Instructor: Prof. A. V. Jayanthan
Book: Atiyah-Macdonald
ROUGH NOTES
Lec-1: Rings, subrings, homomorphisms, ideals
Lec-2: Prime and maximal ideals
Lec-3: Every nonzero ring has a maximal ideal. (Apply this to $A/I$ where ideal $I \subsetneq A.$ From correspondence thm, there is a maximal ideal containing $I$. Especially every non-unit has a maximal ideal containing it).
Let $I \subsetneq A$ be an ideal. Then ($A$ is local with max’l ideal $I$) $\iff$ ($A \setminus I$ consists entirely of units).
$\impliedby$ Every proper ideal avoids containing units, so lies $\subseteq A \setminus (A \setminus I) = I.$ So $A$ is local with maximal ideal $I.$
$\implies$ Suppose there is a nonunit in $A\setminus I.$ Because every nonunit has a max’l ideal containing it, this gives a max’l ideal different from $I.$ Absurd.
$\mathbb{Z}/{p ^n \mathbb{Z}}$ is local with maximal ideal $(\overline{p})$ : Its ideals are ${(\overline{0}) = (\overline{p ^n}) \subseteq (\overline{p ^{n-1}}) \subseteq \ldots \subseteq (\overline{p}) \subseteq (\overline{1})}.$
Let $\mathfrak{m} \subsetneq A$ be a maximal ideal with $1 + \mathfrak{m} \subseteq A ^{\times}$ (We write $A ^{\times}$ for the group of units of $A$). Then $A$ is local with maximal ideal $\mathfrak{m}$ : Suffices to show $A \setminus \mathfrak{m}$ comprises of units. Let $x \in A \setminus \mathfrak{m}.$ Now $\mathfrak{m} + (x) = (1),$ so $m + rx = 1.$ So $rx = 1 - m \in 1 + \mathfrak{m} \subseteq A ^{\times}$ is a unit, making $x$ a unit.
Nilpotents of a ring $A$ form an ideal, called nilradical $\mathfrak{N}$. Turns out ${\mathfrak{N} = \cap \lbrace \text{prime ideals of } A \rbrace.}$
$\subseteq$ : Let $x \in \mathfrak{N},$ and $\mathfrak{p}$ a prime ideal. Now $x ^n = 0 \in \mathfrak{p}$ so $x \in \mathfrak{p}.$
$\supseteq$ : Uses Zorn’s lemma
Lec-4: Q) Think about the zero divisors of $\mathbb{Z} _{n} [x].$
In a ring $A,$ the ideal ${\mathfrak{J} := \cap \lbrace \text{max’l ideals of } A \rbrace}$ is called its Jacobson radical. Notice ${\mathfrak{J} = \lbrace x : 1-Ax \subseteq A ^{\times} \rbrace.}$
$\subseteq$ : Let $x \in \mathfrak{J}.$ Suppose $y \in A$ is such that $1-xy \not\in A ^{\times}.$ Now there is a maximal ideal containing $1 - xy,$ so this maximal ideal contains $(1 - xy) +x(y) = 1$ too, absurd. $\supseteq$ : Let $1 - Ax \subseteq A ^{\times},$ and $\mathfrak{m}$ a max’l ideal. Suppose $x \not\in \mathfrak{m}.$ Now $\mathfrak{m} + (x) = (1),$ so $m + rx = 1.$ So $1 - rx = m \not\in A ^{\times},$ absurd.
Given ideals $I, J \subseteq A,$ we have ideals $I + J = \lbrace i + j : i \in I, j \in J\rbrace$ (generated by union), $I \cap J,$ and $IJ = \lbrace \sum _{\text{finite}} i j : i \in I, j \in J \rbrace$ (generated by set product ${\lbrace ij : i \in I, j \in J \rbrace}$).
In $\mathbb{Z},$ we have $(I+J)(I \cap J) = IJ.$
But this neednt hold everytime : $(I +J)(I \cap J) \subseteq IJ$ is clear, but the inclusion can be proper (eg $(x), (y) \subseteq K[x,y],$ or even $(2), (x) \subseteq \mathbb{Z}[x]$).
