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+\documentclass[../notatki.tex]{subfiles}
+\begin{document}
+  This section will be devoted to the second part of the course, which was
+  preliminaries to module theory, A.K.A. commutative algebra.
+
+  \subsection{Definitions} 
+  Let $R$ be a ring with unity.
+  
+  \begin{definition}
+    A \emph{left module over $R$}, or a \emph{left $R$-module}, is a structure 
+    $M = (M, +, r)_{r\in R}$ satisfying the following axioms:
+    \begin{itemize}
+      \item $(M, +)$ is an abelian group and $0 = 0m$ (for $m\in M$, $0\in R$),
+      \item $r(m_1 + m_2) = rm_1 + rm_2$, for $r\in R, m_1, m_2\in M$,
+      \item $(r_1 + r_2)m = r_1m + r_2m$, for $r_1,r_2\in R, m\in M$,
+      \item $r_1(r_2m) = (r_1r_2)m$,
+      \item $1m = m$.
+    \end{itemize}
+    The definition of a submodule $N\subseteq M$ is natural, like for any other
+    substructure, i.e. $(N,+) < (M,+)$ and it's closed under ring action, i.e.
+    $rN\subseteq N$.
+  \end{definition}
+
+  Module's axioms are the same as the vector space axioms. Module is basically a
+  vector space over a ring rather than a field. There's also a variation of left
+  $R$-modules which is a \emph{right} $R$-modules, it's axioms are analogous,
+  but the order of the ring operations is ''reversed'', i.e. $m(r_1r_2) =
+  (mr_1)r_2$ which is different than $r_1(r_2m)$, unless the ring is
+  commutative.
+
+  \begin{example}
+    \mbox{}
+    \begin{enumerate}
+      \item When $R$ is a field, then $R$-modules are linear spaces over $R$.
+      \item If $(G,+)$ is an abelian group, then $G$ naturally induces a
+        $\bZ$-module structure, with $k\cdot g = \underbrace{g + \ldots +
+        g}_{k \text{ times}}$, for $k\in\bZ, g\in G$. For $k < 0$ take the sum
+        of $-g$.
+      \item Let $G$ be abelian group. Then $\End(G) = \{f\colon G\to G\}$ is the
+        ring of endomorphisms of $G$. Naturally, $G$ is an $\End(G)$-module,
+        for $f\in\End(G)$ and $g\in G$ put $f\cdot g = f(g)$.
+      \item Similarly to fields, if we have $S\subset R$, then $R$ is a
+        $S$-module.
+      \item If we have a ring homomorphism $j\colon R\to S$, then every
+        $S$-module is also an $R$-module with $r\cdot m = j(r)\cdot m$.
+      \item Let $I\subseteq R$ be a left ideal. Then $I$ is a $R$-module, as
+        $RI\subseteq I$.
+    \end{enumerate}
+  \end{example}
+
+  \begin{remark}
+    Every intersection of a nonempty family submodules of $M$ is a submodule of
+    $M$.
+  \end{remark}
+
+  \begin{theorem}[Fundamental $R$-modules homomorphism theorem]
+    Let $f\colon M\to N$ be a homomorphism of $R$-modules. Then:
+
+    \begin{center}
+      \begin{tikzcd}
+        M \arrow[d, "j"'] \arrow[r, "f"] & N \\
+        M/\ker{f} \arrow[ur, dotted, "\exists^^21 h"'] & 
+      \end{tikzcd}
+    \end{center}
+
+    Also, $h$ is an isomorphism of $M/\ker{f}$ and $h[M/\ker{f}]\subseteq N$. 
+  \end{theorem}
+  
+  \begin{theorem}[homomorphism factorisation]
+    Let $f\colon M\to N, g\colon M\to U$ homomorphisms of $R$-modules. Then
+
+    \begin{center}
+      \begin{tikzcd}
+        M \arrow[d, "g"'] \arrow[r, "f"] & N \\
+        U \arrow[ur, dotted, "\exists^^21 h"'] & 
+      \end{tikzcd}
+    \end{center}
+    if and only if $\ker{g}\subseteq \ker{f}$.
