Radicals

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 15 June 2012

Examples.

1. Let $𝔽$ be a field. A finite dimensional vector space is both noetherian and artinian. An infinite dimensional vector space $V$ has $\mathrm{Rad}\left(V\right)=0,$ $\mathrm{soc}\left(V\right)$ and is neither noetherian or artinian.
2. Let $R=\mathbb{Z}.$ Then every submodule of ${}_{R}R$ is generated by one element. The ring $\mathbb{Z}$ is noetherian but not artinian: $\mathbb{Z}\supseteq p\mathbb{Z}\supseteq p^2\mathbb{Z}\supseteq \cdots .$ $$\mathrm{Rad}\left(\mathbb{Z}\right)=\bigcap _{L_{\max }}{L}_{\max }=\bigcap _{p prime}p\mathbb{Z}=0.$$

Radicals and socles

If $m\in M, \mathrm{ann}\left(m\right)=\{r\in R|rm=0\}.$

1. The annihilator of $M$ is $\mathrm{ann}\left(M\right)=\{r\in R|rM=0\}.$
2. The radical of $M$ is $$\mathrm{Rad}\left(M\right)=\bigcap _{P_{\max }}{P}_{\max },$$ the intersection of the maximal proper submodules of $M.$
3. The socle of $M$ is $$\mathrm{soc}\left(M\right)=\sum _{P_{\min }}{P}_{\min },$$ the sum of the simple submodules of $M.$
4. The head of $M$ is $M/\mathrm{Rad}\left(M\right).$
5. The socle series of $M$ is $$0={\mathrm{soc}}^0\left(M\right)\subseteq {\mathrm{soc}}^1\left(M\right)\subseteq \cdots$$ where ${\mathrm{soc}}^1\left(M\right)=M$ and ${\mathrm{soc}}^i\left(M\right)$ is determined by $$\frac{{\mathrm{soc}}^i\left(M\right)}{{\mathrm{soc}}^{i-1}\left(M\right)}=\mathrm{soc}\left(\frac{M}{{\mathrm{soc}}^i\left(M\right)}\right).$$
6. The radical series of $M$ is $$0={\mathrm{Rad}}^0\left(M\right)\supseteq {\mathrm{Rad}}^1\left(M\right)\supseteq \cdots$$ where ${\mathrm{Rad}}^i\left(M\right)=\mathrm{Rad}\left({\mathrm{Rad}}^{i-1}\left(M\right)\right).$
7. The socle length of $M$ is the smallest positive integer $n$ such that ${\mathrm{soc}}^n\left(M\right)=M$ and ${\mathrm{soc}}^{n-1}\left(M\right)\ne M.$
8. The radical length of $M$ is the smallest positive integer $n$ such that ${\mathrm{Rad}}^n\left(M\right)=0$ and ${\mathrm{Rad}}^{n-1}\left(M\right)\ne 0.$
9. The socle layers of $M$ are ${\mathrm{soc}}^k\left(M\right)/{\mathrm{soc}}^{k-1}\left(M\right).$
10. The radical layers of $M$ are ${\mathrm{Rad}}^{k-1}\left(M\right)/{\mathrm{Rad}}^k\left(M\right).$

If $M$ has socle length $n$ then $M$ has radical length $n$ and $${\mathrm{soc}}^j\left(M\right)\supseteq {\mathrm{Rad}}^{n-j}\left(M\right),0\le j\le n.$$

Let $R$ be a ring.

1. $\mathrm{Rad}\left(R\right)=\bigcap _{L_{\max }}{L}_{\max },$ the intersection of the maximal left ideals of $R.$
2. $\mathrm{Rad}\left(R\right)=\bigcap _{I_{\mathrm{prim}}}{I}_{\mathrm{prim}},$ the intersection of the primitive two-sided ideals of $R.$
3. $\mathrm{Rad}\left(R\right)=\{x\in R|1-axbis\; invertible\; for\; alla,b\in R\}.$
4. $\mathrm{Rad}\left(R\right)$ contains all nilpotent ideals.