$IJ \subseteq I \cap J$ always. If $I + J = (1),$ we get ${I \cap J = (I + J)(I \cap J) \subseteq IJ}$ too, giving ${I\cap J = IJ.}$
Lec-5: [CRT] Let ideals ${I _1, \ldots, I _n\subseteq A }.$ The map ${A \overset{f}{\longrightarrow} A /{I _1} \times \ldots \times A /{I _n}}$ sending ${a \mapsto (a \text{ mod } I _1, \ldots, a \text{ mod } I _n) = ([a] _1, \ldots, [a] _n)}$ is a ring homomorphism with kernel ${\ker(f) = I _1 \cap \ldots \cap I _n}.$ Notice $f$ is surjective if and only if $I _j$s are pairwise coprime ie $I _j + I _k = (1)$ for all $j \neq k.$
$\impliedby$ $f ^{-1} ([a _1], \ldots, [a _n])$ contains ${a _1 f ^{-1} ([1], [0], \ldots, [0]) + \ldots + a _n f ^{-1} ([0], \ldots, [0], [1])},$ and each term here is non-empty. For eg we have relations ${i _1 ^{(1)} + i _2 = 1, \ldots, i _1 ^{(n-1)} + i _{n} = 1}$ where ${i _1 ^{(j)} \in I _1}$ and ${i _2 \in I _2, \ldots, i _{n} \in I _n.}$ So ${i _2 , \ldots, i _n}$ are all $1$ mod $I _1,$ giving an element $i _2 \ldots i _{n}$ in $f ^{-1} ([1], [0], \ldots, [0]).$
$\implies$ For eg there is an $a$ in ${f ^{-1} ([1], [0], \ldots, [0]).}$ So $a + i _1 = 1$ and ${a \in I _2, \ldots, I _n,}$ where $i _1 \in I _1.$ So $I _1 + I _j = (1)$ for all $j \neq 1.$
[Prime avoidance] Consider prime ideals ${ \mathfrak{p _1}, \ldots, \mathfrak{p _n} \subseteq A}$ and an ideal $\mathfrak{a}$ contained in $\cup _{1} ^{n} \mathfrak{p _j}.$ Then $\mathfrak{a}$ is contained in some $\mathfrak{p _j}.$
Suffices to show $\mathfrak{a} \subseteq \cup _{j \neq k} \mathfrak{p _j}$ for some $k$ (then done by induction).
Suppose it fails. So there are elems ${a _k \in \mathfrak{a} \setminus (\cup _{j \neq k} \mathfrak{p _j})}$ for every $k.$ So ${a _k \in \mathfrak{p} _k \setminus (\cup _{j \neq k} \mathfrak{p _j}).}$ Now look at ${\sum _{k} (\Pi _{j \neq k} a _j)}$ : in this, for every $k,$ all but one term lie inside $\mathfrak{p} _k$ and the other term lies outside $\mathfrak{p} _k.$ So the sum doesnt lie in any of the $\mathfrak{p} _k$s, contradicting ${\mathfrak{a} \subseteq \cup _{1} ^{n} \mathfrak{p _j}}.$
Let prime ideal $\mathfrak{p}$ contain intersection $\cap _{1} ^{n} \mathfrak{a _j}$ of ideals. Then $\mathfrak{p}$ contains some $\mathfrak{a _j}$ : Suppose not. So there are elems $a _j \in \mathfrak{a _j} \setminus \mathfrak{p}.$ Now $\Pi a _j$ is in $\cap \mathfrak{a _j}$ but outside $\mathfrak{p},$ absurd.
Lec-6: [Colon] For an ideal $I$ and subset $S$ of ring $A,$ the colon ${I : S \overset{\text{def}}{=} \lbrace x \in A : x S \subseteq I\rbrace}.$ This is an ideal containing $I.$ If $S \subseteq I,$ we have $I : S = (1).$ Also if $S _1 \subseteq S _2,$ then $I : S _1 \supseteq I : S _2.$ Especially $I : S \supseteq I : \langle S \rangle,$ where $\langle S \rangle$ is the ideal generated by $S.$ Infact $I : S = I : \langle S \rangle$ (direct).
