$\require{AMScd}$ $ %\begin{CD} %K(X) @>{ch}>> H(X;\mathbb Q);\\ %@VVV @VVV \\ %K(Y) @>{ch}>> H(Y;\mathbb Q); %\end{CD} $ $\newcommand{\C}{\mathcal{C}}$ $\newcommand{\llra}{\Longleftrightarrow}$ $\DeclareMathOperator{\rad}{Rad}$ $\DeclareMathOperator{\Ima}{Im}$ # Part 1: Quivers # Part 2: Representations # Part 3: Relations # Part 4: Finite length & Jordan-Hölder ## Filtrations and composition series ## Finite length ## Jordan-Hölder ## Closed under extension A collection $\mathcal{C}$ is **closed under extensions** if for each exact sequence $$\begin{CD} 0 @>>> A @>>> B @>>> C @>>> 0 \end{CD} $$ $A,C \in \mathcal{C} \implies B \in \mathcal{C}$ > $fl(\Lambda)$ is closed under extensions. Follows immediately from previous results. > If $\mathcal{C}$ is a collection of $\Lambda$-modules, contains the simples and is closed under extensions, then $fl(\Lambda) \subseteq \mathcal{C}$. Sketch: By induction. The simples are in $\mathcal{C}$ and has length $1$. Show that for any $B$ with length $n > 1$, we can find $A \subseteq B$ and create an exact sequence with its factor module $B/A$ of $B$, both of which have length less than $n$. As $\C$ is closed under extensions we get our result. > $A$ $\Lambda$-module. $l(A) < \infty \Longleftrightarrow A$ Artinian and Noetherian Sketch: Showing that Artinian and Noetherian is closed under extensions (and contains the simples), gives the right inclusion. Then let $B$ be Noetherian and Artinian. We can find $S \subseteq B$ which is simple. Then construct a set of subsets of $B$ with $S$ with $l(X) < \infty$. As $B$ is Noetherian it has a maximal element. Show that this maximal element is necessarily equal to $B$. This gives the left inclusion. > The collection $fl(\Lambda) = mod \Lambda$, i.e. precisely the finitely generated $\Lambda$-modules. > $A$ semisimple $\Lambda$-module. Then $A$ Noetherian $\llra$ $A$ Artinian $\llra$ $A$ finite length. ## Summary # Part 5: Radical ## Definition and basic properties The (left) radical of a ring $\Lambda$, $\rad \Lambda$ is defined as the intersection of all maximal ideals of $\Lambda$: $$ \rad \Lambda = \bigcap_{m_i\,\mathrm{maximal\,(left)\,ideal}} m_i $$ > **Proposition 16.** Let $\Lambda$ be a ring. For any $\lambda \in \Lambda$ the following are equivalent: > > (i) $\lambda \in \rad \Lambda$ > (ii) $(1 - x\lambda)$ has a left inverse $x'$, $\forall x \in \Lambda$ > (iii) $\lambda S = (0)$ for all simple $S$ $\Lambda$-modules. Sketch: TODO We then define the Annihilator $\mathrm{Ann}_{\Lambda}(M)$ of a $\Lambda$-module $M$ as being all elements $\lambda \in \Lambda$ such that $\lambda m = 0$. This is clearly a left ideal, but can also be shown to be a right ideal (look at $\lambda m \lambda',\>\> \forall \lambda' \in \Lambda$). We can then redefine the radical, as a consequence of Proposition 16 (i) => (iii), as being the intersection of annihilators of simple modules in $\Lambda$. This yields > **Corollary 17.** $\rad \Lambda$ is a two-sided ideal. Proof follows from the claim above that $\mathrm{Ann}_{\Lambda}(M)$ is a two-sided ideal. > **Theorem 18 (Nakayama Lemma).** Let $M$ be a finitely generated $\Lambda$-module. Then for any ideal $A \subseteq \rad \Lambda$, $aM = M \implies M = (0)$. Proof sketch: Assume $AM = M$ and $M$ is not 0. Then chose a minimal generating set, show that any element in $m \in M$ then can be written as a linear combination of $a_i$ and $m_i$, in $A$ and the generating set respectively. Show then that this implies that $m_1$ is not required in the generating set by exploting that $1-a_i$ has an inverse. ## In relation to nilpotency Before we really can start to employ the radical in interesting ways, we take a brief tour into nilpotency. > **Lemma 19**. Let $\Lambda$ be a left artinian ring. Then > > (a) $\rad \Lambda$ is nilpotent > > (b) If $A \subseteq \Lambda$ is a nilpotent ideal, then $A \subseteq \rad \Lambda$. Proof sketch (a). So $\Lambda$ is left artinian. Thus for the descending chain of ideals $(\rad \Lambda)^i,\, i \in \mathbb{N}$, there is an $n$ such that $(\rad \Lambda)^n = (\rad \Lambda)^{n+1} \dots$. So we wish to prove that this implies that the radical is nilpotent, i.e. that $(\rad \Lambda)^n = 0$. The easiest way to go about this, is the prove that $r^n$ is finitely generated, and then use the Nakayama lemma. Proof (b). We include the entire proof of this, as it is quite nice. So we know that $A$ is nilpotent, and wish to show that $A \in \rad \Lambda$. So let $a \in A$, then $a^n = 0$ for some $n \in \mathbb{N}$. But since $xa \in A,\>\forall x \in \Lambda$, $(xa)^n = 0$. Then consider $$ (1 + xa + (xa)^2 + (xa)^3 + \ldots + (xa)^{n-1})(1 - xa) $$ Every term but the first and last cancel, and we get $$ (1 + xa + (xa)^2 + (xa)^3 + \ldots + (xa)^{n-1})(1 - xa) = 1 - (xa)^n = 1 $$ But since $1 - xa$ has a left inverse for every $x \in \Lambda$, $a \in \rad \Lambda$ and $A \subseteq \rad \Lambda$ as requested. This lemma will be of great use in the following chapter. ## In relation to (left) artinian rings Using what we now have defined and proven, we can start to construct some nice categorization theorems for artinian rings. > **Theorem 20.** $\Lambda$ ring. Then $\Lambda$ semisimple $\Longleftrightarrow$ $\Lambda$ is left artinian and $\rad \Lambda = (0)$. Proof $\Rightarrow$: We have the following chain of implications: $$ \begin{align} \textrm{A is semisimple} &\implies \textrm{A is left Artinian} \overset{\textrm{Lemma 19}}{\implies} \rad \Lambda \textrm{ is nilpotent} \\ \textrm{A is semisimple} &\implies \textrm{A has no non-zero nilpotent ideals} \\ &\implies \rad \Lambda = (0) \end{align} $$ Proof sketch $\Leftarrow$: We simply note that, given $\Lambda$ artinian, $\rad \Lambda = (0)$, and then given **Lemma 19 (b).**, $\Lambda$ has no non-zero nilpotent ideals, and so it is all over. Following this, we have a rather lengthy theorem with an equivalently lengthy corollary. We simply state them without proof. > **Theorem 21.** $\Lambda$ left artinian, $r = \rad \Lambda$. Then > > (a) $\Lambda / r$ is semisimple. > > (b) A left $\Lambda$-module $m$ is semisimple if and only if $rM = 0$. > > (c) There are only finitely many non-isomorphic simple $\Lambda$-modules, and they all occur as direct summands of $\Lambda / r$. > > (d) $\Lambda$ is left Noetherian (as we already know). Clearly, (a) seems to be fairly analagous to **Theorem 20.** Indeed, we prove it by simply showing that $rad(\Lambda / r) = (0)$. To see this, you have to convince yourself that the ideals in $\Lambda / r$ are the ideals in $\Lambda$ which fully contains $r$, as indeed they have to also be ideals under $\Lambda / r$ where all elements of $r$ are factored out. Following this, we get our first result yielding some eqvuialences of artinian rings. > **Corollary 22.** Let $\Lambda$ be a ring. Then the following are equivalent > > (a) $\Lambda$ left artinian. > > (b) Every finitely generated $\Lambda$-module has finite length > > (c) $r = \rad \Lambda$ is nilpotent and $r_i / r_{i+1}$ is finitely generated semisimple $\Lambda$-modules $\forall i \geq 0$. Finally, we get a nice theorem giving us a procedure to check if an ideal is in fact the radical. > **Theorem 23.** Let $\Lambda$ be a left artinian ring and $A \subseteq \Lambda$ a nilpotent ideal. Then > $$ \Lambda / A\,\textrm{semisimple} \llra A = \rad \Lambda$$ Sketch of proof: Clearly theorem 21 gives $\Leftarrow$. For $\Rightarrow$: Convince yourself (as above) that $\rad (\Lambda / A) = \rad(\Lambda) / A$. Then use that $\Lambda / A$ is left artinian (since $\Lambda$ is) and Theorem 20, to show that $\rad (\Lambda / A) = 0$, which by the initial equation implies that $A = \rad(\Lambda)$. ## Admissable quotients of path algebras As a quick interlude, we look at what happens when we have an *admissable* [relation as a] quotient of a path algebra. An admissable relation $\rho$ is such that if $J$ is the ideal generated by all arrows of $k\Gamma$, then $$ J^t \subseteq \left<\rho\right> \subseteq J^2 $$ for some $t \geq 2$. I.e. the relation is such that it is a subset of the ideal generated by all paths of length $2$ but not "smaller" than the ideal of paths of some length $t$. >**Theorem 24.