		Proof.
	This is a restatement of the definition of $\mathrm{Rad}\left(R\right),$ since the submodules of ${}_{R}R$ are the left ideals of $R.$ If $M$ is a simple $R-$module and $m\in M$ then $\mathrm{ann}\left(m\right)=\{r\in R|rm=0\}$ is a maximal left ideal of $R$ because $R/\mathrm{ann}\left(m\right)\cong M.$ The primitive ideal $$\mathrm{ann}\left(M\right)=\{r\in R|rM=0\}=\bigcap _{m\in M}\mathrm{ann}\left(m\right).$$ Let $s\in \mathrm{Rad}\left(R\right).$ Then $R(1-x)=R$ since $1-x$ is not in any maximal left ideal. So $t(1-x)=1$ for some $t\in R.$ So $1-t=-tx\in \mathrm{Rad}\left(R\right).$ So $1-(1-t)=t$ has a left inverse, which must be $1-x.$ So $1-x$ is invertible in $R.$ By (b) $\mathrm{Rad}\left(R\right)$ is an ideal and so $1-axb$ is invertible for every $a,b\in R.$ So $\mathrm{Rad}\left(R\right)=\{x\in R|1-axbis\; invertible\; for\; alla,b\in R\}.$ Assume $1-axb$ is invertible for all $a,b\in R.$ Let $L_{\max }$ be a maximal left ideal not containing $x.$ Then $1=ax+l$ for some $a\in R, p\in L_{\max }.$ So $1-ax\in L_{\max }.$ So $L_{\max }=R$ which is a contradiction. So $x$ is an element of every maximal left ideal. So $\{x\in R|1-axbis\; invertible\; for\; alla,b\in R\}\subseteq \mathrm{Rad}\left(R\right).$ Let $N$ be a nilpotent ideal with $N^k=0.$ If $x\in N$ then $x^k\in N^k=0$ and so $x^k=0.$ Then $(1+x+x^2+\cdots +x^{k-1})(1-x)=1$ and so $1-x$ is invertible. Thus, since $N$ is an ideal, $1-axb$ is invertible for every $a,b\in R.$ Thus, by (c), $N\subseteq \mathrm{Rad}\left(R\right).$ $\square$

The proof of (bb) and (bc) of the following theorem uses:

(Nakayama's lemma) If $M$ is a finitely generated $R-$module and $\mathrm{Rad}\left(R\right)M=M$ then $M=0.$

		Proof.
	Assume $M\ne 0.$ Let $m_1,...,m_k$ be a minimal generating set for $M.$ Since $\mathrm{Rad}\left(R\right)M=M,$ $$m_k=\sum _{i=1}^k{a}_i{m}_i,with{a}_i\in \mathrm{Rad}\left(R\right).$$ So $(1-a_k)m_k=\sum _{i=1}^{k-1}{a}_i{m}_i.$ But $1-a_k$ has a left inverse in $R.$ So $m_k=\sum _{i=1}^{k-1}{(1-a_k)}^{-1}{a}_i{m}_i,$ which contradicts the minimality. So $M=0.$ $\square$

Let $R$ be an artinian ring. Then

1. (a) $\mathrm{Rad}\left(R\right)$ is the largest nilpotent ideal of $R.$
2. (b) If $M$ is a finitely generated $R-$module then
   1. (ba) $M$ is noetherian and artinian,
   2. (bb) $\mathrm{Rad}\left(M\right)=\mathrm{Rad}\left(R\right)M,$
   3. (bc) $\mathrm{soc}\left(M\right)=\{m\in M|\mathrm{Rad}\left(R\right)m=0\}.$
3. (c) $R$ is noetherian.