Eg : In $K[x],$ $(x ^2) : \lbrace x \rbrace = (x).$ In $\mathbb{Z},$ ${(8):(6) = (4)}.$ In general for $m, n \gt 0,$ ${(m) : (n) = \frac{m}{\text{gcd}(m,n)} \mathbb{Z}.}$
For $x \in A,$ $(0) : \lbrace x \rbrace$ ($= 0 : x,$ for brevity) is called the annihilator of $x.$ Eg in $\mathbb{Z} _6,$ ${\overline{0} : \overline{3} = \lbrace \overline{0}, \overline{2}, \overline{4} \rbrace = (\overline{2})}.$
For ideals $I, J _1, J _2$ we have ${ (I : J _1) : J _2 = I : J _1 J _2 }.$ Also ${(\cap I _j) : S = \cap (I _j : S)}$ for any family $\lbrace I _j \rbrace _{j \in J}$ of ideals.
[Radical] For an ideal $I \subseteq A,$ ${r(I) := \lbrace x \in A : x ^n \in I \text{ for some } n \gt 0 \rbrace}$ is called radical of $I.$ This is an ideal containing $I.$
Rewriting, ${ r(I) = \pi ^{-1} (\mathfrak{N} _{A / I}),}$ where $\pi$ is the natural projection $A \overset{\pi}{\twoheadrightarrow} A/I.$ So ${ r(I) = \cap \lbrace \text{prime ideals containing }I \rbrace }.$
Eg: In $\mathbb{Z},$ $r(p _1 ^{\alpha _1} \ldots p _k ^{\alpha _k} \mathbb{Z}) = p _1 \ldots p _k \mathbb{Z},$ where $p _j$s are primes and $\alpha _j$s are $\gt 0.$
For any ideals $I, J \subseteq A,$ ${r(I J) = r(I \cap J) = r(I) \cap r(J)}$ [Direct]. Especially ${r(\mathfrak{p} ^n) = \mathfrak{p} }$ for any prime ideal $\mathfrak{p}$ and int $n \gt 0.$ We also have $r(I + J) = r(r(I) + r(J)).$
${r(I+J) \supseteq r(I) + r(J),}$ so ${r(I+J) \supseteq r(r(I) + r(J)).}$ And ${ I +J \subseteq r(I) + r(J) }$ so ${ r(I+J) \subseteq r(r(I) + r(J))}$ too.
Lec-7: [Extensions, Contractions] Let $A \overset{f}{\to} B$ be a ring homomorphism. For ideal $I \subseteq A,$ $f(I)$ is an ideal of subring $f(A).$ The ideal of $B$ generated by $f(I)$ is called extension of $I,$ denoted $I ^{e}.$
Explicitly, ${ I ^e = \lbrace \sum _{\text{finite}} b f(i) : b \in B, i \in I \rbrace }.$
For any ideal $J \subseteq B,$ $f ^{-1} (J)$ is an ideal of $A$, called contraction of $J$ and denoted $J ^{c}.$
Here $I ^{ec} \supseteq I$ (because for every $i \in I,$ $f(i) \in I ^e$), and $J ^{ce} \subseteq J$ (because $J ^{ce}$ is ideal of $B$ generated by $J \cap f(A)$).
Also $J ^{cec} = J ^c$ (because $(J ^c) ^{ec} \supseteq J ^c,$ and $(J ^{ce}) ^{c} \subseteq J ^{c}$), and similarly $I ^{ece} = I ^e.$
Let $A \overset{f}{\to} B$ be a ring homomorphism. Let $\mathscr{C}$ be the set of contracted ideals in $A$ and $\mathscr{E}$ the set of extended ideals in $B.$
Then i) ${ \mathscr{C} = \lbrace I : I ^{ec} = I \rbrace }$ and ${ \mathscr{E} = \lbrace J : J ^{ce} = J \rbrace }.$ Also ii) $I \mapsto I ^{e}$ is a bijection ${\mathscr{C} \to \mathscr{E}},$ with inverse $J \mapsto J ^{c}.$
Pf: i) ${\mathscr{C} \supseteq \lbrace I : I ^{ec} = I \rbrace }$ is clear. For $\subseteq,$ let $I \in \mathscr{C}.$ So $I = J ^{c}$ for some $J.$ Now ${I ^{ec} = J ^{cec} = J ^c = I},$ ie $I ^{ec} = I$ as needed. Similarly for $\mathscr{E}.$
ii) Clear from above characterisations of $\mathscr{C}, \mathscr{E}.$
Lec-8: Module (informally, “vector space over a ring”). Eg : Vector spaces are modules; For an ideal $I \subseteq A,$ $I$ is an $A-$module. Also $A/I$ is an $A-$module under ${a \bullet [x] := [a] [x];}$ Let $V \overset{T}{\to} V$ be a linear operator on $K-$vector space $V.$ Then $V$ is a $K[x]-$module under ${p(x) \bullet v := p(T) (v);}$ Any abelian group $(G, +)$ is a $\mathbb{Z}-$module
Submodules; Module Homomorphisms. Eg: Any $\mathbb{Z} \to \mathbb{Z},$ $t \mapsto kt$ is a homomorphism of $\mathbb{Z}-$modules; Any linear map, for eg $\mathscr{C} ^1 ([a,b]) \to \mathscr{C} ([a,b])$ sending $f \mapsto f’,$ is a homomorphism of modules.
