MA3203: Ring Theory
Part 1: Quivers
A Quiver is a directed graph, consisting of vertices and arrows between the vertices. We will typically denote the quiver by
For the latter example, we would have
A path in a quiver is a directed walk from one vertex to another, traversing in sequence arrows from the starting arrow to the end arrow. Clearly, not all arrows have paths between them.
We denote the start of a path
We note also that we have a trivial path,
Path algebra of a quiver
Given any field
Let
We define multiplication addition in the following way, with
Multiplication is defined and distributes as expected, but with the following constraints:
Take for instance the quiver
The elements of the path algebra
So just noting now that all paths consisting of subpaths not ending and starting in the same vertex respectively, goes to
So what does this represent, you might ask? A way to interpret it, would be the linear combination of all possible paths one could take of length exactly two. We can then clearly extrapolate this result to paths of length three:
This would be (and you are encouraged to check this) the general result when multiplying three arbitrary elements of
Proposition 1.
$\dim_k k\Gamma < \infty \llra $ the quiver$\Gamma$ has no oriented cycles.
Proof
Proof
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
$fl(\Lambda)$ 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
$A$ $\Lambda$ -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
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
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
We can then redefine the radical, as a consequence of Proposition 16 (i) => (iii), as being the intersection of annihilators of simple modules in
Corollary 17.
$\rad \Lambda$ is a two-sided ideal.
Proof follows from the claim above that
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
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
Proof (b). We include the entire proof of this, as it is quite nice. So we know that
Every term but the first and last cancel, and we get
But since
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
Proof sketch
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
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$r_i / r_{i+1}$ are 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
For
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
for some
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
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
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
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:
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:
Radicals of representations
The theorem above, gives us a nice way to think about radicals of representations. Indeed, let
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
We clearly had some opinions a priori, but what can we say a posteriori? Well,
We make a selection
We note then that the sum can be written in terms of elements of
Now consider the submodule generated by
Why is this? Remember that we can write any element in
Thus, we have found a generating set of
Essential epimorphisms
Let
then
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
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)
(b)
is a composition of isomorphisms.
(c)
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 | |||
Finitely many non-isom simple |
|||
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 honing in our attention on some subclass of modules is reasonable.
Let
In other words, P is projective if for every epimorphism between two modules
An important point, is to note that every free module is projective.
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
Projective covers
We now combine the notion of an essential epimorphism, with the notion of a projective module.
Let
So what exactly are these? So, they are essential epimorphisms from projective modules. This means first of all that for any
So what does this tell us combined? Well, it tells us that not only does
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
For (b), show that
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
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.
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.
Sketch of proof: We know that
This now, gives us a procedure to construct a projective cover, as if we can just find a projective cover onto top of the module - well then we can easily extend the cover onto the module itself by making sure the above diagram commutes.
Local rings
We say that
Proposition 35. Let
$\Lambda$ be a local ring. Then$0$ and$1$ are the only idempotents.
Proof: Show that an invertible idempotent necessarily is
Indecomposable projectives
We know that if the endomorphism ring
Corollary 36. Let
$\Lambda$ be a ring and$M$ be a$\Lambda$ -module, then$\End_{\Lambda}(M) \textrm{ local }\implies M$ indecomposable.
We are now ready to state yet another equivalency theorem.
Proposition 37.
$\Lambda$ left artinian,$P$ finitely generated projective. Then the following are equivalent:(a)
$P$ indecomposable.(b)
$\End_{\Lambda}(P)$ local.(c)
$rP$ is the only maximal submodule of$P$ .(d)
$P/rP$ is simple.
Sketch of proof:
(a) => (d): Follows directly from Proposition 32 (c).
(d) => (c): So we have that
(c) => (b): We need to show that all non-invertible endomorphisms form an ideal. The first task is to classify these non-invertible endomorphisms. Then use a direct argument to show that they are closed under differences and multiplication with general endomorphisms.
(b) => (a): Directly from corollary 36.
Corollary 38. Let
$\Lambda$ be a left artinian ring. Then the following are equivalent:(a)
$\Lambda$ local(b)
$r = \rad\Lambda$ is a maximal left ideal(c)
$\Lambda/r$ simple left$\Lambda$ -module.
Sketch of proof: We know that
Krull-Remak-Schmidt — Projectives
Finally we have enough machinery to start on one of the main results of the course, namely the Krull-Remak-Schmidt theorem. For starters, we prove the result in the context of projectives.