		Proof.
	(a) Let $n$ be such that $\mathrm{Rad}{\left(R\right)}^n=\mathrm{Rad}{\left(R\right)}^{2n}.$ If $\mathrm{Rad}{\left(R\right)}^n\ne 0$ then there is a minimal left ideal with $\mathrm{Rad}{\left(R\right)}^{n}I\ne 0$ (since $\mathrm{Rad}{\left(R\right)}^n\mathrm{Rad}{\left(R\right)}^n\ne 0$ ). Let $x\in I,$ $x\ne 0,$ be such that $\mathrm{Rad}{\left(R\right)}^{n}x\ne 0.$ By minimality, $I=\mathrm{Rad}{\left(R\right)}^{n}x=\mathrm{Rad}{\left(R\right)}^n\mathrm{Rad}{\left(R\right)}^{n}x.$ So $x=ax,$ with $a\in \mathrm{Rad}\left(R\right).$ So $(1-a)x=0.$ Since $1-a$ is invertible in $R,$ $x=0.$ But this is a contradiction. So $\mathrm{Rad}{\left(R\right)}^n=0.$ So $\mathrm{Rad}\left(R\right)$ is a nilpotent ideal. (ba) Let $M_i=\mathrm{Rad}{\left(R\right)}^{i}M.$ Then, since $M$ is finitely generated and $R$ is artinian, there is a surjective homomorphism $$R\oplus \cdots \oplus R\to M.$$ Thus $M$ is artinian. So $M_i/M_{i+1}$ is artinian and $\mathrm{Rad}\left(R\right)$ acts by 0. So $M_i/M_{i+1}$ is a $R/\mathrm{Rad}\left(R\right)-$module and thus $M_i/M_{i+1}$ is a finite direct sum of simple submodules. So, by (a), $M$ has a composition series and is both noetherian and artinian. (bb) By Nakayama's lemma, $\mathrm{Rad}\left(R\right)(M/N_{\max })=0$ for every maximal proper submodule $N_{\max }\subseteq M.$ So $\mathrm{Rad}\left(R\right)M\subseteq N_{\max }$ for every $N_{\max }.$ So $\mathrm{Rad}\left(R\right)M\subseteq \mathrm{Rad}\left(M\right).$ Since $M/M_1$ is a finite direct sum of simple modules, $\mathrm{Rad}(M/M_1)=0.$ So $\mathrm{Rad}\left(M\right)\subseteq M_1=\mathrm{Rad}\left(R\right)M.$ (bc) The set $$N=\{m\in M|\mathrm{Rad}\left(R\right)m=0\}$$ is a submodule of $M$ and $\mathrm{Rad}\left(R\right)N=0.$ Since $N$ is artinian, $N$ is a finite direct sum of simple submodules. So $\mathrm{soc}\left(M\right)\supseteq N.$ Nakayama's lemma implies that if $S$ is a simple module, then $\mathrm{Rad}\left(R\right)S=0.$ So $\mathrm{Rad}\left(R\right)\mathrm{soc}\left(M\right)=0.$ So $\mathrm{soc}\left(M\right)\subseteq N.$ (c) follows from (ba). $\square$

Semisimplicity

Let $M$ be an $R-$module. Then $M$ has a simple submodule.

(Schur's lemma)

1. If $R^{\lambda }$ are $R^{\mu }$ are simple $R-$modules then $${\mathrm{Hom}}_R(R^{\lambda },R^{\mu })=0,if\; R^{\lambda }\ncong R^{\mu },and{\mathrm{End}}_R\left(R^{\lambda }\right)={𝔻}_{\lambda }is\; a\; division\; ring.$$
2. If $M=\underset{\lambda \in \hat{A}}{⨁}{\left(R^{\lambda }\right)}^{\oplus m_{\lambda }}$ is a finite direct sum of simple modules then $${\mathrm{End}}_R\left(M\right)=\underset{\lambda \in \hat{A}}{⨁}{M}_{m_{\lambda }}\left({𝔻}_{\lambda }\right),$$ where ${𝔻}_{\lambda }={\mathrm{End}}_R\left(R^{\lambda }\right)$ are division rings.

		Proof.
	Let $\phi :R^{\lambda }\to R^{\mu }$ be a homomorphism. Then, since $R^{\lambda }$ and $R^{\mu }$ are simple, $\ker \phi$ is either 0 or $R^{\lambda },$ and $\mathrm{im}\phi$ is either 0 or $R^{\mu }.$ So $\phi$ is either 0 or an isomorphism. If $M=\underset{\lambda \in \hat{A}}{⨁}\underset{i=1}{\overset{m_{\lambda }}{⨁}}{R}^{\lambda ,i},$ with $R^{\lambda ,i}\cong R^{\lambda },$ for $1\le i\le m_{\lambda },$ then $${\mathrm{End}}_R\left(M\right)=\underset{\lambda \in \hat{A}}{⨁}\underset{i,j=1}{\overset{m_{\lambda }}{⨁}}{\mathrm{End}}_R(R^{\lambda ,i},R^{\lambda ,j})=\underset{\lambda \in \hat{A}}{⨁}{M}_{m_{\lambda }}\left({𝔻}_{\lambda }\right).$$ $\square$

Let $M$ be an $R-$module.