For $A-$modules $M,N,$ the set of $A-$module homomorphisms from $M$ to $N$ is written $\text{Hom} _{A} (M,N).$ This is an $A-$module too.
Eg: ${\text{Hom} _{\mathbb{Z}} (\mathbb{Z}, \mathbb{Z}) \simeq \mathbb{Z} }$ under $f \mapsto f(1).$ In gen for an $A-$module $M,$ ${\text{Hom} _{A} (A,M) \simeq M}$ under $f \mapsto f(1).$
Let $M \overset{f}{\to} M’$ be a homom (implicitly, of $A-$modules) and $N,N’$ modules. This induces a homom ${\text{Hom}(M’, N’) \to \text{Hom}(M, N’)},$ $u \mapsto u \circ f,$ and a homom ${ \text{Hom}(N,M) \to \text{Hom}(N, M’)},$ $v \mapsto f \circ v.$
Quotient (wrt a submodule); Let $M \overset{f}{\to} N$ be a module homom. Then $f(M),$ $\ker(f)$ are submodules, and $M/{\ker(f)} \simeq f(M)$ under $t + \ker(f) \mapsto f(t).$ We call $N / {f(M)}$ the cokernel $\text{coker}(f).$ So $f$ is injective if and only if its kernel is $\lbrace 0\rbrace,$ and surjective if and only if its cokernel is $\lbrace 0 \rbrace.$
Eg: For integer $k \gt 0,$ $\mathbb{Z} \to k\mathbb{Z}$ with $t \mapsto kt$ is an isomorphism of $\mathbb{Z}-$modules.
Lec-9: For submodules $\lbrace M _i \rbrace _{i \in I}$ of $M,$ we have ${\sum _{i \in I} M _i := \lbrace \sum _{i \in I} x _i \text{ } \vert \text{ } x _i \in M _i \text{ with } x _i = 0 \text{ for all but finitely many }i \rbrace},$ the submodule generated by $\lbrace M _i \rbrace _{i \in I}.$ Intersection $\cap _{i \in I} M _i$ is also a submodule.
For an $A-$module $M$ and $S \subseteq A,$ ${SM := \lbrace \sum _{\text{finite}} sm : s \in S, m \in M \rbrace }$ is a submodule. Notice $SM = \langle S \rangle M,$ where $\langle S \rangle$ is the ideal generated by $S.$
Given submodules $N, P \subseteq M,$ we have ideal ${ (N : P) \overset{\text{def}}{=} \lbrace a \in A \text{ } \vert \text{ } aP \subseteq N \rbrace }.$ Especially $(0 : M)$ is called annihilator ideal of $M,$ written $\text{Ann}(M).$ We call $M$ faithful if $\text{Ann}(M) = (0).$
Consider $A-$module $M$ and ideal $I \subseteq A.$ When is $M$ an $A/I-$module under $[a] \bullet m := am$ ?
For $\bullet$ to be well-defined, we must have ${``[a _1] = [a _2] \implies a _1 m = a _2 m \text{ for all } m”}$ ie ${I \subseteq \text{Ann} _{A} (M).}$ This constraint suffices.
Here notice ${\text{Ann} _{A/I} (M)}$ ${= \lbrace [a] : [a] \bullet m = 0 \text{ for all } m \rbrace }$ ${ = \text{Ann} _{A} (M) / I }.$ So $M$ is always faithful as an $A/{A\text{nn}(M)} -$module.
