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MA3203: Ring Theory

$\require{AMScd}$ $ %\begin{CD} %K(X) @>{ch}>> H(X;\mathbb Q);\\ %@VVV @VVV \\ %K(Y) @>{ch}>> H(Y;\mathbb Q); %\end{CD} $ $\newcommand{\C}{\mathcal{C}}$ $\newcommand{\llra}{\Longleftrightarrow}$ $\DeclareMathOperator{\rad}{Rad}$ # 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. # 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 implication $$ A + X = B \implies X = B $$ >**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 ## Essential epimorphisms