** Let $\Lambda = (\Gamma, \rho) = k\Gamma / \left<\rho\right>$ be an admissible quotient of the path algebra $k\Gamma$. Then $$\rad (k\Gamma / \left<\rho\right>) = J / \left<\rho\right> \overset{def}{=} \bar{J}$$ Sketch of proof: Show first that $\Lambda$ is a factor of $k\Gamma/J^t$, and so we get that $\Lambda$ is left artinian (through finitely dimensional $k$-algebra). Show then that $J$ is nilpotent, and finally that $\Lambda / \bar{J} \cong k\Gamma / J \cong k^{\left|\Gamma_0\right|}$, so it is semisimple. This implies through theorem 23 that $\bar J = \rad \Lambda$. ## Radicals of modules We are now ready to extend the definition of radicals to modules, which is done in precisely the way one would expect. Let $B$ be a $\Lambda$-module. Then the radical over $B$ over $\Lambda$ is defined by $$ \rad_{\Lambda} B = \bigcap_{\textrm{All maximal submodules } B' \textrm{ of } B} B' $$ Clearly, for a ring as a module over itself, this definition coincides with the definition of the radical over a ring. To get some more mileage out of the definition, we need to introduce the concept of a submodule being *small* in another. So let $A, B$ be two $\Lambda$-submodules, with $A \subseteq B$. Then **$A$ is small in $B$** if for every $X \subseteq B$ we have the equivalency $$ A + X = B \llra X = B $$ >**Proposition 25.** Let $\Lambda$ be a ring, and $B$ a finitely generated $\Lambda$-module. Then >$$A \subset B \textrm{ small in }B \llra A \subseteq \rad B$$. Sketch of proof: The proof is (perhaps surprisingly) involved, and invokes Zorn's lemma. For a rough idea: $\Rightarrow$ can be shown contrapositively by showing that if $A$ is not a subset of $\rad B$, then there exists a maximal subset not containing $A$, which then violates the condition of smallness due to its maximality. The other (hard) way $\Leftarrow$: Construct a set for each $X$ in the smallness condition which consists of all proper submodules of $B$ containing $X$. Then show that all chains in this set has an upper bound, due to $B$ being finitely generated. Invoking Zorn's lemma gives us a maximal element $B'$ in $B$ such that $A+X \subseteq B' \subset B$ (proper subset). Thus for $A+X = B$, $X$ has to equal $B$. The next theorem gives us a nice way to associate the radical of a module with the radical of its constituent ring, under some (strict) conditions. >**Theorem 26.** Let $\Lambda$ be a left artinian ring, and let $A$ be a finitely generated $\Lambda$-module. Then $\rad A = r\cdot A$, where $r = \rad \Lambda$. In other words, we can associate the radical of a module with the radical of its ring, multiplied by itself. Sketch of the proof: $\subseteq$: We only have to show that $rA$ is small in $A$ by Proposition 25. But then consider $rA + X = A$. Multiply on both sides by $r$ to obtain $r^2A + rX = rA$. Reinsert back into the first expression. Convince yourself that $rX + X = X$, and continue by induction. Then by the Artinianness of $\Lambda$ we get the result. $\supseteq$: Use that $A / rA$ is semisimple to show that its radical is $0$. Then by the familiar construction, show that $\rad(A / rA) = \rad(A) / rA$, yielding $A \subseteq rA$. ## Radicals of representations [#] TODO (If/when we get some better commutative diagram package here, we can show some examples) The theorem above, gives us a nice way to think about radicals of representations. Indeed, let $(\Gamma, \rho)$ be a representation