Lec-10: Consider an $A -$module $M.$ If $M = \sum _{i \in I} A x _i,$ we say $\lbrace x _i \rbrace _{i \in I}$ generate $M.$ A module having a finite set of generators is called finitely generated.
Eg: In $\mathbb{Z}$ (as a $\mathbb{Z}-$module), both $\lbrace 1 \rbrace$ and $\lbrace 2,3 \rbrace$ are minimal generating sets. (Minimal in the sense no proper subset can generate the module. Such a thing can’t happen in vector spaces).
Consider $A-$modules $\lbrace M _i \rbrace _{i \in I}.$ Their direct sum ${ \oplus _{i \in I} M _i = \lbrace (x _i) _{i \in I} : \text{each } x _i \in M _i, \text{ and } x _i = 0 \text{ for all but finitely many } i\rbrace },$ and direct product ${ \Pi _{i \in I} M _i = \lbrace (x _i) _{i \in I} : \text{each } x _i \in M _i \rbrace }$ are also $A-$modules (under usual componentwise operations). When $I$ is finite these are the same objects (When $I = \emptyset$ its conventionally the zero module). An $A-$module $M$ is free if its of the form $M \simeq \oplus _{i \in I} A.$
An $A-$module $M$ is finitely generated (f.g.) if and only if it is a quotient of $A ^n$ for some $n \gt 0.$
$\implies$ Say $M = \sum _{1} ^{n} Ae _i.$ Now $A ^n \overset{f}{\twoheadrightarrow} M$ sending $(a _1, \ldots, a _n) \mapsto a _1 e _1 + \ldots + a _n e _n$ gives ${M \simeq A ^n /{\ker(f)}.}$
$\impliedby$ The quotient is generated by ${[(1, 0, \ldots, 0)], \ldots, [(0,0,\ldots, 1)].}$
EDIT (30/10/21)
Lec-11: Let $M$ be an f.g. $A-$module, and $\mathfrak{a} \subseteq A$ an ideal. Let $\varphi \in \text{End} _{A} (M)$ with $\varphi(M) \subseteq \mathfrak{a} M.$ Then ${\varphi ^n + a _{n-1} \varphi ^{n-1} + \ldots + a _1 \varphi + a _0 = 0}$ for some ${ a _0, \ldots, a _{n-1} \in \mathfrak{a} }.$
Pf: Let $M = \sum _{1} ^{n} Ae _j.$ We have ${ (\varphi(e _1), \ldots, \varphi(e _n)) = (e _1, \ldots, e _n) \begin{pmatrix} a _{11} &\ldots &a _{1n} \\\ \vdots &\ddots &\vdots \\\ a _{n1} &\ldots &a _{nn} \end{pmatrix} }$ for some $a _{ij}$s in $\mathfrak{a}.$
So, in terms of matrices over $\text{End} _{A} (M),$ we have ${ \begin{pmatrix} \varphi - a _{11} &-a _{21} &\ldots &-a _{n1} \\\ -a _{12} &\varphi - a _{22} &\ldots &-a _{n2} \\\ \vdots &\vdots &\ddots &\vdots \\\ -a _{1n} &-a _{2n} &\ldots &\varphi - a _{nn} \end{pmatrix} \begin{pmatrix} e _1 \\\ e _2 \\\ \vdots \\\ e _n \end{pmatrix} = \begin{pmatrix} 0 \\\ 0 \\\ \vdots \\\ 0 \end{pmatrix} .}$ So the matrix (in $M _n (R)$ where $R = \text{End} _{A} (M)$) has determinant $0,$ giving a relation of required form.
We can think of this as left multiplying with $\text{Adj}(P)$ and using ${ P \text{Adj}(P) = \text{Adj}(P) P = \det(P) I }.$ The latter identity holds for matrices over arbitrary rings.
Cor: Let $M$ be an f.g. $A-$module, and $\mathfrak{a} \subseteq A$ an ideal with $\mathfrak{a}M = M.$ Then there exists ${x \equiv 1 (\text{mod } \mathfrak{a})}$ with $xM = 0.$ (Take $\varphi = \text{id}$ above to get $a _0, \ldots, a _{n-1}.$ Then look at ${1 + a _{n-1} + \ldots + a _{0}}$).