1. $\mathrm{soc}\left(M\right)=M$ if and only if for every submodule $N\subseteq M$ there is a submodule $N\prime \subseteq M$ with $M=N\oplus N\prime .$
2. Let $N$ be a submodule of $M.$ If $\mathrm{soc}\left(M\right)=M$ then $\mathrm{soc}\left(N\right)=N$ and $\mathrm{soc}(M/N)=M/N.$

		Proof.
	(a) ⇐) If $\mathrm{soc}\left(M\right)\ne M$ then $M=\mathrm{soc}\left(M\right)\oplus N\prime .$ Let $N$ be a simple submodule of $N\prime$ (the existence of $N$ is nontrivial and uses Zorn's lemma, see Theorem ???). Then $\mathrm{soc}\left(M\right)+N\ne \mathrm{soc}\left(M\right),$ but this is a contradiction to the definition of $\mathrm{soc}\left(M\right).$ ⇒) Let $N$ be a submodule of $M$ and let $N\prime =\sum _{P\cap N=0}P$ be the sum of the simple submodules $P$ of $M$ such that $P\cap N=0.$ Then $N\cap N\prime =0$ since, for a simple submodule $P$ of $M,$ $P\cap N=P$ or $P\cap N=0.$ Since $N+N\prime \supseteq \mathrm{soc}\left(M\right)=M, N+N\prime =M.$ So $M=N\oplus N\prime .$ $\square$

The following are equivalent:

1. $M$ is a finite direct sum of simple submodules.
2. $M$ is artinian and $\mathrm{soc}\left(M\right)=M.$
3. $M$ is noetherian and $\mathrm{soc}\left(M\right)=M.$
4. $M$ has a finite composition series and $\mathrm{soc}\left(M\right)=M.$
5. $M$ is finitely generated and $\mathrm{soc}\left(M\right)=M.$
6. $M$ is artinian and $\mathrm{Rad}\left(M\right)=0.$

		Proof.
	The implications (a)⇔(b), (a)⇔(c), (a)⇔(d) follow directly from Proposition ??? and Proposition ???a. (a)⇒(f) follows directly from the definitions. (f)⇒(a): Let $N_i$ be a finite (by DCC) number of maximal propoer submodules such that $$\mathrm{Rad}\left(M\right)=\bigcap N_i=0.$$ Then $$\begin{array}{rrcl}\phi :& M& \to & M/N_1\oplus \cdots \oplus M/N_k\\ & m& \mapsto & (m+N_1,...,m+N_k)\end{array}$$ has $\ker \phi =0.$ So $M\cong \mathrm{im}\left(M\right)$ is finite length and $\mathrm{soc}\left(M\right)=M.$ So $M$ is a direct sum of simple submodules. (c)⇒(e) since $M$ is noetherian implies that $M$ is finitely generated. (e)⇒(c): Let $N$ be a submodule of $M$ and let $N\prime$ be a complement. Then $N\cong M/N\prime$ and thus, since $M$ is finitely generated. Thus every submodule of $M$ is finitely generated. So $M$ is noetherian. $\square$

(Artin-Wedderburn) The following are equivalent:

1. $R$ is artinian and $\mathrm{Rad}\left(R\right)=0,$
2. ${}_{R}R$ is a finite direct sum of simple modules,
3. $R\cong \underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left({𝔻}_{\lambda }\right),$ where $\hat{A}$ is a finite index, $d_{\lambda }$ are positive integers, and ${𝔻}_{\lambda }$ are division rings.

