Path algebra of a quiver

Given any field $k$, we can use a quiver $\Gamma$ to construct an algebra over the quiver. We call this the path algebra over $\Gamma$. We construct this as follows:

Let $k$ be a field, and $\Gamma = (\Gamma_0, \Gamma_1)$ a quiver. Then the elements of the path algebra $k\Gamma$ are linear combinations of all the paths of $\Gamma$:

$$ x = a_1 p_1 + a_2 p_2 + \ldots + a_t p_t,\quad a_i \in k,\>p_i \textrm{ path in } \Gamma $$

We define multiplication addition in the following way, with $x = a_1 p_1 + \ldots + a_t p_t$, $y = b_1 p_1 + \ldots + b_t p_t$.

$$ \begin{align} (x+y) &= (a_1 + b_1) p_1 + \dots + (a_t + b_t) p_t \end{align} $$

Multiplication is defined and distributes as expected, but with the following constraints:

$$ p \cdot q = \left\{\begin{array}{ll} pq & \textrm{if q ends where p starts} \\ 0 & \textrm{else}\end{array} \right. $$

Take for instance the quiver

$$ \begin{CD} \Gamma: 1 @>\alpha>> 2 \end{CD} $$

The elements of the path algebra $k\Gamma$ are on the form $x = a_1 e_1 + a_2 e_2 + a_3 \alpha$. So what would the multiplication of two elements of this path algebra yield?

$$ \begin{align} x\cdot y &= (a_1 e_1 + a_2 e_2 + a_3)(b_1 e_1 + b_2 e_2 + b_3 \alpha) \\ &= (a_1 b_1 e_1^2 + a_1 b_2 e_1 e_2 + a_1 b_3 e_1 \alpha + a_2 b_1 e_2 e_1 + a_2 b_2 e_2^2 + a_2 b_3 e_2 \alpha + a_3b_1 \alpha e_1 + a_3 b_2 \alpha e_2 + a_3 b_3 \alpha^2) \end{align} $$

So just noting now that all paths consisting of subpaths not ending and starting in the same vertex respectively, goes to $0$, for instance $e_1 \alpha$, or $\alpha^2$, we get the following:

$$ \begin{align} x\cdot y &= (a_1 b_1 e_1^2 + a_2 b_2 e_2^2 + a_2 b_3 e_2 \alpha + a_3b_1 \alpha e_1) \end{align} $$

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:

$$ a_1e_1^3 + a_2e_2^3 + a_3\alpha e_1^2 + a_4 e_2\alpha e_1 + a_5e_2^2 \alpha $$

This would be (and you are encouraged to check this) the general result when multiplying three arbitrary elements of $k\Gamma$.

Proposition 1. $\dim_k k\Gamma < \infty \llra $ the quiver $\Gamma$ has no oriented cycles.

Proof $\Rightarrow$. Suppose $k\Gamma$ has finite dimension (as a vector space over $k$). It then permits a finite basis (since $k$ is a divison ring). But clearly the basis of a path algebra is its paths, so the quiver only has a finite number of paths. Suppose then that $\Gamma$ has an oriented cycle. Then we could construct a path not in the basis by taking any path starting in any vertex of our oriented cycle, and traverse this $n$ times until we have a new path not in the basis. But this is a contradiction as our basis can not generate our new path.

Proof $\Leftarrow$. Suppose $\Gamma$ has no oriented cycles. Then it contains a path of maximal length, say $n$, since eventually all paths will terminate - considering our finite vertex-set.Since we have no directed cycles, we can topologically sort the graph. Then clearly, the $i$th vertex of the graph can contain at most $n-i$ forward arrows. The possible ways to construct paths at each arrow is $(n-i)!$. And so we get the number of paths to be the sum of all these; a finite number. And so the basis of any element in $k\Gamma$ is finite, so the dimension is also finite.

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 have 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

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 $r_i / r_{i+1}$ 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 $\Leftarrow$.

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

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.

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 $\Lambda$ 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 $g$.

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 $\Rightarrow$: Any module can be written as a factor of a free module. Then construct a map form such a free module to an arbitrary module 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 $B'$.

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.

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 $f'=\pi_M f$ is an epimorphism, and that $\pi_M$ is an essential epimorphism. Thus $f$ is an epimorphism. Extend the commutative diagram to include $P \overset{\bar f}{\to} P/rP$. Show that $\bar f$ is an isomorphism, implying that $f$ is a projective cover (Proposition 32).

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 $\Lambda$ is a local ring if the non-invertible elements of $\Lambda$ is an ideal.

Proposition 35. Let $\Lambda$ be a local ring. Then $0$ and $1$ are the only idempotents.

Proof: Show that an invertible idempotent necessarily is $1$. Then show that any non-invertible idempotent leads to a contradiction by considering $e$ and $1-e$ (both non-invertible).

Indecomposable projectives

We know that if the endomorphism ring $\End_{\Lambda}(M)$ contains idempotents not $0$ or $1$, then $M$ is indecomposable, and vica versa. This leads us immediately to the following corollary.

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 $P/rP$ is simple, which means that $rP$ is maximal. But as $rP$ is the intersection of all maximal submodules, it must be the only one.