[Nakayama] Let $M$ be an f.g. $A-$module, and $\mathfrak{a} \subseteq A$ an ideal contained in Jacobson radical $\mathfrak{J}.$ If $\mathfrak{a} M = M,$ then $M = 0.$
Pf: The $x$ in prev cor is a unit (Else there is maximal ideal containing it. This also contains elems of $\mathfrak{a},$ so contains $1$ too, absurd). So $xM = 0$ gives $M = 0.$
Containment in $\mathfrak{J}$ is crucial. Eg: $\mathbb{Z}-$module $\mathbb{Z} / {5\mathbb{Z}}$ and ideal $3\mathbb{Z}.$
Lec-12: Let $M$ be an f.g. $A-$module with submodule $N,$ and $\mathfrak{a} \subseteq \mathfrak{J}$ an ideal. If $M = \mathfrak{a}M + N,$ then $M = N.$
Pf: Nakayama on $M/N$ will do : Taking ${M \overset{\pi}{\twoheadrightarrow} M /N},$ we get $\pi(M) = \pi(\mathfrak{a}M +N)$ ie $M/N = \mathfrak{a} (M/N).$
Let $(A, \mathfrak{m})$ be a local ring with max’l ideal $\mathfrak{m},$ and $M$ an f.g. module over it.
Since $M/{\mathfrak{m}M}$ is annihilated by $\mathfrak{m},$ this $M/{\mathfrak{m}M}$ is an $(A/{\mathfrak{m}})-$module, ie an $(A/{\mathfrak{m}})-$vector space.
This vect sp is f.g., so has a basis. Pick $e _1, \ldots, e _n \in M$ such that $\lbrace \pi (e _1), \ldots, \pi (e _n) \rbrace$ is a basis (where ${M \overset{\pi}{\twoheadrightarrow} M/{\mathfrak{m} M}}$). Now $\lbrace e _1, \ldots, e _n \rbrace$ generate $M.$
Let $N$ be the submodule $\sum _{1} ^{n} A e _j$ generated by $e _j$s. Now ${N \hookrightarrow M \overset{\pi}{\twoheadrightarrow} M / {\mathfrak{m} M} }$ has image ${ \pi (N) = \pi(\sum _{1} ^{n} A e _j) = \sum _{1} ^{n} (A/{\mathfrak{m}}) \pi (e _j) = M/{\mathfrak{m} M}.}$ So $\pi (N) = \pi(M),$ making $N + \mathfrak{m} M = M.$ Finally Nakayama gives $M = N.$
Lec-13: [Exact Sequences] A sequence of $A-$modules and $A-$homomorphisms ${ \text{ } \cdots \longrightarrow M _{i-1} \overset{f _i}{\longrightarrow} M _i \overset{f _{i+1}}{\longrightarrow} M _{i+1} \longrightarrow \cdots \text{ } }$ is called exact at $M _i$ if ${\text{im}(f _i) = \ker(f _{i+1}).}$ The sequence is called exact if it is exact at each module.
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| An exact sequence (Pic from Wikipedia) |
An exact sequence of the form ${ 0 \to M’ \overset{f}{\to} M \overset{g}{\to} M’’ \to 0 }$ is called a short exact sequence. In other words $f$ is injective, $g$ surjective, and ${ \text{im}(f) = \ker(g) }.$ That is, $f$ is injective, $g$ surjective, and $M/{f(M’)} (=\text{coker}(f)) \simeq M’’$ via $g.$
An exact sequence ${ \text{ } \cdots \longrightarrow M _{i-1} \overset{f _i}{\longrightarrow} M _i \overset{f _{i+1}}{\longrightarrow} M _{i+1} \longrightarrow \cdots \text{ } }$ splits up into short exact sequences ${ 0 \to \text{im}(f _i) \hookrightarrow M _i \twoheadrightarrow \text{im}(f _{i+1}) \to 0 }.$
Th (Throughout, we consider $A-$modules and $A-$homomorphisms)
(a) ${ M’ \overset{u}{\to} M \overset{v}{\to} M’’ \to 0 }$ a sequence of homomorphisms. This is exact if and only if ${ 0 \to \text{Hom}(M’’, N) \overset{\overline{v}}{\to} \text{Hom}(M, N) \overset{\overline{u}}{\to} \text{Hom}(M’, N) }$ is exact for every $N.$
(b) ${ 0 \to N’ \overset{u}{\to} N \overset{v}{\to} N’’ }$ a sequence of homomorphisms. This is exact if and only if ${ 0 \to \text{Hom}(M, N’) \overset{\overline{u}}{\to} \text{Hom}(M,N) \overset{\overline{v}}{\to} \text{Hom}(M, N’’) }$ is exact for every $M.$
Pf $\color{purple}{\textbf{(a)}}$ $\implies$ $\ker(\overline{v})$ ${= \lbrace f \in \text{Hom}(M’’, N) : f \circ v = 0 \rbrace }$ which is $0$ because of surjectivity of $v.$
Also $\text{im}(\overline{v})$ $= \lbrace f \circ v : f \in \text{Hom}(M ‘’, N) \rbrace$ and $\text{ker}(\overline{u})$ $=\lbrace g \in \text{Hom}(M,N) : g \circ u = 0 \rbrace$ are equal.