+  \end{theorem}
+
+  \begin{remark}
+    A homomorphism $f\colon M\to N$ is 1-1 $\Leftrightarrow$ $\ker f = \{0\}$.
+  \end{remark}
+
+  \begin{definition}
+    $M$ is a \emph{simple} $R$-module when $M\neq\{0\}$ and it has no proper
+    submodules, i.e. its only submodules are $\{0\}$ and $M$.
+  \end{definition}
+
+  For $R$-module $M$ by $\End_R(M)$ denote the set of endomorphisms of $M$.
+
+  \begin{lemma}[Schur's]
+    If $M$ is a simple $R$-module, then $\End_R(M)$ is a division ring.
+  \end{lemma}
+
+  \begin{proof}
+    Let $0\neq f\in\End_R(M)$. Then $f[M] = M$ and $\ker{f} = \{0\}$, hence $f$
+    is an automorphism and it's invertible.
+  \end{proof}
+
+  \begin{fact}
+    If $R$ is a commutative ring and $M$ is a simple $R$-module, then 
+    $\End_R(M)$ is a field.
+  \end{fact}
+
+  Moreover, a simple module $M$ is actually cyclic, as for any $0\neq m\in M$ $Rm = M$.
+  Hence, every endomorphism $f\in\End_R(M)$ is determined by $f(m) = rm$.
+  Thus, every endomorphism is of the shape $\phi_r\colon m\mapsto rm$ for some
+  $r\in R$. We get a natural ring epimorphism $\Phi\colon R\to\End_R(M)$ with
+  $r\mapsto\phi_r$, hence $R/\ker{\Phi}\cong \End_R(M)$, so
+  $\ker{\Phi}\triangleleft R$ is a
+  maximal ideal.
+  \subsection{Direct sum}
+  Let $M$ be an $R$-module, $N_1,N_2\subseteq M$ submodules. Then $N_1 + N_2 =
+  \{n_1 + n_2\mid n_1\in N_1, n_2\in N_2\}$ is a submodule of $M$. We can define
+  it similarly for $N_1 + N_2 + \ldots + N_k$.
+
+  \begin{definition}
+    Let $N_1,\ldots,N_k\subset M$ be submodules of $M$. We say that $M$ is a
+    \emph{direct sum} of $N_1,\ldots, N_k$ and write $M = N_1\oplus
+    \ldots\oplus N_k$, when every element of $m$ is uniquely generated in
+    $N_1+\ldots+N_k$, i.e. $\forall m\in M$ $\exists! n_1\in N_1,\ldots, n_k\in
+    N_k$ $m = n_1+\ldots+n_k$.
+  \end{definition}
+
+  The direct sum can be as well defined for a larger family of submodules $\cN =
+  \{N_i, i\in I\}$. In this case $M = \bigoplus_{i\in I} N_i$ when every element
+  can be uniquely described as a sum of finitely many elements of the
+  submodules in $\cN$. 
+
+  It turns out that direct sum can be also defined abstractly, 
+  as the coproduct in the category of $R$-modules.
+
+  \begin{definition}
+    Let $\{M_i, i\in I\}$ be a family of $R$-modules. Then the \emph{direct sum}
+    of this family is defined as
+    \[
+      \bigoplus_{i\in I}M_i = \bigsqcup_{i\in I}M_i = \{f\in\prod_{i\in
+      I}M_i\mid \{i\mid f(i) \neq 0\} \text{ is finite}\}
+    \]
+  \end{definition}
+
+  Associate all $M_i$ with the $i$-th axis of $\bigoplus_{i\in I}M_i$ and let
+  $f_i\colon M_i\to M$ be embedding to the $i$-th axis. The
+  direct sum has the following universal property:
+
+  \begin{remark}
+    Let $M$: $R$-module and $g_i\colon M_i\to M$ for every $i\in I$. Then
+    $\exists!h\colon\bigsqcup_{i\in I}M_i\to M$ such that:
+    \begin{center}
+      \begin{tikzcd}
+        M_i \arrow[d, "f_i"'] \arrow[r, "g_i"] & N \\
+        \displaystyle\bigsqcup_{i\in I}M_i \arrow[ur, dotted, "h"'] & 
+      \end{tikzcd}
+    \end{center}
+  \end{remark}
+
+  I abused the notation a little bit, as technically this is an alternative 
+  definition of the direct sum and it's not obvious that it is coherent with the
+  previous definition. However, it easily follows from the universal property of
+  the coproduct.