		Proof.
	(a)⇔(b) is a consequence of Proposition ???. (a)⇐(c) is a consequence of the fact that the simple $M_{d_{\lambda }}\left({𝔻}_{\lambda }\right)$ module is ${𝔻}_{\lambda }^{d_{\lambda }}$ the vector space of column vectors of length $d_{\lambda }.$ (a)⇒(c): The map $$\begin{array}{rcl}{R}^{\mathrm{op}}& \to & {\mathrm{End}}_R\left({}_{R}R\right)\\ r& \mapsto & {\phi }_r\end{array}where{\phi }_r\left(x\right)=xr,for\; x\in R,$$ is a ring isomorphism. Thus, by Schur's lemma, $$R^{\mathrm{op}}\cong {\mathrm{End}}_R\left({}_{R}R\right)\cong \underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left({𝔻}_{\lambda }\right),and\; thusR\cong {\left(\underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left({𝔻}_{\lambda }\right)\right)}^{\mathrm{op}}\cong \underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left({𝔻}_{\lambda }\right).$$ $\square$

Radicals and finiteness conditions for rings

Let $R$ be a ring.

1. The ring $R$ is noetherian if ${}_{R}R$ is noetherian.
2. The $R$ is artinian if ${}_{R}R$ is artinian.
3. A left ideal of $R$ is a submodule of ${}_{R}R.$
4. An ideal $I$ of $R$ is primitive if $I=\mathrm{ann}\left(M\right)$ for a simple $R-$module $M.$

Simple and almost simple rings

1. The ring $R$ is primitive if 0 is a primitive ideal.
2. The ring $R$ is semiprimitive if $\mathrm{Rad}\left(R\right)=0.$
3. The ring $R$ is simple if its only ideals are 0 and $R.$
4. The ring $R$ is prime if $A,B$ are ideals with $AB=0$ then $A=0$ or $B=0.$
5. An ideal $P$ is prime if $R/P$ is a prime ring.
6. The ring $R$ is semiprime if 0 is the only nilpotent ideal.

Let $R$ be a ring and let $\mathrm{Spec}\left(R\right)$ be the set of prime ideals of $R.$

1. $R$ is semiprime if and only if $\bigcap _{𝔭\in \mathrm{Spec}\left(R\right)}𝔭=0.$
2. $R$ is primitive if and only if $R$ is a dense subring of ${\mathrm{End}}_{𝔻}\left(U\right)$ for some $𝔻-$vector space $U.$
3. $R$ is artinian and semiprime if and only if $R$ is artinian and semiprimitive.
4. $R$ is artinian and primitive if and only if $R$ is artinian and simple.
5. $R$ is artinian and primitive if and only if $R\cong M_n\left(𝔻\right),$ for some $n\in {\mathbb{Z}}_{\ge 1},$ $𝔻$ a division ring.

Burnside's theorem and Jacobson density

A subring $R$ of ${\mathrm{End}}_{𝔻}\left(U\right)$ is dense if for every $\alpha \in {\mathrm{End}}_{𝔻}\left(U\right)$ and every finitely generated $V\subseteq U$ there is an $r\in R$ with ${\mathrm{Res}}_V^U\left(r\right)={\mathrm{Res}}_V^U\left(\alpha \right).$ Define a topology on ${\mathrm{End}}_{𝔻}\left(U\right)$ by making $$U(\alpha ,V)=\{\beta \in {\mathrm{End}}_{𝔻}\left(U\right)|{\mathrm{Res}}_V^U\left(\beta \right)={\mathrm{Res}}_V^U\left(\alpha \right)\}open$$ for each $\alpha \in {\mathrm{End}}_{𝔻}\left(U\right)$ and each finitely generated $V\subseteq U.$ Then $R$ is dense is ${\mathrm{End}}_{𝔻}\left(U\right)$ if ${\mathrm{End}}_{𝔻}\left(U\right)$ is the closure of $R,$ $\stackrel{\_}{R}={\mathrm{End}}_{𝔻}\left(U\right).$

Example. Consider an infinite dimensional vector space $U$ with basis $u_1,u_2,....$ Then $$\begin{array}{rcl}{\mathrm{End}}_{ℂ}\left(U\right)& \cong & M_{\infty }\left(ℂ\right)\\ & =& \left\{infinite\; matrices\; with\; a\; finite\; number\; of\; nonzero\; entries\; in\; each\; column\right\}.\end{array}$$ Let $$\begin{array}{rcl}I& =& \left\{finite\; rank\; elements\; of{M}_{\infty }\left(ℂ\right)\right\}\\ & =& \{\alpha \in {\mathrm{End}}_{ℂ}\left(U\right)|\mathrm{im}\left(\alpha \right)is\; finite\; dimensional\}\end{array}$$ and let $$R=\{n\cdot 1+l|n\in \mathbb{Z},l\in I\}.$$ Then $R$ is a dense subring of ${\mathrm{End}}_{ℂ}\left(U\right),$ $$ℂ={\mathrm{End}}_R\left(U\right)andR\ne {\mathrm{End}}_{ℂ}\left(U\right).$$

