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port initial notes on commutative algebra (ideals, modules, Noetherian rings) (#1)
* port initial notes on commutative algebra: ideals & modules, Nakayama's lemma, etc * port notes on Noetherian rings&modules * add ideals related definitions * improve Cayley-Hamilton proof (specially determinant trick explanation) * polishing * add typos detection * add some exercises, and proof of Z and K[X} being PID
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.github/workflows/typos.toml
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.github/workflows/typos.toml
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[default.extend-words]
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thm = "thm"
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# equations stuff
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ba = "ba"
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nd = "nd"
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# names
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Strang = "Strang"
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Bootle = "Bootle"
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.github/workflows/typos.yml
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.github/workflows/typos.yml
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name: typos
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on:
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pull_request:
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branches: [ main ]
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types: [ready_for_review, opened, synchronize, reopened]
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push:
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branches:
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- main
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jobs:
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typos:
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if: github.event.pull_request.draft == false
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name: Spell Check with Typos
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runs-on: ubuntu-latest
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steps:
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- uses: actions/checkout@v4
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- name: Use typos with config file
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uses: crate-ci/typos@master
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with:
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config: .github/workflows/typos.toml
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.gitignore
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.gitignore
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*.snm
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*.snm
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*.vrb
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*.vrb
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galois-theory-notes.bib
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galois-theory-notes.bib
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commutative-algebra-notes.bib
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@@ -7,6 +7,7 @@ Notes, code and documents done while reading books and papers.
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- [Notes on "Abstract Algebra" book, by Charles C. Pinter](abstract-algebra-charles-pinter-notes.pdf)
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- [Notes on "Abstract Algebra" book, by Charles C. Pinter](abstract-algebra-charles-pinter-notes.pdf)
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- [Notes on Weil pairing](weil-pairing.pdf)
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- [Notes on Weil pairing](weil-pairing.pdf)
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- [Notes on Galois Theory](galois-theory-notes.pdf)
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- [Notes on Galois Theory](galois-theory-notes.pdf)
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- [Notes on Commutative Algebra](commutative-algebra-notes.pdf)
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In-between math & crypto:
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In-between math & crypto:
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commutative-algebra-notes.pdf
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commutative-algebra-notes.pdf
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commutative-algebra-notes.tex
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commutative-algebra-notes.tex
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\end{thm}
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\end{thm}
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\begin{thm}{9.10}
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\begin{thm}{9.10}
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An irreducible polynomial $f \in K[t]$ ($K \subseteq \mathbb{C}$) is \emph{separable over} $K$ if it has simple zeros in $\mathbb{C}$, or equivelently, simple zeros in its splitting field.
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An irreducible polynomial $f \in K[t]$ ($K \subseteq \mathbb{C}$) is \emph{separable over} $K$ if it has simple zeros in $\mathbb{C}$, or equivalently, simple zeros in its splitting field.
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\end{thm}
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\end{thm}
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\begin{lemma}{9.13}
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\begin{lemma}{9.13}
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Let $f_i$ be the minimal polynimal of $\alpha_i$ over $K$.
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Let $f_i$ be the minimal polynimal of $\alpha_i$ over $K$.
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Then, $M \supseteq L$ is spliting field of $\Prod_{i=1}^r f_i$, since $M$ is normal enclosure of $L:K$.
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Then, $M \supseteq L$ is splitting field of $\Prod_{i=1}^r f_i$, since $M$ is normal enclosure of $L:K$.
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For every zero $\beta_{ij}$ of $f_i$ in $M$,\\
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For every zero $\beta_{ij}$ of $f_i$ in $M$,\\
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$\exists$ an isomorphism $\sigma: K(\alpha_i) \longrightarrow K(\beta_{ij})$ by Corollary \ref{5.13}.
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$\exists$ an isomorphism $\sigma: K(\alpha_i) \longrightarrow K(\beta_{ij})$ by Corollary \ref{5.13}.
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@@ -1065,7 +1065,7 @@ For $n \geq 3, ~~\mathbb{D}_n \subseteq \mathbb{S}_n$ (subgroup of the Symmetry
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then, $\exists$ at least one $c$ in $(a,b)$ such that $Df(c)=0$.
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then, $\exists$ at least one $c$ in $(a,b)$ such that $Df(c)=0$.
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\end{thm}
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\end{thm}
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\begin{proof} (proof source: cue math website)
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\begin{proof} (proof source: cue math website)
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Notice that when $Df(x_i)=0$ occours, is a maximum or minimum (extrema) value of $f$.
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Notice that when $Df(x_i)=0$ occurs, is a maximum or minimum (extrema) value of $f$.
