 
   Every element in an ideal is a linear combination of its generators:\\
   $J'$ is generated by $\{ g_1, \ldots, g_m, 1-Yf \}$
 
-  $$\Longrightarrow~ \forall j \in J',~~ j=(\sum \text{(polynomial)}~ g_i) + \text{(polynomial)} \cdot (1 - Yf)$$
+  $$\Longrightarrow~ \forall j \in J',~~ j=(\sum \text{(polynomial)}~ g_i) + \text{(polynomial)} \cdot (1 - Yf).$$
 
   which, since $1 \in J'$,
-  $$1= \left( \sum_{i=1}^m p_i(X,Y) g_i(X) \right) + q(X,Y) \cdot (1 - Y f(X))$$
+  $$1= \left( \sum_{i=1}^m p_i(X,Y) g_i(X) \right) + q(X,Y) \cdot (1 - Y f(X)).$$
 
-  substitute $Y=1/f$,
-  $$1= \left( \sum_{i=1}^m p_i(X,\frac{1}{f}) g_i(X) \right) + q(X,\frac{1}{f}) \cdot \underbrace{(1 - \frac{1}{f} f(X))}_{0}$$
+  Now work in the localization
+  $$A_f=A[f^{-1}],$$
+  where $f$ is invertible, and apply the homomorphism
+  $$A[Y]\longrightarrow A_f,\qquad Y\longmapsto f^{-1}.$$
+
+  Then
+  $$1= \left( \sum_{i=1}^m p_i(X,f^{-1}) g_i(X) \right)
+  +q(X,f^{-1})\cdot \underbrace{(1-f^{-1}f(X))}_{0},$$
   thus
-  $$1= \sum_{i=1}^m p_i(X,\frac{1}{f}) g_i(X)$$
-  multiply by $f^n$,
-  $$f^n= \sum_{i=1}^m A_i(X) g_i(X)$$
+  $$1= \sum_{i=1}^m p_i(X,f^{-1}) g_i(X).$$
 
-  thus $f^n$ is a linear combination of $g_i$.\\
-  Therefore $f^n \in J$, so $f \in rad~J$.
+  Since there are finitely many polynomials $p_i$, $\exists~ N\geq 0$ such that
+  $$f^N p_i(X,f^{-1})\in A ~~\forall~ i.$$
+
+  Multiplying by $f^N$,
+  $$f^N= \sum_{i=1}^m A_i(X) g_i(X),$$
+  where
+  $$A_i(X)=f^N p_i(X,f^{-1})\in A.$$
+
+  Thus $f^N$ is a linear combination of $g_i$.\\
+  Therefore $f^N \in J$, so $f \in \sqrt{J}$.
+
+  Hence
+  $$I(V(J))\subseteq\sqrt{J}.$$
+
+  Combining both inclusions,
+  $$I(V(J))=\sqrt{J}.$$
   \end{enumerate}
 \end{proof}
+