+
+  \subsection{Free modules}
+  \begin{definition}
+    Subset $\{m_i\}_{i\in I}\subseteq M$ is \emph{linearly independent}, when
+    the only (finite) solution to the equation $\bigcup_{i\in I}r_im_i = 0$ is $r_i = 0$
+    for all $i\in I$. If a subset is not linearly independent, the it is
+    \emph{linearly dependent}.
+  \end{definition}
+
+  \begin{definition}
+    A subset $B\subseteq M$ is a \emph{basis} of $M$ when $B$ is linearly
+    independent and generates $M$ as an $R$-module.
+  \end{definition}
+
+  \begin{definition}
+    A module $M$ is \emph{free} if it has a basis.
+  \end{definition}
+
+  For example, $\{0\}$ is linearly dependent, as $1\cdot 0 = 0$. Also $\{m_0,
+  m_0\}$ is linearly dependent. Look at $\bQ$ as a $\bZ$-module, any $a\neq
+  b\in\bQ$ are linearly dependent, hence it has no basis as a $\bZ$-module.
+  But, as $(\bZ_n,+)$ is cyclic, then it has a basis over $\bZ$. If $M_i,i\in
+  I$ are free, then so is the coproduct $\bigsqcup_i M_i$. 
+  
+  \begin{remark}
+    \label{remark:basis-alt}
+    Let $A = \{a_i, i\in I\}\subseteq M$ be a subset of module $M$. Then the
+    following conditions are equivalent:
+    \begin{enumerate}
+      \item $A$ is a basis of $M$,
+      \item $\forall m\in M$ $m = \sum_i r_ia_i$ is the unique representation,
+      \item $\underset{\text{$R$-module}}{\forall N}\forall g\colon A\to N$
+        $\exists!g'\colon M\to N$ $R$-linear s.t. $g\subseteq g'$.
+    \end{enumerate}
+  \end{remark}
+
+  Hence, free abelian groups are the same as free $\bZ$-modules.
+  The third item basically says that we can say everything about an embedding of
+  a free module just by understanding what happens on the basis.
+
+  \begin{remark}
+    Let $A=\{a_i,i\in I\}$ be a basis of $M$. Then $M = \bigoplus_i Ra_i$.
+  \end{remark}
+  \begin{remark}
+    Let $A=\{a_i,i\in I\}$ be set. Then $A$ is a basis of the module
+    $\bigsqcup_{a\in A} R_a$, where $R_a\cong R$.
+  \end{remark}
+
+  \begin{theorem}
+    When $R$ is a commutative ring, then all basis of an $R$-module $M$ are
+    equinumerous.
+  \end{theorem}
+
+  \begin{remark}
+    All $R$-modules are homomorphic image of some free $R$-module.
+  \end{remark}
+
+  \begin{proof}[Proof (sketch)]
+    Look at the free module $N=\bigsqcup_{m\in M} R_m$.
+  \end{proof}
+
+  \begin{fact}
+    \label{fact:free-modules-proj}
+    Let $M, N$ be $R$-modules, $N$ free and $f\colon M\to N$ be an epimorphism.
+    Then $M\cong \ker f\oplus N$, i.e. there is a submodule $N'\subseteq M$ such
+    that $N'\cong N$ and $M=\ker f\oplus N'$.
+  \end{fact}
+
+  \begin{proof}
+    Let $B\subseteq N$ be a basis. For every $b$ choose any $b'\in f^{-1}[b]$.
+    Let $g\colon B\to M$. By remark \ref{remark:basis-alt}(3) $g$ extends to a
+    homomorphism $g'\colon N\to M$. Moreover, $f\circ g'\colon N\to N$ and
+    $\restr{(f\circ g')}{B} = \id_B$, hence $f\circ g = \id_N$ by
+    \ref{remark:basis-alt}(2). Hence $g'$ is a monomorphism, $g'[N]\cong N$ and
+    $M = \ker f\oplus g'[N]$, because for any $m\in M$ we have that $m = (m -
+    (g'f)(m)) + (g'f)(m)$, but $f(m - (g'f)(m)) = f(m) - fg'f(m) = f(m) - f(m) =
+    0$, hence $m - (g'f)(m)\in\ker f$ and obviously $(g'f)(m)\in g'[N]$.