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$\Longrightarrow$ if a function is continuous, it is guaranteed to have both a maximum and a minimum point in the interval.\\
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$\Longrightarrow$ if a function is continuous, it is guaranteed to have both a maximum and a minimum point in the interval.\\
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Two possibilities:
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Two possibilities:
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@@ -1076,9 +1076,9 @@ For $n \geq 3, ~~\mathbb{D}_n \subseteq \mathbb{S}_n$ (subgroup of the Symmetry
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since $f$ not constant, must change directions in ordder to start and end at the same $y$-value ($f(a)=f(b)$).\\
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since $f$ not constant, must change directions in ordder to start and end at the same $y$-value ($f(a)=f(b)$).\\
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Thus at some point between $a$ and $b$ it will either have a minimum, maximum or both.
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Thus at some point between $a$ and $b$ it will either have a minimum, maximum or both.
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\begin{enumerate}[a.]
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\begin{enumerate}[a.]
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\item does the maximum occour at a point where $Df > 0$?\\
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\item does the maximum occur at a point where $Df > 0$?\\
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No, because if $Df > 0$, then $f$ is increasing, but it can not increase since we're at its maximum point.
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No, because if $Df > 0$, then $f$ is increasing, but it can not increase since we're at its maximum point.
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\item does the maximum occour at a point where $Df < 0$?\\
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\item does the maximum occur at a point where $Df < 0$?\\
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No, because if $Df < 0$, then $f$ is deccreasing, which means that at our left it was larger, but we're at a maximum point, so a contradiction.
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No, because if $Df < 0$, then $f$ is deccreasing, which means that at our left it was larger, but we're at a maximum point, so a contradiction.
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\end{enumerate}
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\end{enumerate}
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Same with minimus.\\
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Same with minimus.\\
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\end{enumerate}
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\end{enumerate}
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%/// TODO tabulate this next lines
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%/// TODO tabulate this next lines
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With high probablility, $\alpha$ will not cancel the coeffs with $deg \geq d+1$. % TODO check which is the name of this theorem or why this is true
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With high probability, $\alpha$ will not cancel the coeffs with $deg \geq d+1$. % TODO check which is the name of this theorem or why this is true
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Let $g(x)=a \cdot x^{d+1}, ~~ h(x)=b \cdot x^{d+1}$, and set $f(x) = g(x) + \alpha h(x)$.
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Let $g(x)=a \cdot x^{d+1}, ~~ h(x)=b \cdot x^{d+1}$, and set $f(x) = g(x) + \alpha h(x)$.
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Imagine that P can chose $\alpha$ such that $a x^{d+1} + \alpha \cdot b x^{d+1} = 0$, then, in $f(x)$ the coefficients of degree $d+1$ would cancel.
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Imagine that P can chose $\alpha$ such that $a x^{d+1} + \alpha \cdot b x^{d+1} = 0$, then, in $f(x)$ the coefficients of degree $d+1$ would cancel.
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@@ -314,14 +314,15 @@ $$g(z) = (f(z)-f(r))\cdot (z-r)^{-1}$$
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\section{STIR (main idea)}
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\section{STIR (main idea)}
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\emph{Update from 2024-03-22, notes from Héctor Masip Ardevol (\href{https://hecmas.github.io/}{https://hecmas.github.io}) explanations.}
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\emph{Update from 2024-03-22, notes from Héctor Masip Ardevol\\
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(\href{https://hecmas.github.io/}{https://hecmas.github.io}) explanations.}
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\vspace{0.3cm}
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\vspace{0.3cm}
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Let $p \in \mathbb{F}[x]^{<n}$.
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Let $p \in \mathbb{F}[x]^{<n}$.
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In FRI we decompose $p(x)$ as
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In FRI we decompose $p(x)$ as
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$$p(x) = p_e(x^2) + x \cdot p_o(x^2)$$
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$$p(x) = p_e(x^2) + x \cdot p_o(x^2)$$
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with $p_e, p_o \in \mahtbb{F}[x]^{<n}$ containing the even and odd powers respectively.
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with $p_e, p_o \in \mathbb{F}[x]^{<n}$ containing the even and odd powers respectively.
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The next FRI polynomial is
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The next FRI polynomial is
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$$p_1(x) = p_e(x) + \alpha p_o(x)$$
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$$p_1(x) = p_e(x) + \alpha p_o(x)$$
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@@ -329,7 +330,7 @@ for $\alpha \in^R \mathbb{F}$.
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In STIR, this would be $q(x)=x^2$,
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In STIR, this would be $q(x)=x^2$,
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$$Q(z,y) = p_e(y) + z \cdot p_o(y)$$
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$$Q(z,y) = p_e(y) + z \cdot p_o(y)$$
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and then, $p(x) = Q(x, q(x))$. And $Q$ fullfills the degree from Fact 4.6 from the STIR paper.
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and then, $p(x) = Q(x, q(x))$. And $Q$ fulfills the degree from Fact 4.6 from the STIR paper.
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We can generalize to a $q$ with bigger degree, or with another shape, and adapting $Q$ on the choice of $q$.
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We can generalize to a $q$ with bigger degree, or with another shape, and adapting $Q$ on the choice of $q$.
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