M .github/workflows/typos.toml → .github/workflows/typos.toml

@@ -13,6 +13,7 @@ extenaion = "extension"
 
 # strings that are not a typo:
 thm = "thm"
+MSIS = "MSIS"
 
 # equations stuff
 ba = "ba"

M .gitignore → .gitignore

@@ -15,3 +15,4 @@ galois-theory-notes.bib
 commutative-algebra-notes.bib
 algebraic-coding-theory-notes.bib
 pinocchio-groth16.bib
+lattices.bib

A lattices.pdf

A lattices.tex

@@ -0,0 +1,303 @@
+\documentclass{article}
+\usepackage[utf8]{inputenc}
+\usepackage{amsfonts}
+\usepackage{amsthm}
+\usepackage{amsmath}
+\usepackage{enumerate}
+\usepackage{hyperref}
+\usepackage{amssymb}
+\usepackage{tikz} % diagram
+\usepackage{changepage} % for adjustwidth
+
+\begin{filecontents}[overwrite]{lattices.bib}
+@misc{chris-peikert,
+  author = {Chris Peikert},
+  title = {{A decade of lattice cryptography}},
+  year = {2016}
+}
+@misc{matan,
+  author = {Matan Prasma},
+  title = {{Lattices, an elementary and rigorous account}},
+  year = {2024}
+}
+@misc{slides-micciano,
+  author = {Daniele Micciancio},
+  title = {{Introduction to Lattices}},
+  year = {2024}
+}
+@misc{micciano-regev,
+  author = {Daniele Micciancio and Oded Regev},
+  title = {{Lattice-based Cryptography}},
+  year = {2008}
+}
+@misc{menezes,
+  author = {Alfred Menezes},
+  title = {{A Gentle Introduction to Lattice-Based Cryptography}},
+  year = {2026}
+}
+@misc{cynthia-dwork,
+  author = {Cynthia Dwork},
+  title = {{Lattices and Their Application to Cryptography}},
+  year = {1998}
+}
+\end{filecontents}
+\nocite{*}
+
+
+\theoremstyle{definition}
+
+\newtheorem{thm}{Theorem}[section]
+\newtheorem{lemma}[thm]{Lemma}
+\newtheorem{prop}[thm]{Proposition}
+\newtheorem{cor}[thm]{Corollary}
+\newtheorem{defn}[thm]{Definition}
+\newtheorem{remark}[thm]{Remark}
+\newtheorem{ex}[thm]{Example}
+
+\newcommand{\Rr}{\mathbb{R}}
+\newcommand{\Zz}{\mathbb{Z}}
+\newcommand{\Ll}{\mathcal{L}}
+
+\title{Notes on Lattices}
+\author{arnaucube}
+\date{2026}
+
+\begin{document}
+
+\maketitle
+
+\begin{abstract}
+    Notes taken while studying lattice-based cryptography.
+
+    Usually while reading books and papers I take handwritten notes in paper notebooks, this document contains some of them re-written to $LaTeX$.
+\end{abstract}
+
+\tableofcontents
+
+\section{Lattices}
+\subsection{Hard problems}
+
+\begin{defn}[SVP (Shortest Vector Problem)]
+find a shortest non-zero vector $x \in \Ll$, ie. that minimizes the Euclidean norm $||x||$, that is $\lambda_1(\Ll)=||x||$.
+\end{defn}
+
+\begin{defn}[CVP (Closest Vector Problem)]
+given a vector $y \in \Rr^n$ with $y \not\in \Ll$, find a nonzero vector $x \in \Ll$ that is closest to $y$, ie. that minimizes the Euclidean norm $||y-x||$.
+\end{defn}
+
+\begin{defn}[NTRU key recovery problem]
+Given $h(x)$, find ternary polynomials $f, g \in \mathcal{T}$ satisfying
+$$f(x) \cdot h(x) = g(x) \pmod q \in R_q$$
+Where $R_q = \Zz_q[X]/(X^n-1)$.
+\end{defn}
+
