diff --git a/.github/workflows/typos.toml b/.github/workflows/typos.toml new file mode 100644 index 0000000..0c9ac88 --- /dev/null +++ b/.github/workflows/typos.toml @@ -0,0 +1,5 @@ +[default.extend-words] +thm = "thm" +ba = "ba" +Strang = "Strang" # name +Bootle = "Bootle" # name diff --git a/.github/workflows/typos.yml b/.github/workflows/typos.yml new file mode 100644 index 0000000..de7bc65 --- /dev/null +++ b/.github/workflows/typos.yml @@ -0,0 +1,21 @@ +name: typos +on: + pull_request: + branches: [ main ] + types: [ready_for_review, opened, synchronize, reopened] + push: + branches: + - main + +jobs: + typos: + if: github.event.pull_request.draft == false + name: Spell Check with Typos + runs-on: ubuntu-latest + steps: + - uses: actions/checkout@v4 + - name: Use typos with config file + uses: crate-ci/typos@master + with: + config: .github/workflows/typos.toml + diff --git a/commutative-algebra-notes.pdf b/commutative-algebra-notes.pdf index 2f56e1a..8134c3f 100644 Binary files a/commutative-algebra-notes.pdf and b/commutative-algebra-notes.pdf differ diff --git a/commutative-algebra-notes.tex b/commutative-algebra-notes.tex index cf3d1ae..dea6725 100644 --- a/commutative-algebra-notes.tex +++ b/commutative-algebra-notes.tex @@ -174,7 +174,7 @@ \begin{defn}{local ring} A \emph{local ring} has a unique maximal ideal. - Notation: locall ring $A$, its maximal ideal $\mM$, residue field $K=A/\mM$: + Notation: local ring $A$, its maximal ideal $\mM$, residue field $K=A/\mM$: $$A \supset \mM ~\text{or}~ (A, \mM) ~\text{or}~ (A, \mM, K)$$ \end{defn} @@ -230,7 +230,7 @@ A subset $S \subset \Sigma$ is \emph{totally ordered} if for every pair $s_1,s_2 Conversely:\\ Suppose $x \not\in \mM$ for some maximal ideal $\mM$. - Then $\mM$ and $x$ generte the unit ideal $(1)$, so that we have $u + xy = 1$ for some $u \in \mM$ and some $y \in A$. + Then $\mM$ and $x$ generate the unit ideal $(1)$, so that we have $u + xy = 1$ for some $u \in \mM$ and some $y \in A$. Hence $1 -xy \in \mM$, and is therefore not a unit. \end{proof} @@ -632,7 +632,7 @@ $A[\psi] \subset End M$ is the subring geneerated by $A$ and the action of $\psi \begin{prop}{AM.2.8} \label{2.8} - Let $x_i \forall i \in [n]$ be elements of $M$ whose images $\frac{M}{m M}$ from a basis of this vecctor space. Then the $x_i$ generate $M$. + Let $x_i \forall i \in [n]$ be elements of $M$ whose images $\frac{M}{m M}$ from a basis of this vector space. Then the $x_i$ generate $M$. \end{prop} \begin{proof} Let $N$ submodule $M$, generated by the $x_i$. @@ -722,7 +722,7 @@ As in with rings, it is equivalent to say that \section{Exercises} -For the exercises, I follow the assignements listed at \cite{mit-course}. +For the exercises, I follow the assignments listed at \cite{mit-course}. The exercises that start with \textbf{R} are the ones from the book \cite{reid}, and the ones starting with \textbf{AM} are the ones from the book \cite{am}. diff --git a/galois-theory-notes.pdf b/galois-theory-notes.pdf index 69bf452..dbb2932 100644 Binary files a/galois-theory-notes.pdf and b/galois-theory-notes.pdf differ diff --git a/galois-theory-notes.tex b/galois-theory-notes.tex index 40155ae..e989811 100644 --- a/galois-theory-notes.tex +++ b/galois-theory-notes.tex @@ -277,7 +277,7 @@ \end{thm} \begin{thm}{9.10} - 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. + 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. \end{thm} \begin{lemma}{9.13} @@ -809,7 +809,7 @@ Let $f_i$ be the minimal polynimal of $\alpha_i$ over $K$. - Then, $M \supseteq L$ is spliting field of $\Prod_{i=1}^r f_i$, since $M$ is normal enclosure of $L:K$. + Then, $M \supseteq L$ is splitting field of $\Prod_{i=1}^r f_i$, since $M$ is normal enclosure of $L:K$. For every zero $\beta_{ij}$ of $f_i$ in $M$,\\ $\exists$ an isomorphism $\sigma: K(\alpha_i) \longrightarrow K(\beta_{ij})$ by Corollary \ref{5.13}. @@ -1065,7 +1065,7 @@ For $n \geq 3, ~~\mathbb{D}_n \subseteq \mathbb{S}_n$ (subgroup of the Symmetry then, $\exists$ at least one $c$ in $(a,b)$ such that $Df(c)=0$. \end{thm} \begin{proof} (proof source: cue math website) - Notice that when $Df(x_i)=0$ occours, is a maximum or minimum (extrema) value of $f$. + Notice that when $Df(x_i)=0$ occurs, is a maximum or minimum (extrema) value of $f$. $\Longrightarrow$ if a function is continuous, it is guaranteed to have both a maximum and a minimum point in the interval.\\ Two possibilities: @@ -1076,9 +1076,9 @@ For $n \geq 3, ~~\mathbb{D}_n \subseteq \mathbb{S}_n$ (subgroup of the Symmetry since $f$ not constant, must change directions in ordder to start and end at the same $y$-value ($f(a)=f(b)$).\\ Thus at some point between $a$ and $b$ it will either have a minimum, maximum or both. \begin{enumerate}[a.] - \item does the maximum occour at a point where $Df > 0$?\\ + \item does the maximum occur at a point where $Df > 0$?\\ No, because if $Df > 0$, then $f$ is increasing, but it can not increase since we're at its maximum point. - \item does the maximum occour at a point where $Df < 0$?\\ + \item does the maximum occur at a point where $Df < 0$?\\ 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. \end{enumerate} Same with minimus.\\ diff --git a/notes_fri_stir.pdf b/notes_fri_stir.pdf index 4f16693..4eb5c83 100644 Binary files a/notes_fri_stir.pdf and b/notes_fri_stir.pdf differ diff --git a/notes_fri_stir.tex b/notes_fri_stir.tex index 6baf16c..e729985 100644 --- a/notes_fri_stir.tex +++ b/notes_fri_stir.tex @@ -76,7 +76,7 @@ Consider the following protocol: \end{enumerate} %/// TODO tabulate this next lines -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 +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 Let $g(x)=a \cdot x^{d+1}, ~~ h(x)=b \cdot x^{d+1}$, and set $f(x) = g(x) + \alpha h(x)$. 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. @@ -314,14 +314,15 @@ $$g(z) = (f(z)-f(r))\cdot (z-r)^{-1}$$ \section{STIR (main idea)} -\emph{Update from 2024-03-22, notes from Héctor Masip Ardevol (\href{https://hecmas.github.io/}{https://hecmas.github.io}) explanations.} +\emph{Update from 2024-03-22, notes from Héctor Masip Ardevol\\ +(\href{https://hecmas.github.io/}{https://hecmas.github.io}) explanations.} \vspace{0.3cm} Let $p \in \mathbb{F}[x]^{