Notes taken while studying Commutative Algebra, mostly from Atiyah \& MacDonald book \cite{am} and Reid's book \cite{reid}.
Notes taken while studying Commutative Algebra, mostly from Atiyah \& MacDonald book \cite{am} and Reid's book \cite{reid}. For the exercises, I follow the assignments listed at \cite{mit-course}.
Usually while reading books and papers I take handwritten notes in a notebook, this document contains some of them re-written to $LaTeX$.
Usually while reading books and papers I take handwritten notes in a notebook, this document contains some of them re-written to $LaTeX$.
if $m\in M_2$, then $\beta(m)\in\beta(M_1)=\beta(M_2)$, so that there is an $n \in M_1$ such that $\beta(m)=\beta(m)$.
if $m\in M_2$, then $\beta(m)\in\beta(M_1)=\beta(M_2)$, so that there is an $n \in M_1$ such that $\beta(m)=\beta(n)$.
Then $\beta(m-n)=0$, so that
Then $\beta(m-n)=0$, so that
$$m - n \in M_2\cap ker(\beta)=M_1\cap ker(\beta)$$
$$m - n \in M_2\cap ker(\beta)=M_1\cap ker(\beta)$$
@ -773,6 +886,114 @@ As in with rings, it is equivalent to say that
Hence $m \in M_1$, thus $M_1= M_2$.
Hence $m \in M_1$, thus $M_1= M_2$.
\end{proof}
\end{proof}
\begin{cor}{R.3.5}{Properties of Noetherian modules.}\label{R.3.5}
\begin{enumerate}[i.]
\item if $\forall i \in[r],~~M_i$ are Noetherian modules, then
$\bigoplus_{i=1}^r M_i$ is Noetherian.
\item if $A$ a Noetherian ring, then an $A$-module $M$ is Noetherian iff it is finite over $A$.
\item if $A$ a Noetherian ring, $M$ a finite module, then any submodule $N \subset M$ is again finite.
\item if $A$ a Noetherian ring, and $\psi: A \longrightarrow B$ a ring homomorphism such that $B$ is a finite $A$-module, then $B$ is a Noetherian ring.
\end{enumerate}
\end{cor}
\begin{proof}
\begin{enumerate}[i.]
\item a direct sum $M_1\oplus M_2$ is a particular case of an exact sequence.
Then, Proposition \ref{R.3.4.P} proves this statement when $r=2$. The case $r>2$ follows by induction.
\item if $M$ finite, then $\exists~$ surjective homomorphism
$$A^r \longrightarrow M \longrightarrow0$$
for some $r$, so that $M$ is a quotient
$$M \cong A^r / N$$
for some submodule $N \subset A^r$.
$A^r$ is a Noetherian module by i., so $M$ is Noetherian due Proposition \ref{R.3.4.P}.
Conversely, $M$ Noetherian implies $M$ finite.
item as in previous implications:\\
$M$ finite and $A$ Noetherian $\Longrightarrow$$M$ is Noetherian,\\
$\Longrightarrow$ since $N \subseteq M$, then $N$ is Noetherian too\\
$\Longrightarrow$ which implies that $N$ is a finite $A$-module.
\item$B$ is Noetherian as an $A$-module; but ideals of $B$ are submodules of $B$ as an $A$-submodule, so that $B$ is a Noetherian ring.
Claim: this finite set ($\{ f_{m,j}\}$) generates $I$.
$\forall f \in I$, if $\deg f =m$, then its leading coefficient is $a \in J_m$,
hence if $m \geq n$, then $a \in J_m=J_n$, so that
$$a =\sum b_i a_{n,i} ~~ \text{with}~ b_i \in A$$
and
$$f -\sum b_i X^{m-n}\cdot f_{n, i}$$
has degree $<m$.
Similarly, if $m \leq n$, then $a \in J_m$, so that
$$a =\sum b_i a_{m, i} ~~\text{with}~ b_i \in A$$
and
$$f -\sum b_i f_{n, i}$$
has degree $<m$.
\vspace{0.3cm}
By induction on $m$, $f$ can be written as a linear combination of finitely many elements.
Thus, any ideal of $A[x]$ is finitely generated.