+\begin{defn}[SIS (Short Integer Solution)]
+Given $q \in \Zz$ and matrix $A \in \Zz_q^{m \times n}$, find a non-zero short $z \in \Zz^n$ s.th.
+$$A \cdot z = 0 \pmod q$$
+
+\small{"short": $z \in \{B, B\}^n$ for some bound $B$, often $z \in \{-1, 0, 1 \}^n$.}
+\end{defn}
+
+The set of all solutions to the SIS problem forms the (q-ary) lattice
+$$\Lambda^{\perp}_q(A) = \{ z \in \Zz^n ~|~ A z = 0 \in \Zz_q^m \} \supseteq \Zz_q^n$$
+where the SIS problem becomes the SVP for $\Lambda_q^{\perp}(A)$
+
+\begin{itemize}
+\item LLL algorithm is used to solve SVP and CVP.
+
+  \item Both SVP and CVP become computationally difficult (NP hard) as the lattice dimension $n$ grows.
+\end{itemize}
+
+
+
+\begin{defn}[Ring SIS (Ring Short Integer Solution)]
+given a uniformly random matrix $A \in^R R_q^n$, with elements $a_i \in R_q$, find a non-zero integer vector $z \in R^n$ of norm $||z|| \leq \beta$ such that
+
+$$A z = \sum_i a_i \cdot z_i = 0 \in R_q$$
+\end{defn}
+
+\begin{defn}[MSIS (Module Short Integer Solution)]
+given a uniformly random matrix $A \in^R R_q^{m \times n}$, with columns $a_i \in R_q^m$, find a non-zero integer vector $z \in R^n$ of norm $||z|| \leq \beta$ such that
+
+$$A z = 0 \in R_q^m$$
+$$ie.~ \sum_i a_{ij} \cdot z_j = 0 \in R_q, ~~for~ i=1, \ldots, m$$
+\end{defn}
+
+Notice that Module-SIS is a Ring-SIS with $rank=1$ ($m=1$).
+
+The next form is useful to compress the size of a SIS matrix by $n$ columns: 
+\begin{defn}[Hermite Normal Form (HNF)]
+  Assume the leftmost $n$ columns $A_1 \in \Zz_q^{n \times n}$ of $A=[A_1|A_2] \in \Zz_q^{n \times m}$ form an invertible matrix.
+
+  Then replace $A$ by
+  $$A_1^{-1} \cdot A = [ I_n ~|~ \overline{A} = A_1^{-1} \cdot A_2]$$
+  and threat the $I_n$ submatrix as implicit.
+
+  Then, $A$ and $[I_n ~|~ \overline{A}]$ have exactly the same set of short SIS solutions.
+\end{defn}
+
+\begin{defn}[LWE distribution]
+  $s \in \Zz_q^n$ (secret), $a \in \Zz_q^n$, $e \in \Xi$, output
+  $$(a, b= \langle s, a \rangle + e \pmod q) \in (\Zz_q^n \times \Zz_q)$$
+\end{defn}
+
+\begin{defn}[Search-LWE]
+  given $m$ independent samples $(a_i, b_i) \in \Zz_q^n \times \Zz_q$, for a uniform $s \in \Zz_q^n$ (fixed for all samples), find $s$.
+\end{defn}
+\begin{defn}[Decision-LWE]
+  given $m$ independent samples $(a_i, b_i) \in \Zz_q^n \times \Zz_q$, distinguish which one is not random and it is an actual instance.
+\end{defn}
+
+
+
+\subsection{Dual lattices}
+
+From previous section, we have LWE-Lattice and SIS-Lattice:
+
+
+Search-LWE can be seen as a Bounded-Distance Decoding (BDD) problem on q-ary m-dimensional integer lattices:\\
+for LWE samples, vector $b$ is relatively close to exactly one vector in the LWE lattice:
+\begin{defn}[LWE Lattice] \label{lwe-lattice}
+  A LWE-Lattice $\Ll$ is defined
+  \begin{align*}
+    \Ll &= \{ A^T s ~|~ s \in \Zz_q^n \} + q \Zz^m\\
+	&= \{ A^T s + qz ~|~ s \in \Zz_q^n, z \in \Zz^m \}
+  \end{align*}