with admissable relations. Then by theorem 24, its radical is $\bar J (k\Gamma / \rho)$. If you think about it long enough (or watch the curriculum videos), this implies that we can associate the radical of such a representation in each vertex, with the sum of the image of all its paths. ## Top of a module We might at some point ask ourselves what happens when we quotient out the radical of a module. As the radical is clearly defined in terms of the maximal submodules, we should expect the radical to contain some of the most "covering elements". Thus, factoring out these, we could imagine that the residue classes we are left with, respresent the more "outlying" elements in the module, or the *topmost* elements. We call this the **top of a module**. So let $A$ be a left artinian finitely generated $\Lambda$-module with radical $r$. Then the **top of the $A$** is defined as $ A / rA$. We clearly had some opinions a priori, but what can we say a posteriori? Well, $A / rA$ is semisimple by Theorem 21, and so can be written as a direct sum of simple summands: $$ A / rA = \bigoplus_{i=1}^t S_i $$ We make a selection ${x_1',\ldots,x_t'}$ from these, we send them back to $A$ by the reverse inclusion and end up with $\{x_1,\ldots,x_t\}$. So what do we know about these elements? Well, first of all, we can clearly linearly combine them to send any element to the radical, as they represent the different elements of the aforementioned quotient over the radical. So we can write $$ a - \sum_{i=0}^t \lambda_i x_i \in rA,\>\lambda_i \in \Lambda $$ We note then that the sum can be written in terms of elements of $a$ and $r$: $$ \sum_{i=0}^t \lambda_i x_i = \sum_{j=0}^n r_j a_j $$ Now consider the submodule generated by $A' = \Lambda\{x_1, x_2,\ldots, x_n\} \subseteq A$. Then consider $$ r(A'/A) = A'/A $$ Why is this? Remember that we can write any element in $A'$ as a combination of elements of $r$, so $A'$ is clearly closed under $r$. But now we can use Nakayama's lemma to get that $$ A'/A = 0 \implies A' = A $$ Thus, we have found a generating set of $A$ by the means of its *top*. ## Essential epimorphisms Let $f: A \to B$ be an epimorphism (onto). Then let $g: X \to A$. If for all such maps $$ fg: X \to B \textrm{ onto } \implies g \textrm{ onto, } $$ then $f$ is called **an essential epimorphism**. Put into plain english, we thus require that for $f$ to be an essential epimorphism; if it's composition with any other map is onto, then the other component necessarily is onto. So what does this mean? If $g$ is not onto, it means that $g(X) \subset A$, i.e. it is a proper subset of $A$. Then as this subset does necessarily not map onto $B$ by means of $f$, we clearly see that $f$ 'requires' the full underlying set $A$ to map onto. Indeed, we can say that $f$'s domain contains no superfluous elements for $f$ to be onto. Before proving our main result in relation to this definition, we first state the following >**Lemma 27.** Let $\lambda$ be a left artinian ring and $A,B$ finitely generated $\Lambda$-modules. Then >$$ f: A \to B \textrm{ onto } \llra \bar f: A/rA \to B/rA \textrm{ onto }, r = \rad{\Lambda}$$ Sketch of proof: To show $\Rightarrow$, simply draw the commutative diagram and use a direct argument. For $\Leftarrow$, show first that $B = \Ima f + rB$. Show then that $rB$ is small in $B$, yielding $\Ima f = B$. Now we can get a nice classification theorem for an epimorphism to be essential: >**Theorem 28.