@ -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}.
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.
