Noetherian rings: add ex. 3.2 & 3.5

commutative-algebra-notes.tex

@@ -1158,7 +1158,81 @@ $0 \longrightarrow L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\long
 
   So, $L$ provides $k$ generators for the kernel part of $M$, $N$ provides $p$ "lifts" for the quotient part of $M$; thus $M$ is generated by $k+p$ elements.\\
   Thus $M$ is finitely generated over $A$.
+\end{proof}
 
+
+\subsection{Exercises Chapter 3}
+
+\begin{ex}{R.3.2}
+  $K$ a field, $A \supset K$ a ring which is finite dimensional as a $K$-vector space.
+   Prove that $A$ is Noetherian and Artinian.
+\end{ex}
+\begin{proof}
+  $dim(A)=n < \infty$, so every ideal $\aA$ of $A$ is a $K$-subspace of $A$, because if $x \in \aA$ and $c \in K$, then $c \cdot x \in \aA$.
+
+  \begin{enumerate}
+    \item Noetherian:\\
+      let $I_1 \subseteq I_2 \subseteq \ldots$ be an ascending chain of ideals in $A$.
+
+      Since each $I_i$ is a subspace, we have
+      $$dim_K(I_1) \leq dim_K(I_2) \leq \ldots \leq n$$
+      where at some $i=m$ we have $dim_K(I_m)=dim_K(I_{m+1})$; then since $I_m \subseteq I_{m+1}$, we have $I_m = I_{m+1}$. So $A$ is Noetherian.
+      
+    \item Artinian:\\
+      Similarly, if $I_1 \supseteq I_2 \supseteq \ldots$ a descending chain of ideals in $A$.
+
+      then
+      $$n \geq dim_K(I_1) \geq dim_K(I_2) \geq \ldots \geq 0$$
+      where at some $i=m$ we have $dim_K(I_m)=dim_K(I_{m+1})$; then since $I_m \subseteq I_{m+1}$, we have $I_m = I_{m+1}$. So $A$ is Artinian.
+  \end{enumerate}
+\end{proof}
+
+\begin{ex}{R.3.5}
+  Let $0 \longrightarrow L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N \longrightarrow 0$ an exact sequence. Let $M_1, M_2 \subseteq M$ be submodules of $M$.
+
+  Prove if the following holds or not:
+  $$\beta(M_1)=\beta(M_2) ~\text{and}~ \alpha^{-1}(M_1)=\alpha^{-1}(M_2)~~\Longrightarrow~ M_1=M_2$$
+\end{ex}
+\begin{proof}
+  Counterexample showing that it does not hold:
+
+  Let $K$ a field, $M = K \oplus K~, L=K,~N=K$.
+
+  Set, for $l \in L,~ (m_1, m_2) \in M$,
+  \begin{align*}
+    \alpha:~ &l \longmapsto (l, 0)\\
+    \beta:~ &(m_1, m_2) \longmapsto m_2
+  \end{align*}
+
+  So we have
+  $$0 \longrightarrow K \stackrel{\alpha}{\longrightarrow} K^2 \stackrel{\beta}{\longrightarrow} K \longrightarrow 0$$
+
+  Then,
+  \begin{align*}
+    M_1 &= \{ (x, x) ~|~ x \in K\} ~~~~\sim\text{(diagonal line)} \\
+    M_2 &= \{ (0, x) ~|~ x \in K\} ~~~~\sim\text{(y-axis)}
+  \end{align*}
+
+  (Geometric interpretation: $M_1,~ M_2$ are the \emph{diagonal line} and \emph{y-axis} respectively; and $\alpha,~\beta$ capture information about the \emph{vertical} components (x-axis, y-axis respectively), but not about the \emph{diagonal} way a submodule is embedded in $M$).
+
+  Then,
+  \begin{align*}
+    \beta(M_1) &= \{ x ~|~ x \in K\} = K \\
+    \beta(M_2) &= \{ x ~|~ x \in K\} = K
+  \end{align*}
+  thus, $\beta(M_1) = \beta(M_2)$.
+
+  For $M_1,~~ (l,0)\in M$ iff $l=0$, thus $\alpha^{-1}(M_1) = \{0\}$,\\
+  for $M_2,~~ (l,0)\in M$ iff $l=0$, thus $\alpha^{-1}(M_2) = \{0\}$,\\
+  thus $\alpha^{-1}(M_1)=\alpha^{-1}(M_2)$.
+
+
+  So we've seen that
+  \begin{align*}
+    \beta(M_1) = \beta(M_2)\\
+    \alpha^{-1}(M_1)=\alpha^{-1}(M_2)
+  \end{align*}
+  while having $M_1 \neq M_2$.
 \end{proof}
 
 \bibliographystyle{unsrt}

galois-theory-notes.tex

@@ -461,8 +461,12 @@
   \begin{enumerate}[i.]
     \item show that $\eta$ is a \emph{well defined} map:
 
-      if $g_1 K=g_2 K$, then for some $k \in K$, $g_1 k =g_2$, so
-      $$\eta(g_1K)=\psi(g_1) = \psi(g_1)\psi(k) = \psi(g_1 k) = \psi(g_2) = \eta(g_2 k)$$
+      if we have two representatives of the same coset, ie. $g_1 K=g_2 K$, we want to show that $\eta(g_1 K) = \eta(g_2 K)$, so that $\eta$ is a well-defined map.
+      
+      \vspace{0.3cm}
+      By the coset properties for some $k \in K$, $g_1=g_2 k$, so
+
+      $$\eta(g_1K)=\psi(g_1) = \psi(g_2 k) = \eta(g_2 k K) = \eta(g_2 K)$$
 
       Thus, $\eta$ does not depend on the choice of coset representatives, and
       the map $\eta: G/ker(\psi) \longrightarrow \psi(G)$ is uniquely defined