Proposition 39. Let
$\Lambda$ be a left artinian ring. Then(a)
$1 = e_1 + \dots + e_n$ , where$e_i$ are primitive, orthogonal idempotents.(b) If
$e_i$ are orthogonal idempotents, then let$e = e_1 + \dots + e_n$ . Then
$$\Lambda e = \bigoplus_{i=i}^n \Lambda e_i$$ (c) Let
$e \neq 0$ be an idempotent. Then
$$\Lambda e \textrm{ indecomposable } \llra \textrm{ e is primitive }$$
Proof: TODO
Finally, we get the following result:
Proposition 40. Let
$\Lambda$ be a left artinian ring.$P$ finitely generated projective$\Lambda$ -module. Then$$P \cong \bigoplus_{i=1}^n P_i$$ with$P_i$ indecomposable. The decomposition is unique up to isomorphism and ordering.
In other words, we can deconstruct any
Sketch of proof: The top of
Summary
Definitions
- A projective module
$P$ is one where for any epimorphism$B \to C$ between modules, and any homomorphism between$P \to C$ , we can lift the homomorphism up to$P \to B$ . - A projective cover is an essential epimorphism
$P \to X$ where$P$ is a projective module. In a sense it is the best approximation of$X$ by$P$ . - A local ring is a ring where the non-invertible elements form an ideal.
List of results
- Projective modules can be identified as the summand of a free module.
- Every finitely generated module over an artinian ring has a projective cover.
- Projective covers of a module are unique up to isomorphism.
$f: P \to A$ projective$\llra$ $\bar f: P/rP \to A/rA$ isomorphism.- The direct sum of projective covers are a projective cover.
$f: P \to P/rP$ is a projective cover. This makes tremendously much sense in the light of our informal definition of a projective cover above.- Let
$f: P \to rP$ be a projective cover. Then the following are equivalent: (a)$P \cong Q \llra P/rP \cong Q/rQ$ ,$P,Q$ projective (b)$P$ indecomposable$\llra P/rP$ simple (c) two different direct sums of$P$ are equivalent under permutation. - The projective modules of a left artinian ring can be associated with the simple decomposition of the top of ring.
- The projective cover of a top extends back to the module.
- Local rings only have
$0$ and$1$ as idempotents. - For a left artinian ring, with
$P$ finitely generated projective module, the following are equivalent: (a)$P$ indecomposable (b)$\End_{\Lambda}(P)$ local (c)$rP$ is the only maximal submodule of$P$ . (d)$P/rP$ simple We then showed a similar result for a left artinian ring itself. - For a left artinian ring, the identity can be written as a sum of orthogonal and primitive idempotents.
- The submodule generated by the sum of an orthogonal set of idempotents is decomposable into submodules generated by the individual idempotents.
- The submodule generated by an idempotent is indecomposable if and only if this idempotent is primitive.
Part 7: Krull-Remak-Schmidt — General case
We now proceed to generalise the last result of the previous part, namely the decomposability of projective finitely generated modules over an artinian ring. We relax the requirements, and now only demand that the module in question is finitely generated (over an artinian ring).
Before we state the general result, we require some fitting preliminaries.
Lemma 41 (Fitting Lemma). Let
$\Lambda$ be a ring, and$M$ be a$\Lambda$ -module of finite length. Then$\exists n \geq 1$ such that$$ M = \Ima\phi^n \oplus \ker\phi^n$$
Sketch of proof: By finite length, we know that
Theorem 42. Let
$\Lambda$ be a left artinian ring,$M$ finitely generated. Then$$M \textrm{ indecomposable } \llra \End_{\Lambda}(M) \textrm{ local}$$
Sketch or proof: We have seen
Finally, we arrive at,
Theorem 43 (Krull-Remak-Schmidt). Let
$\Lambda$ be a left artinian ring, and let$M$ be a finitely generated$\Lambda$ -module. Then(a)
$M$ can be written as a finite direct sum of indecomposable modules$$ M = \bigoplus_{i=1}^n M_i$$ (b) This decomposition is unique up to isomorphism and ordering.
(Full proof): TODO.
Part 8: Artin Algebras
"All" the previous work has lead up to the Krull-Remak-Schmidt theorem, and we now need new ventures. We start with Artin algebras. Before this can be defined, we have to extend the notion of an algebra over a field slightly.
Let
The notions of
As before, let
Proposition 44. Let
$\Lambda$ be an artin$R$ -algebra. And$A,B$ finitely generated$R$ -modules. Then(a)
$\Hom_{\Lambda}(A,B)$ is a finitely generated$R$ -module.(b)
$\End_{\Lambda}(A)$ is an artin$R$ -algebra.(c)
$\Lambda$ is a left artinian ring.