\end{proof}
\begin{cor}{R.3.6.C}
if $A$ a Noetherian ring, and $\psi: A \longrightarrow B$ a ring homomorphism such that $B$ is a fingen extension ring of $\psi(A)$, then $B$ is Noetherian.
In particular, any fingen algebra over $\mathbb{Z}$ or over a field $K$ is Noetherian.
\end{cor}
\begin{proof}
the assumption is that $B$ is a quotient of a polynomial ring,
$$B \cong A[x_1, \ldots, x_n]/ I$$
for some ideal $I$.
By the Hilbert basis theorem \ref{hilbert-basis} and induction,\\
$A$ being Noetherian implies that $A[x_1, \ldots, x_n]$ is Noetherian.
And by Corollary \ref{R.3.5}(iv),\\
$A[x_1, \ldots, x_n]$ being Noetherian implies that $A[x_1, \ldots, x_n]/I$ is Noetherian.
\end{proof}
\newpage
\newpage
\section{Exercises}
\section{Exercises}
@ -813,7 +1034,7 @@ The exercises that start with \textbf{R} are the ones from the book \cite{reid},
Let
Let
$$\psi^{-1}(P)=\{ a \in A | \psi(a)\in P \}= A \cap P$$
$$\psi^{-1}(P)=\{ a \in A | \psi(a)\in P \}= A \cap P$$
The claim is that $\psi^{-1}(P)$ is prime iddeal of $A$.
The claim is that $\psi^{-1}(P)$ is prime ideal of $A$.
\begin{enumerate}[i.]
\begin{enumerate}[i.]
\item show that $\psi^{-1}(P)$ is an ideal of $A$:\\
\item show that $\psi^{-1}(P)$ is an ideal of $A$:\\
@ -1001,6 +1222,195 @@ The exercises that start with \textbf{R} are the ones from the book \cite{reid},
\subsection{Exercises Chapter 2}
\subsection{Exercises Chapter 2}
\begin{ex}{R.2.9}
$0\longrightarrow L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N \longrightarrow0$ is a s.e.s. of $A$-modules. Prove that if $N, L$ are finite over $A$, then $M$ is finite over $A$.
\end{ex}
\begin{proof}
Denote the generators of $L$ and $N$ respectively as
\begin{align*}
\{l_1, \ldots, l_k \}&\subseteq L\\
\{n_1, \ldots, n_p \}&\subseteq N
\end{align*}
By s.e.s. definition,
\begin{itemize}
\item[-]$\alpha$ is injective (one-to-one), so
$$\forall l_i \in L,~ \exists~ x_i \in M ~\text{s.th.}~ \alpha(l_i)=x_i$$
\item[-]$\beta$ is surjective (onto), so
$$\forall n_j \in N,~ \exists~ y_j \in M ~\text{s.th.}~ \beta(y_j)=n_j$$
\end{itemize}
We will show that $\{x_1, \ldots, x_k, y_1, \ldots, y_p \}$ generate $M$, and thus $M$ is finite:
So, $L$ provides $k$ generators for the kernel part of $M$, $N$ provides $p$ "lifts" for the quotient part of $M$; thus $M$ is generated by $k+p$ elements.\\
Thus $M$ is finitely generated over $A$.
\end{proof}
\subsection{Exercises Chapter 3}
\begin{ex}{R.3.2}
$K$ a field, $A \supset K$ a ring which is finite dimensional as a $K$-vector space.
Prove that $A$ is Noetherian and Artinian.
\end{ex}
\begin{proof}
$dim(A)=n < \infty$, so every ideal $\aA$ of $A$ is a $K$-subspace of $A$, because if $x \in\aA$ and $c \in K$, then $c \cdot x \in\aA$.
\begin{enumerate}
\item Noetherian:\\
let $I_1\subseteq I_2\subseteq\ldots$ be an ascending chain of ideals in $A$.
Since each $I_i$ is a subspace, we have
$$dim_K(I_1)\leq dim_K(I_2)\leq\ldots\leq n$$
where at some $i=m$ we have $dim_K(I_m)=dim_K(I_{m+1})$; then since $I_m \subseteq I_{m+1}$, we have $I_m = I_{m+1}$. So $A$ is Noetherian.