+    Moreover, $\ker f\cap g'[N] = \{0\}$, so the representation is unique.
+  \end{proof}
+
+  \subsection{Projectivity and injectivity}
+  Last fact from the previous subsection raises a natural generalization.
+
+  \begin{definition}
+    An epimorphism $f\colon M\to N$ \emph{splits} when $M = \ker f \oplus M'$
+    for some $M'\subset M$.
+  \end{definition}
+
+  \begin{remark}
+    An epimorphism $f\colon M\to N$ splits $\Leftrightarrow$ $\exists g\colon
+    N\to M$ $fg = \id_N$.
+  \end{remark}
+
+  \begin{proof}
+    $\Rightarrow$ It's clear that $M'\cong N$ with $\restr{f}{M'}$ being the
+    isomorphism. Hence, $g = \restr{f}{M'}^{-1}$ is such that $gf = \id_N$.
+    $\Leftarrow$ Check that $M = \ker f \oplus \im g$.
+  \end{proof}
+
+  \begin{remark}
+    Let $g\colon M\to N$ be a monomorphism. Then $g[M]$ is a direct summand of
+    $N$ $\Leftrightarrow$ $\exists f\colon N\to M$ $fg = \id_M$.
+  \end{remark}
+
+  \begin{definition}
+    An $R$-module $N$ is \emph{projective} when $\forall M$ $\forall f\colon
+    M\overset{\text{epi}}{\longrightarrow}N$ $f$ splits. 
+  \end{definition}
+
+  From the fact \ref{fact:free-modules-proj}, a free module is projective. There is also a dual
+  notion:
+
+  \begin{definition}
+    An $R$-module $M$ is \emph{injective}, when $\forall N$ $\forall g\colon
+    M\overset{\text{mono}}{\longrightarrow} N$ $N = \im(g)\oplus N'$ for some
+    $N'\subseteq N$.
+  \end{definition}
+
+  \subsubsection{Exact sequences}
+  In literature notion of injective and projective modules are often described
+  with
+  exact sequences. On this lecture we did not introduce it, however I think that
+  it gives a nice visual representation to think about projective and injective
+  modules, so I'll give these alternative definitions here.
+
+  \begin{definition}
+    A sequence of morphisms 
+    \[ G_0\overset{f_1}{\longrightarrow} G_1 \overset{f_2}{\longrightarrow}
+    \ldots \overset{f_n}{\longrightarrow} G_n \]
+    is \emph{exact} when $\im f_i = \ker{f_{i+1}}$ for every $1\le i < n$. In
+    other words, ''jumping'' by to morphisms always leads to $0$.
+  \end{definition}
+
+  \begin{definition}
+    An $R$-module $N$ is \emph{projective}, if every exact sequence of morphisms
+    \[ 0\longrightarrow A\longrightarrow M\longrightarrow N\longrightarrow 0\]
+    is a \emph{split exact} sequence, i.e. the middle module is isomorphic to
+    the direct sum of the other.
+  \end{definition}
+    
+  Try to check for yourself that this definition is equivalent to the previous
+  one, it's quite straightforward. Injectivity is very similar:
+
+  \begin{definition}
+    An $R$-module $M$ is \emph{injective}, if every exact sequence of morphisms
+    \[ 0\longrightarrow M\longrightarrow N\longrightarrow A\longrightarrow 0\]
+    is a split exact sequence, i.e. the middle module is isomorphic to
+    the direct sum of the other.
+  \end{definition}
+
+  \subsubsection{Properties}
+  There are nice properties of projective and injective modules, portrayed by
+  the following theorems.
+
+  \begin{theorem}
+    The following conditions are equivalent:
+    \begin{enumerate}
+      \item Module $P$ is projective,
+      \item For every morphism $f\colon M\to N$ of arbitrary modules $M$ and $N$
+        and every $g\colon P\to N$ there is $h\colon P\to M$ such that $f\circ h = g$
+      \item There is a module $L$ such that $P\oplus L$ is free.