Proof: TODO
Part 9: Categories and functors
Basic definitions
For a while, we have worked under the assumption that modules over a quiver and representations of the same quiver, are for all intents and purposes the same thing. We have seen how we can associate the objects of these, and how we can translate the morphisms. We are now going to develop the machinery required to formalise this, and justify the equivalency.
We wish now to generalise the notion of different categories of mathematical objects. For instance, we are very familiar with the fact that there are such different types of objects, e.g. matrices, vectors, functions etc..
A Category
Examples of categories:
- Ring. Category of all rings.
- mod
$\Lambda$ . Category of all finitely generated modules over a ring$\Lambda$ . - Mod
$\Lambda$ . Category of all modules over a ring$\Lambda$ . - Rep(
$\Lambda$ ,$\rho$ ). Category of all representations over a quiver with an admissable relation. $\add M$ ,$M$ a$\Lambda$ -module. The direct summands in a finite number of copies of$M$ :$X \in \add;\>\>M^n = X \oplus Y,\>\> n \leq \infty$ .
One could say that we've in sense moved up one level of abstraction. It is now of interest to define notions we have previously seen, but for categories in a general setting.
Let
A subcategory
Our motivation for these definitions to begin with, was to compare categories. It is then only natural that we describe their morphisms:
A (covariant) functor
If instead the functor is contravariant, we require a morphism
Let
If we have two preadditive categories, a functor between them is said to be additive if
As a final definition, we consider morphisms of functors.
Let
commutes.
Equivalencies
We are now ready to state and show what is required to prove the main goal we set out to prove: The equivalency of modules over path algebras and representations of quivers.
Let again
Note the resemblance to the generalized definition of an isomorphism.
Using this definition to find equivalencies may or may not prove difficult. We can simplify the task of proving a functor an equivalency through the means of the following definitions:
Let
Then
Using these definitions, we get a handy tool in the next proposition.
Proposition 45. Let
$\CC$ and$\DD$ be (R)-categories and$F: \CC \to \DD$ a(n) functor. Then$$ F \textrm{ is an equivalence } \llra F \textrm { full, faithful and dense} $$
Proof. Omitted
Theorem 46. Let
$(\Gamma, \rho)$ be a quiver with admissiable relations$\left<\rho\right>$ . Let$\Lambda$ be the module$\Lambda = k\Gamma / \left<\rho\right>$ . Then the functors$F: \textrm{mod }\Lambda \to \Rep(\Gamma, \rho),\>\> G: \Rep(\Gamma, \rho)\to\textrm{mod }\Lambda$ defined as the association we have seen before, is an equivalency of$k$ -categories.
Proof. TODO
Part 10: Projectivization
We now introduce a few more results, yielding in the end, another method of comparing rings. We first start by reiterating and stating some notation, and state some basic results without proof. In the following
$\Gamma = \End_{\Lambda}(A)^{op}$ is an artin$R$ -algebra (Proposition 44).$A$ is a left$\End_{\Lambda}(A)$ -module.$_{\Lambda}A_{\Gamma}$ is a$\Lambda-\Gamma$ bimodule.$\Hom_{\Lambda}(_{\Lambda}A_{\Gamma}, _{\lambda}X),\>\> x \in \smod \Lambda$ is a left$\Gamma$ -module.
Lemma 47. Let
$\Lambda$ be a ring. Then(a)
$Hom_{\Lambda}(A, B_1 \oplus B_2) \to \Hom_{\Lambda}(A, B_1) \oplus \Hom_{\Lambda}(A, B_2)$ , given by$$ \alpha(f,g)(a) \overset{def}{=} (f(a),g(a)),\>\> a \in A $$ is an isomorphism.(b)
$Hom_{\Lambda}(A_1 \oplus A_2, B) \to \Hom_{\Lambda}(A_1, B) \oplus \Hom_{\Lambda}(A_2, B)$ , defined similarily, is an isomorphism.
Proof. TODO
Proposition 48. Let
$\Lambda$ be an artin$R$ -algebra, and$A \in \smod \Lambda$ ,$\Gamma = \End_{\Lambda}(A)^{op}$ . Define the (R-)functor
$$e_A = \Hom(A,-): \smod \Lambda \to \smod \Gamma$$ It has the following properties(a)
$\Hom_{\Lambda}(Z, X) \to \Hom_{\Gamma}(e_A(Z), e_A(X))$ is an$R$ -isomorphism for all$Z \in \add A$ .(b)
$X \in add A \implies e_A(x)$ is a projective$\Gamma$ -module.(c)
$\left.e_A\right|_{\add A} : \add A \to \mathscr{P(\Gamma)}\,(\textrm{ = finitely generated projective } \Gamma\textrm{-modules})$ is an equivalence of (R-)categories.
Sketch of proof: TODO
Lemma 49.