\item Artinian:\\
Similarly, if $I_1\supseteq I_2\supseteq\ldots$ a descending chain of ideals in $A$.
where at some $i=m$ we have $dim_K(I_m)=dim_K(I_{m+1})$; then since $I_m \subseteq I_{m+1}$, we have $I_m = I_{m+1}$. So $A$ is Artinian.
\end{enumerate}
\end{proof}
\begin{ex}{R.3.5}
Let $0\longrightarrow L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N \longrightarrow0$ an exact sequence. Let $M_1, M_2\subseteq M$ be submodules of $M$.
M_2 &= \{ (0, x) ~|~ x \in K\} ~~~~\sim\text{(y-axis)}
\end{align*}
(Geometric interpretation: $M_1,~ M_2$ are the \emph{diagonal line} and \emph{y-axis} respectively; and $\alpha,~\beta$ capture information about the \emph{vertical} components (x-axis, y-axis respectively), but not about the \emph{diagonal} way a submodule is embedded in $M$).
Then,
\begin{align*}
\beta(M_1) &= \{ x ~|~ x \in K\} = K \\
\beta(M_2) &= \{ x ~|~ x \in K\} = K
\end{align*}
thus, $\beta(M_1)=\beta(M_2)$.
For $M_1,~~ (l,0)\in M$ iff $l=0$, thus $\alpha^{-1}(M_1)=\{0\}$,\\
for $M_2,~~ (l,0)\in M$ iff $l=0$, thus $\alpha^{-1}(M_2)=\{0\}$,\\
thus $\alpha^{-1}(M_1)=\alpha^{-1}(M_2)$.
So we've seen that
\begin{align*}
\beta(M_1) = \beta(M_2)\\
\alpha^{-1}(M_1)=\alpha^{-1}(M_2)
\end{align*}
while having $M_1\neq M_2$.
\end{proof}
\begin{ex}{R.3.3}
Let $A$ a ring, $I_1, \ldots, I_k$ ideals such that each $A/I_i$ is a Noetherian ring.
Prove that $\bigoplus A/I_i$ is a Noetherian $A$-module, and deduce that if $\bigcap I_i =0$ then $A$ is also Noetherian.
\end{ex}
\begin{proof}
\begin{enumerate}[i.]
\item by Corollary \ref{R.3.5} (i), if $M_i$ Noetherian modules, then $\bigoplus M_i$ is Noetherian.
$\Longrightarrow$ thus $\bigoplus A/I_i$ is Noetherian.
\item Take the canoncial homomorphism
$$\phi: A \longrightarrow\bigoplus_{i=1}^n A/ I_i$$
by $\phi(a)=(a+I_1, a+I_2, \ldots, a+I_n)$.
$\phi$ is injective: $ker(\phi)=\{ a \in A | a \in I_i \forall i \}$.
Since we're given $\cap I_i =0$, then $ker(\phi)=\cap I_i$, and $\phi$ is injective.
Thus, $\phi$ is the isomorphism $A \cong im(\phi)$, where $im(\phi)$ is an $A$-submodule of $\bigoplus A/I_i$.
We know that any submodule of a Noetherian module is Noetherian, thus, since
\begin{itemize}
\item$A/I_i$ is Noetherian by hypothesis of the exercise
\item$A \cong im(\phi)$
\item$im(\phi)$ is an $A$-submodule of $\bigoplus A/I_i$
\end{itemize}
then, $A$ is Noetherian.
\end{enumerate}
\end{proof}
\begin{ex}{R.3.4}
Prove that if A is a Noetherian ring and M a finite A-module, then there
exists an exact sequence $A^q \stackrel{\alpha}{\longrightarrow} A^p \stackrel{\beta}{\longrightarrow} M \longrightarrow0$.
That is, M has a presentation as an A-module in terms of finitely many generators and relations.
if we have two representatives of the same coset, ie. $g_1 K=g_2 K$, we want to show that $\eta(g_1 K)=\eta(g_2 K)$, so that $\eta$ is a well-defined map.
\vspace{0.3cm}
By the coset properties for some $k \in K$, $g_1=g_2 k$, so
$$\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
Thus, $\eta$ does not depend on the choice of coset representatives, and
the map $\eta: G/ker(\psi)\longrightarrow\psi(G)$ is uniquely defined
the map $\eta: G/ker(\psi)\longrightarrow\psi(G)$ is uniquely defined
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