$\subseteq$ Clear that $f \circ v \circ u = 0,$ because $v$ kills all $u(x)$s.
$\supseteq$ [Drawing a commutative diagram helps] Given $g \in \text{Hom}(M,N)$ with $g \circ u = 0,$ we need to exhibit an $f \in \text{Hom}(M’’, N)$ with $g = f \circ v.$
Since $v$ is surjective, definining $f ( \text{ } v(t) \text{ }) := g(t)$ will assign (we didnt yet show unique) values of $f$ at every elem of $M’’.$ This $f$ is a well defined map : Suppose $v(t _1) = v(t _2)$ in $M’’.$ So $(t _1 - t _2) \in \ker(v) = \text{im}(u),$ so $t _1 - t _2 = u(p)$ for some $p.$ Hence $g(t _1 - t _2) = g(u(p)) = 0,$ ie $g(t _1) = g(t _2)$ as needed. Similarly check $f$ is a homomorphism.
$\impliedby$ $\ker(\overline{v})=0$, ie for any module $N$ and ${f \in \text{Hom}(M’’, N),}$ $f \circ v = 0$ implies $f = 0.$ That is, any $M ‘’ \overset{f}{\to} N$ with $\text{im}(v) \subseteq \ker(f)$ has $M’’ = \ker(f).$ Setting $N := M’’ /{\text{im}(v)}$ and $f$ as the projection map, ${ M’’ = \text{im}(v) },$ ie $v$ is surjective.
We know ${ \text{im}(\overline{v}) = \ker(\overline{u}) },$ ie ${ \lbrace f \circ v \text{ } \vert \text{ } M ‘’ \overset{f}{\to} N \rbrace = \lbrace M \overset{g}{\to} N \text{ } \vert \text{ } g \circ u = 0 \rbrace,}$ for every $N.$ This gives ${ \text{im}(u) = \ker(v) }.$
$\subseteq$: Take $N = M’’$ and $f = \text{id}.$
$\supseteq$: Say $v(t) = 0.$ We want an $x$ with $t = u(x).$ Taking $N = M/{\text{im}(u)}$ and $g$ as the projection ${M \twoheadrightarrow M/{\text{im}(u)},}$ ${ g = f \circ v }$ for some ${M’’ \overset{f}{\to} N}.$ So ${g(t) = f(0) = 0 },$ ie ${ t \in \ker(g) = \text{im}(u)},$ as needed.
$\color{purple}{\textbf{(b)}}$ Similar (straightforward but tedious).
Lec-14: Consider a class $\mathscr{C}$ of $A-$modules. A map ${\lambda : \mathscr{C} \to \mathbb{Z}}$ is called additive if ${ \lambda(M’) - \lambda(M) + \lambda(M’’) = 0 }$ for every short exact sequence ${ 0 \to M’ \to M \to M’’ \to 0 }$ with the modules in $\mathscr{C}.$ Eg $\dim$ is additive on the class of finite dimensional $K-$vector spaces.
Let ${0 \to M _0 \to M _1 \to \ldots \to M _n \to 0}$ be an exact sequence, and $\lambda$ an additive function on a class $\mathscr{C}$ containing the $M _j$s and kernels of all the above homomorphisms. Then $\sum _{j = 0} ^{n} (-1) ^{j} \lambda(M _j) = 0.$
Pf: Split it into short exact sequences, and add the corresponding equations.
[Tensor Product of Modules] Still figuring this out.