+\end{defn}
+
+SIS can be seen as a SVP on q-ary m-dimensional lattices:
+\begin{defn}[SIS-Lattice] \label{sis-lattice}
+  A SIS-Lattice $\Ll$ is defined
+$$\Ll^\perp = \{ z \in \Zz^m ~|~ Az=0 \in \Zz_q^n \} \supseteq q \Zz^m$$
+\end{defn}
+
+
+\begin{defn}[Dual of a lattice] \label{dual-lattice}
+    For a lattice $\Ll$, its dual is
+    $$\Ll^* = \{ y \in \Rr^m ~|~ \langle y, x \rangle \in \Zz ~\forall~ x \in \Ll \}$$
+\end{defn}
+The dual lattice consists of all vectors that measure lattice points with integer-valued linear measurements.
+
+\begin{remark}
+  For a lattice basis $B=(b_1, \ldots, b_n)$, the dual basis $B^*=(b_1^*, \ldots, b_n^*)$ is the unique basis satisfying
+  $$
+    \langle b_i^*, b_j \rangle
+    = \delta_{ij} =
+      \begin{cases}
+	1, & \text{if } i = j, \\
+	0, & \text{if } i \neq j
+      \end{cases}
+  $$
+  where $\delta_{ij}$ is the Kronecker delta.
+\end{remark}
+
+\begin{lemma}
+    $$\Ll = q \Ll^{\perp *} ~\text{ie.}~ \frac{1}{q} \Ll = \Ll^{\perp *}$$
+\end{lemma}
+\begin{proof}
+    \begin{enumerate}[i.]
+    \item
+      Take $y \in \frac{1}{q} \Ll$, then by \ref{lwe-lattice}, $\exists~ s \in \Zz^n, z \in \Zz^m$ such that
+    $$y = \frac{1}{q} (A^Ts + qz) = \frac{1}{q} A^T s +z$$
+
+    Let $x \in \Ll^\perp$. By definition (\ref{sis-lattice}) $Ax=0 \pmod q$, therefore $\exists~ k \in \Zz^n$ such that $Ax = qk$.
+
+    Compute
+    \begin{align*}
+	\langle y, x \rangle &= \langle \frac{1}{q} A^T s + z, x \rangle\\
+			     &= \frac{1}{q} \langle A^T s, x \rangle + \langle z, x \rangle\\
+			     &= \frac{1}{q} \langle s, Ax \rangle + \langle z, x \rangle\\
+			     &= \frac{1}{q} \langle s, qk \rangle + \langle z, x \rangle\\
+			     &= \langle s, k \rangle + \langle z, x \rangle
+    \end{align*}
+    and $\langle s, k \rangle + \langle z, x \rangle \in \Zz$, since each of $s, k, z, x \in \Zz$ separately.
+
+    Thus $\langle y, x \rangle \in \Zz$, hence by \ref{dual-lattice}, $y \in \Ll^{\perp *}$.
+
+    We started from $y \in \frac{1}{q}\Ll$, so, now we have
+    $$y \in \Ll^{\perp *} ~\forall~ y \in \frac{1}{q} \Ll$$
+    therefore
+    $$\Ll^{\perp *} \supseteq \frac{1}{q} \Ll$$
+    
+    \item
+    Now we prove the reverse inclusion. Take $y \in \Ll^{\perp *}$.
+
+    Observe that $q \Zz^m \subseteq \Ll^\perp$, because
+    $$\forall z \in \Zz^m,~ A(qz) = qAz = q 0 = 0 \pmod q$$
+
+    Since $y \in \Ll^{\perp *}$, we must have $\langle y, qz \rangle \in \Zz ~\forall~ z \in \Zz^m$.
+
+    Let $e_i \in \Zz^m$ be the $i-th$ standard basis vector, ie. $e_i=(0, \ldots, 0, 1, 0, \ldots, 0)$ (where one is at the i-th position).
+
+    We know that $\langle y, qz \rangle \in \Zz ~\forall z \in \Zz^m$ (including $z=e_i$), thus, for
+    $q e_i = (0, \ldots, 0, q, 0, \ldots, 0)$,
+    $$\langle y, q e_i \rangle = q \langle y, e_i \rangle = q \cdot y_i$$
+    since $\langle y, e_i \rangle = y_i$ (extracts the i-th position from $y$).
+