+    \end{enumerate}
+  \end{theorem}
+
+  \begin{proof}
+    Try showing (2)$\Rightarrow$(1) and (1)$\Rightarrow$(3) on your own. 
+    We'll only show (3)$\Rightarrow$(2), as it is the most interesting one.
+
+    Assume there is $L$ such that $P\oplus L$ is free. Take any $M, N$,
+    epimorphism $g\colon M\to N$ and morphism $g\colon P\to N$. We have a
+    natural morphism $j\colon P\oplus L\to P$. Let's find $h\colon P\oplus L\to
+    M$ such that the following diagram commutes:
+    \begin{center}
+      \begin{tikzcd}
+        M \arrow[rrr, "f"] & & & N \\
+        & & P \arrow[ur, "g"] & \\
+        & P\oplus L \arrow[ur, "j"] \arrow[dotted, uul, "h"] & &
+      \end{tikzcd}
+    \end{center}
+
+    Let $B = \{p_i + l_i, i\in I\}$ be the basis of $P\oplus L$. Then let
+    $h'\colon B\to M$ be such that $h'(p_i + l_i) \in f^{-1}[g(p_i)]$. By remark
+    \ref{remark:basis-alt} this extends to $h\colon P\oplus L\to M$. It's easy
+    to check that $fh = gj$ and $\restr{h}{P}$ is what we're looking for.
+  \end{proof}
+
+  \begin{theorem}
+    A module $Q$ is injective $\Leftrightarrow$ for every monomorphism $f\colon
+    M\to N$ of arbitrary modules $M, N$ and a morphism $g\colon M\to Q$ there is
+    $h\colon N\to Q$ such that $hf = g$.
+  \end{theorem}
+
+  \begin{proof}
+    $\Leftarrow$ is easy. For $\Rightarrow$, consider $(Q\oplus N)/L$, where $L$
+    is a submodule generated by $\{(g(m), -f(m))\mid m\in M\}$. We have a
+    natural morphism $j\colon Q\to (Q\oplus N) /L$. Let $k\colon N\to (Q\oplus
+    N)$ be also natural, i.e. $k(n) = n + L$. Then we have $kf = jg$, as for any
+    $m\in M$ $ j(g(m)) - k(f(m)) = g(m) - f(m) + L = 0 + L$. But $j$ is a
+    monomorphism (why?) and $Q$ is injective, so there must be $j'\colon
+    (Q\oplus N)/L\to Q$ such that $jj' = \id_Q$. Then $h = j'k$ is good.
+
+    \begin{center}
+      \begin{tikzcd}
+        M \arrow[rrr, "f"] \arrow[dr, "g"] & & & N \arrow[ddl, dotted, "h'"] \\
+        & Q \arrow[dr, bend right, "j"'] & & \\
+        & & (Q\oplus N)/ L \arrow[ul, bend right, "j'"']  &
+      \end{tikzcd}
+    \end{center}
+  \end{proof}
+
+  \begin{theorem}[Baer's criterion]
+    $R$-module $Q$ is injective $\Leftrightarrow$ for every ideal $I\subset R$
+    and homomorphism $f\colon I\to Q$, $f$ extends to a homomorphism $R\to Q$.   
+  \end{theorem}
+
+  \begin{corollary}
+    If $R$ is PID, then $Q$ is injective $\Leftrightarrow$ $\forall r\in
+    R\setminus \{0\}$ $\forall m\in Q$ $\exists m'\in Q$ $rm' = m$ (i.e. every
+    element is divisible by every non-zero $r\in R$).
+  \end{corollary}
+
+  Hence, $\bQ$ is injective module. Moreover, every field is an injective
+  $\bZ$-module.
+
+  \begin{corollary}
+    Let $R$ be a Noetherian ring and $\{Q_j\}_\{j\in J\}$ be a family of
+    injective $R$-modules. Then the direct sum $\bigoplus_j Q_j$ is also
+    injective.
+  \end{corollary}
+\end{document}