$\Lambda$ R-algebra,$A \in \smod \Lambda$ ,$\Gamma = \End_{\Lambda}(A)^{op}$ .$e_A: \add A \to \mathscr{P}(\Gamma)$ . Then(a)
$X \neq (0)$ in$\add A$ $\llra$ $e_A(X) \neq (0)$ in$\mathscr(\Gamma)$ .(b)
$X \in \add A$ :$X$ indecomposable$\llra$ $e_A(X)$ indecomposable(c)
$X,Y \in \add A$ .$e_A(X) \cong e_A(Y) \llra X \cong Y$ .
Proof. TODO.
Part 11: Basic artin algebras
We now introduce yet another concept that helps us in creating an equivalence theorem. In fact, we will end up with defining a new type of equivalence.
Let
So say we have
Let now
Proposition 49. Let
$\Lambda$ and$\Sigma$ be as above. Then$$e_P = \Hom_{\Lambda}(P, -): \smod \Lambda \to \smod \Sigma$$ is an equivlance of$R-$ categories.
Proof: TODO.
Morita equivalence
The previous result and its preceding definitions, leads us to the following definition:
Let
Theorem 50. Let
$\Lambda$ be a finite dimensional$k$ -algebra over an algebraically closed field$k$ . Then(a) There exists a basic
$k$ -algebra$\Sigma$ such that$\Lambda$ and$\Sigma$ are morita equivalent.(b) Suppose
$\Lambda$ is basic. Then there exists a quiver$(\Gamma, \rho)$ with admissable relations over$k$ , such that$\Lambda \cong k\Gamma / \left<\rho\right>$ .
Sketch of proof: (a) Follows directly from proposition 49. (b) TODO
Part 12: Duality
To present the very final results of the course, we first need to define the concept of duality.
So let
Recall that a contravariant functor takes morphisms in
Proposition 51.
$\Lambda$ finite dimensional$k$ -algebra, with$k$ a field. Then$D = \Hom_k(-, k): \smod \Lambda \to \smod \Lambda^{op}$ is a duality.
Proof. TODO
There is particularily one duality construction we take notion to, namely that of quivers and representations. To construct the dual of a quiver, all you need to do is to create a new quiver where the direction of the arrows are switched. If you have a representation, over finite dimensional vector spaces in each vertex, transposing the matrix-representation of the morphism-representations of the arrows will complete the dual construction.
Lemma 52. Let
$\Lambda$ be a finite dimensional$k-$ algebra and$k$ a field.(a) Then for every exact sequence in
$\Lambda$ , there exists a dual one in$\Lambda^{op}$ .(b) S simple
$\Lambda$ -module$\llra$ $D(S)$ simple$\Lambda^{op}$ .(c)
$\ell(A) = \ell(D(A))$
Proof. TODO
Finally, we introduce a definition nicely leading into the final chapter of the course.
Given a statement
Part 13: Injective modules
Definition
Let
Looks familiar? Clearly it is the dual statement of a projective module:
Proposition 53. Let
$\Lambda$ be a finite dimensional$k$ -algebra, with$k$ a field. Let$P \in \smod \Lambda$ . Then(a)
$P$ projective$\Lambda$ -module$\llra$ $D(P)$ injective$\Lambda^{op}$ -module.(b) Any
$\Lambda$ -module$m \in \smod \Lambda$ is a submodule of an injective$\Lambda$ -module in$\smod \Lambda$ .
Sketch of proof: TODO
Injective envelopes
Let
A monomorphism
Socle
Let
Final results
We now have the terminology in place to state the final results of the course.
Lemma 54.
$\Lambda$ artin$R-$ algebra,$A \subseteq X \in \lmod \Lambda$ . Then the following are equivalent:(a)
$A$ is an essential submodule of$X$ .(b)
$\Soc X \subseteq A$ (c)
$\Soc A = \Soc X$
Proof. TODO
Propositon 55. Let
$\Lambda$ be an artin$R-$ algebra,$(0) \neq A \in \lmod \Lambda$ . Then
$$A \xrightarrow{} I \textrm{ injective envelope} \llra \Soc A = \Soc I $$
Proof: TODO
Lemma 56. Let
$\Lambda$ be an artin algebra. Then$\Soc A = \{a \in A \mid r\cdot a = (0)\}$ ,$r = \Rad \Lambda$ .
Proof: TODO
Proposition 57. Let
$\Lambda$ be a finite dimensional$k$ -algebra,$k$ field. Then$$P\overset{f}{\to}A\textrm{ projective cover } \llra D(A) \overset{D(f)}{\to} D(P)\textrm{ injective envelope.}$$
Proof. TODO