+    So, since $\langle y, qz \rangle \in \Zz ~\forall z \in \Zz^m$, and $\langle y, q e_i \rangle = q y_i$, then $q y_i \in \Zz$
+
+    This argument works $\forall i \in [m]$, that is, $q y_1, \ldots, q y_m \in \Zz$. Therefore
+    $$qy \in \Zz^m ~\forall~ y \in \Ll^{\perp *}$$
+
+    For $\overline{x} \in ker(A) \subseteq \Zz_q^m$, since $A \overline{x}=0$, then $Ax =0 \pmod q$, thus $x \in \Ll^\perp$ (by its definition \ref{sis-lattice}).
+
+    Since $y \in \Ll^{\perp *}$, by the definition of $\Ll^{\perp *}$ (\ref{dual-lattice}), $\langle y, x \rangle \in \Zz$. Multiply it by $q$,
+    $$ q \langle y, x \rangle = \langle qy, x \rangle \in q \Zz$$
+    hence $\langle qy, x \rangle = 0 \pmod q$, thus $qy \pmod q \in ker(A)^\perp$.
+
+    \begin{adjustwidth}{4em}{0pt}
+      We'll now prove the following identity, valid for every integer $q\geq 2$, which we will use later:
+      $$(\ker A)^\perp=im(A^T)$$
+      where $A:\Zz_q^m \to \Zz_q^n$.
+
+      Let $y \in im(A^T)$. Then $\exists~ s \in \Zz_q^n$ such that $y=A^Ts$.
+
+      Then, $\forall~ x \in \ker(A)$, by definition of the kernel, $Ax=0$. Want to prove that $y$ is orthogonal to every $x \in \ker(A)$:
+      $$\langle y,x\rangle = \langle A^Ts,x\rangle = \langle s,Ax\rangle = 0$$
+
+      thus, $\langle y,x\rangle =0 ~\forall~ x \in \ker(A)$, hence $y$ is orthogonal to every $x$, that is, $y \in (\ker(A))^\perp$.
+
+      Therefore,
+      $$im(A^T) \subseteq (\ker(A))^\perp$$
+
+      For the reverse inclusion, it suffices to show that both sets have the same number of elements:
+
+      By the First Isomorphism Theorem,
+      $$|im(A)| = \frac{ |\Zz_q^m|}{|\ker(A)|} = \frac{q^m}{|\ker(A)|}$$
+
+      Notice that for every subgroup $K \subseteq \Zz_q^m$, the dot product on $\Zz_q^m$ is nondegenerate, therefore
+      $$|K^\perp| = \frac{q^m}{|K|}$$
+
+      so, applied to $\ker(A)$,
+      $$|\ker(A)^\perp| = \frac{q^m}{|\ker(A)|}= |im(A)|$$
+
+      which, by transposition properties, $|im(A)|=|im(A^T)|$, thus
+      $$|\ker(A)^\perp| = |im(A^T)|$$
+
+      hence,
+      $$(\ker(A)^\perp) = im(A^T)$$
+    \end{adjustwidth}
+
+    Continuing from $qy \pmod q \in (ker(A))^\perp$, we've just seen that $(ker(A))^\perp = im(A^T)$, thus
+    $$\exists s \in \Zz^n ~\text{s.th.}~ qy=A^Ts \pmod q$$
+
+    Therefore, for some
+    $$z \in \Zz^m, ~ qy=A^Ts +qz$$
+    which means that $qy \in \Ll$, by definition of $\Ll$ (\ref{lwe-lattice}). Divide that by $q$, obtaining
+    $$y \in \frac{1}{q} \Ll$$
+
+    Thus, $\forall y \in \Ll^{\perp *}$, we have $y \in \frac{1}{q} \Ll$.
+
+    Hence $\Ll^{\perp *} \subseteq \frac{1}{q} \Ll$.
+    \end{enumerate}
+
+    \vspace{0.6cm}
+    Putting together $(i).~ \Ll^{\perp *} \supseteq \frac{1}{q} \Ll$, and $(ii).~ \Ll^{\perp *} \subseteq \frac{1}{q} \Ll$, we have
+    $$\Ll^{\perp *} = \frac{1}{q} \Ll$$
+\end{proof}
+
+
+\bibliographystyle{unsrt}
+\bibliography{lattices.bib}
+
+\end{document}