(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 $\Lambda \cong \End_{\Lambda}(_\Lambda\Lambda)^{op}$, and so it follows directly from proposition 37.

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 $P$ finitely generated projective $\Lambda$-module into a direct sum of $P_i$ indecomposable projective $\Lambda$-modules.

Sketch of proof: The top of $P$ is isomorphic to a direct sum of simple submodules. Construct a projective cover for each of these submodules, and then by uniqueness of covers we have our result.

Summary

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 $\CC$ contains a set of objects, $\Obj(\CC)$, and for each pair of objects in $\CC$, a set of morphisms between these: $\Hom_{\CC}(A, B)$. For such a morphism, we write $f:A\to B$. We furthermore require that there exists identity morphisms for each object, and that composition of morphisms are associative.

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 $\CC$ be a category and $A, B$ objects in $\CC$. Then a morphism $f: A \to B$ is an isomorphism if there exists a morphism $g: B \to A$ in $\Hom_{\CC}(A,B)$, such that $fg = 1_B$ and $gf = 1_A$. In other words, for an isomorphism, we require that there exists another morphism that composes to the identity. It is (perhaps) not hard to show that this definition is in coherence with definitions we have used for groups, rings and modules.

A subcategory $\DD$ of $\CC$, is a category such that $\Obj(\DD) \subseteq \Obj(\CC)$ and $\Hom_{\DD}(A,B) \subseteq \Hom_{\CC}(A,B)$. We also need that composition in $\DD$ is just the restriction of the composition in $\CC$. If $\Hom_{\DD}(A,B) = \Hom_{\CC}(A,B)$, then the subcategory is called a full 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 $F: \CC \to \DD$ associates each object $C \in \Obj(\CC)$ with an object $D \in \Obj(\DD)$, and each morphism $f:A\to B$ in $\Hom_{\CC}(A,B)$ a morphism $F(f): F(A) \to F(B)$ in $\Hom_{\DD}(F(A), F(B))$. We also require that the identity morphism for $A$ maps to the identity morphism in $1_A$. Finally, we require $F(gf) = F(g)F(f)$.

If instead the functor is contravariant, we require a morphism $f: A\to B$ to be mapped to $g: F(B) \to F(A)$, and $F(gf) = F(f)F(g)$.

Let $R$ be a commutative ring, and $\CC$ a category. We say that the category $\CC$ is (a)preadditive(R-Category) if $\Hom_{\CC}(A,B)$ is an abelian group(R-Module) for all $A, B$ in $\CC$, and if morphism-compositions are bilinear (R-Bilinear).

If we have two preadditive categories, a functor between them is said to be additive if $\Hom_{\CC}(A,B)\to \Hom_{\DD}(F(A), F(B))$ is a homomorphism between groups.

As a final definition, we consider morphisms of functors.

Let $\CC, \DD$ be categories, with two functors $F, G: \CC \to \DD$. Then $\phi: F \to G$ is a morphism of functors if for each object $C \in \CC$, there is a morphism $$\phi_C: F(C) \to G(C),\quad \textrm{ in } \DD$$ such that for each morphism $f:A\to B$ in $\CC$, the diagram

$$ \begin{CD} F(A) @>\phi_A>> G(A)\\ @VF(f)VV @VVG(f) V \\ F(B) @>\phi_B>> G(B) \end{CD} $$

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 $\CC$ and $\DD$ be (R-)categories, and $F: \CC \to \DD$ an (R-)functor. $F$ is an equivalence if there exists an (R-)functor $G: \DD \to \CC$ such that

$$ HF \cong id_{\CC} \\ FH \cong id_{\DD} $$

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 $F: \CC \to \DD$ be a (R-)functor.

Then $F$ is full if $F: \Hom_{\CC}(A,B) \to \Hom_{\DD}(A,B)$ is surjective for all $A,B$.

$F$ is faithful if $F: \Hom_{\CC}(A,B) \to \Hom_{\D}(F(A), F(B)$ is injective for all $A, B$.

$F$ is dense if for each object $D \in \DD$, there exists an object $C \in \CC$ such that $F(C) \cong D$.

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

Morita equivalence

The previous result and its preceding definitions, leads us to the following definition:

Let $\Lambda$ and $\Sigma$ be two (arbitrary) rings. $\Lambda$ and $\Sigma$ are morita equivalent if $\lmod \Lambda$ and $\lmod \Sigma$ are equivalent categories.

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

Definition

Let $\Lambda$ be a ring, and $I \in \lmod \Lambda$. Then $I$ is injective if for every morphism $i: A \to B$ in $\lmod \Lambda$ and every homomorphism $f: A\to I$ there exists an $h: B \to I$ such that $hi = f$.

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 $A \subset X$ both be $\Lambda$-modules. Then $A$ is an essential submodule of $X$, if for each non-zero submodule $B$ of $X$, the following holds:

$$ A \cap B \neq (0)$$

A monomorphism $i: A \to X$ is an essential monomorphism if $i(A)$ is an essential submodule of $X$. If $I$ is injective also, $i$ is said to be an injective envelope. This is clearly the dual notion of a projective cover.

Socle

Let $\Lambda$ be an artin $R-$algebra, and $A \in \lmod \Lambda$. The socle of $A$, denoted $\Soc A$, is the sum of all simple submodules of $A$.

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