@@ -0,0 +1,5 @@
+[default.extend-words]
+thm = "thm"
+ba = "ba"
+Strang = "Strang" # name
+Bootle = "Bootle" # name

@@ -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
+

@@ -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 @@
   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 @@
 
 
 \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 @@ \newpage
 
 \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}.
 

@@ -277,7 +277,7 @@ TODO
 \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 $L=K(\alpha_1, \ldots, \alpha_r)$ with $\alpha_i^{n_i} \in K(\alpha_1, \ldots, \alpha_{j-1})$ (by definition of $L:K$ being a radical extension).
 
   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 @@ \end{itemize}
   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 @@ \item $f$ is not constant:\\
     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.\\

@@ -76,7 +76,7 @@ \item If V is convinced that $deg(f) \leq d$, V believes that both $g, h$ have $deg \leq d$.
 \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 $p \in \mathbb{F}[x]^{<n}$.
 
 In FRI we decompose $p(x)$ as
 $$p(x) = p_e(x^2) + x \cdot p_o(x^2)$$
-with $p_e, p_o \in \mahtbb{F}[x]^{<n}$ containing the even and odd powers respectively.
+with $p_e, p_o \in \mathbb{F}[x]^{<n}$ containing the even and odd powers respectively.
 
 The next FRI polynomial is
 $$p_1(x) = p_e(x) + \alpha p_o(x)$$
@@ -329,7 +330,7 @@ for $\alpha \in^R \mathbb{F}$.
 
 In STIR, this would be $q(x)=x^2$,
 $$Q(z,y) = p_e(y) + z \cdot p_o(y)$$
-and then, $p(x) = Q(x, q(x))$. And $Q$ fullfills the degree from Fact 4.6 from the STIR paper.
+and then, $p(x) = Q(x, q(x))$. And $Q$ fulfills the degree from Fact 4.6 from the STIR paper.
 
 We can generalize to a $q$ with bigger degree, or with another shape, and adapting $Q$ on the choice of $q$.