** Let $\lambda$ be a left artinian ring and $A,B$ finitely generated $\Lambda$-modules. Then the following are equivalent: > > (a) $f$ is an essential epimorphism. > > (b) $\ker f \subseteq rA$, $r = \rad \Lambda$. > > (c) $\bar f: A/rA \to B/rA$ is an isomorphism. Sketch of proof: (a) $\implies$ (b): Show that if $f$ is essential, then $\ker f$ is small in $rA$. (b) $\implies$ (c): Show that if $\ker f \subseteq rA$, then $rA / (\ker f) = rB$. Then show that $$ \begin{CD} A/rA @>>> (A/\ker f)/(rA / \ker f) @>>> (B / rB) \end{CD} $$ is a composition of isomorphisms. (c) $\implies$ (a): Consider $g: X \to A$. Draw the commutative diagram for $f$, $g$, $\bar f$ and $\bar g$. Show by using Lemma 27 that $g$ is necessarily onto. ## Summary * The radical of a ring $\rad \Lambda$ is the intersection of its maximal ideals. * The radical of a module $\rad_{\Lambda} M$ is the intersection of its maximal submodules. * The radical is itself a two-sided ideal (submodule). * A module $A$ is small in another $B$ if $A + X = B \implies X = B\quad$. * The top of a module is the quotient of the module with its radical: $A/rA$. * An epimorphism $f$ is essential, if for every epimorphic composition $fg \implies g$ is an epimorphism. ### Table of equivalencies || Conditions || $\llra$ || $\llra$ ||$\llra$ || || ||$\lambda \in \rad \Lambda$ || $(1 - x\lambda)$ has a left inverse for all $x \in \Lambda$ || $\lambda S = (0)$, $S$ simple || || ||$A \subseteq \rad \Lambda$ || $AM = M \implies M = 0$ || || || ||$\rad \Lambda$ nilpotent || $A$ nilpotent ideal, then $A \subseteq \rad \Lambda$ || $ $ || || ||$\Lambda$ semisimple || $\Lambda$ left artinian and $\rad \Lambda = (0)$ || || || $\Lambda$ left artinian || $\Lambda /r$ semisimple || $M = \Lambda$-module semisimple iff $rM = 0$ || Finitely many non-isom simple $\Lambda$-modules || || $\Lambda$ left artinian || $A \subset \Lambda$ nilpotent, $\Lambda/A$ semisimple || $A = \rad \Lambda$|| || || || $\rad_{\Lambda} M$ || $\rad\Lambda \cdot M$ || || ||$\Lambda$ left artinian, $A,B$ fin gen || $f: A \to B$ onto || $\bar f: A/rA \to B/rB \textrm{ onto }$ || ||$\Lambda$ left artinian, $A,B$ fin gen || $f$ essential epi || $\ker f \subseteq rA$ || $\bar f: A/rA \to B/rB$ isomorphism|| # Part 6: Projective Modules ## Projective modules The motivation for the following chapter is to look at a class of modules where we for sure can classify all indecomposable submodules — in general we will indeed struggle with such a classification, so we will have to look at some subclass. Let $\Lambda$ be a ring, and $P$ a $\Lambda$-module. Then $P$ is **a projective module**, if for every epimorphism $g: B \to C,\>B,C\>\Lambda\textrm{-submodules}$, and for every homomorphism $f: P \to C$, we can find a homomorphism $h:P \to B$ such that $gh = f$. In other words, P is projective if for every epimorphism between two modules $B$ and $C$, of which there exists a homomorphism from P to the latter, we can extend the homomorphism "onto" the first. And this for all such homomorphisms from $P$ to $C$. Another equivalent statement is that for any projection $g$ of one $\Lambda$-module onto another, then any homomorphism from a projective module into the latter correlates with a homomorphism into the former. These homomorphisms are similar in the sense that they are "preserved" under the projection $g$. >**Proposition 29.** Let $\Lambda$ be a ring, and $P$ a $\Lambda$-module. Then >$$P \textrm{ projective } \llra \exists \textrm{ a free module } F \textrm{ and a } \Lambda\textrm{-module } Q \textrm{ such that } F \cong P \oplus Q $$ Sketch of proof $\Rightarrow$: Any module can be written as a factor of a free module. Then construct a map form such a free module to an arbitrary moduke and show that it necessarily is onto. Apply this to $P$. Use that $P$ is projective to show that $\exists h$ such that $F \cong \Ima g \oplus \ker h$, where $g$ is an onto map from $F$ to $P$. $\Leftarrow$: Construct maps $h': P\oplus Q \to B, \pi: P\oplus Q \to P, \nu: P \to P \oplus Q$ and show that we can for any $g: B \to C$ go from $P$ through $\nu$ and $h'$ to get the desired effect. ## Projective covers We now combine the notion of an essential epimorphism, with the notion of a projective module. Let $f: P \to M$ be a $\Lambda$-homomorphism, then $f$ is a **projective cover of $M$**, if $P$ is projective and $f$ is an essential epimorphism. So what exactly are these? So, they are essential epimorphisms from projective modules. This means first of all that for any $g: B \to P$, and $fg$ is onto, then $g$ necessarily is onto. As noted, we can think of this as saying that $f$ is not superfluous in any sense over $P$, i.e. unless it can work on the entirity of $P$, it will no longer preserve it's epimorphism-structure. Furthermore, we know that $P$ is a projective $\Lambda$-homomorphism. Which means that for any epimorphism $h: B' \to C'$, any homomorphism from $P$ to $C'$ induces a homomorphism to $B'$. [#] This is perhaps not rigorous enough and may contain some general logical mistakes? So what does this tell us combined? Well, it tells us that not only does $P$ map nicely onto a set of epimorphic $\Lambda$-modules, but also that if $P$ does so epimorphically; it "covers" the projections. >**Theorem 30.** Let $\Lambda$ be a left artinian ring, and $A$ a finitely generated $\Lambda$-module. > > (a) $\exists$ a projective cover $f:P \to A$ ($P$ finitely generated) > > (b) All projective covers of $A$ are isomorphic; in the sense that if $f_1: P_1 \to A$ and $f_2: P_2 \to A$, then there exists an isomorphism $g: P_1 \to P_2$ such that $f_2g = f_1$. The proof of this Theorem, mostly part (a), is (very) lenghty. The idea is to let $f:P \to A$ be an epimorphism with $\ell(a)$ minimal, $P$ projective. Then show that $\ker f \subseteq rP$ which by Theorem 28 implies that $f$ is an essential epimorphism. For (b), show that $\ell(P1) \subseteq \ell(P2) \subseteq \ell(P1)$. >**Proposition 31.** Let $\Lambda$ be a left artinian ring, $f: P \to A$ an epimorphism, $A$ finitely generated, and $P$ projective. Then > > (a) $f:P\to A$ projective cover $\llra$ $\bar f: P/rP \to A/rA$ isomorphism. > > (b) If $\{f_i: P_i \to A_i\}_{i=1}^m$ is a set of maps from submodules of $P$ to $A$, then the function defined on their direct sum in the natural way is a projective cover iff. each map $f_i$ is a projective cover. Sketch of proof: (a) goes almost directly from Proposition 28. (b) Use that $r(P_1 \oplus P_2 \oplus \dots \oplus P_m) = rP_1 \oplus rP_2 \dots \oplus rP_m$, and (a). ## Projective modules II We return to projective modules, and prove two nice results. >**Proposition 32.** Let $\Lambda$ be a left Artinian ring. All modules in the following are finitely generated. $P, Q$ are projective modules. > (a) $P \to P/rP$ is a projective cover. > > (b) $P \cong Q \llra P/rP \cong Q/rP$. > > (c) $P$ indecomposable $\llra$ $P/rP$. > > (d) Assume $P = \oplus_{i=^1}^{n} P_i \cong \oplus_{j=1}^m Q_j$, and $P_i, Q_j$ indecomposable. Then > $n = m$ and there exists a permutation $\pi$ such that $P_i = Q_{\pi(i)}$. Proof. [#] TODO Following this, we get another result: >**Corollary 33.** Let $\Lambda$ be an Artinian ring. Then $\Lambda / r$ is semisimple and >$$\Lambda /r = \bigoplus_i S_i$$ with $S_i$ simple. Let $P_i \to S_i$ be a projective cover of $S_i$. The only indecomposable projective $\Lambda$-modules up to isomorphism, > are $P_i$. Sketch of proof: Follows from Proposition 31, 32 and from uniqueness of projective covers. ## Projective covers II >**Lemma 34.** Let $\Lambda$ be a left Artinian ring, and $M$ finitely generated. Let $f':P \to M/rM$ be a projective cover. Then if a homomorphism $f: P \to M$ exists such that >$$ >\begin{CD} >P @>f>> M;\\ >@| @VV\pi_M V \\ >P @>f'>> M/rM; >\end{CD} >$$ > >commutes, then $f$ is a projective cover.

## Local rings ## Indecomposable projectives ## Krull-Remak-Schmidt — Projectives ## Summary # Part 7: Krull-Remak-Schmidt — General case # Part 8: Artin Algebras # Part 9: Categories and functors # Part 10: Projectivization # Part 11: Basic artin algebras # Part 12: Duality # Part 13: Injective modules