Let $U$ be a simple $R-$module and let $\mathrm{Im}\left(R\right)$ be the image of $R$ in $\mathrm{End}\left(U\right).$ Let $𝔻={\mathrm{End}}_R\left(U\right),$ a division ring.

1. $\mathrm{Im}\left(R\right)$ is a dense subring of ${\mathrm{End}}_{𝔻}\left(U\right).$
2. If $R$ is artinian then $\mathrm{Im}\left(R\right)={\mathrm{End}}_{𝔻}\left(U\right).$

We will show that if $x_1,...,x_n\in U$ and $\alpha \in {\mathrm{End}}_{𝔻}\left(U\right)$ then there is an $r\in R$ with $r{x}_i=\alpha x_i$ for $1\le i\le n.$ The proof is by induction on $n$ using the following lemma.

Let $u\in U.$ Then $$\mathrm{ann}(x_1,...,x_n)u=0\iff u\in 𝔻-span\{x_1,...,x_n\}.$$

		Proof of Theorem 5.1.
	(a) Assume the lemma and assume that $x_1,...,x_n\in U$ and $\alpha \in {\mathrm{End}}_{𝔻}\left(U\right)$ are given. By the induction assumption, there is $r\prime \in R$ such that $$r\prime x_i=\alpha x_i,for\; 1\le i\le n-1.$$ If $x_n\notin 𝔻-span\{x_1,...,x_n\}$ then, by the lemma, $\mathrm{ann}(x_1,...,x_{n-1})x_n\ne 0.$ Since $\mathrm{ann}(x_1,...,x_{n-1})x_n$ is a nonzero $R-$submodule of $U$ and $U$ is simple $\mathrm{ann}(x_1,...,x_{n-1})x_n=U.$ So $$l{x}_n=(\alpha -r\prime )x_n,for\; some\; l\in \mathrm{ann}(x_1,...,x_{n-1}).$$ Then $$(r\prime -l)x_i=x_i,for\; 1\le i\le n-1,and(r\prime +l)x_n=x_n.$$ (b) Let $R$ be artinian and let $U$ be a simple module. Let $I$ be a minimal element of $$\left\{\mathrm{ann}(x_1,...,x_k)\right|x_1,...,x_k\in U\}$$ and let $x_1,...,x_k$ be the finite subset of $U$ such that $I=\mathrm{ann}(x_1,...,x_k).$ Let $u\in I.$ If $\mathrm{ann}(x_1,...,x_k)u\ne 0$ then $\mathrm{ann}(x_1,...,x_k,u)\subseteq I$ and $\mathrm{ann}(x_1,...,x_k,u)\ne I,$ a contradiction to the minimality of $I.$ So $\mathrm{ann}(x_1,...,x_n)u=0.$ So $u\in 𝔻-span\{x_1,...,x_n\}.$ So $U$ is finite dimensional. Now (c) follows from (b). $\square$

		Proof of Lemma 5.2.
	⇐) Trivial. ⇒) Assume $\mathrm{ann}(x_1,...,x_n)u=0.$ The proof is by induction on $n.$ Case 1. If $\mathrm{ann}(x_1,...,x_{n-1})x_n=0$ then $x_n\in \mathrm{span}\{x_1,...,x_{n-1}\}$ and so $\mathrm{ann}(x_1,...,x_{n-1})u=0.$ So $u\in \mathrm{span}\{x_1,...,x_{n-1}\}.$ Case 2. If $\mathrm{ann}(x_1,...,x_{n-1})x_n\ne 0$ then $\mathrm{ann}(x_1,...,x_{n-1})x_n=U.$ Define an $R-$module homomorphism $$\begin{array}{rrcl}\alpha :& U& \to & U\\ & l{x}_n& \mapsto & lu,\end{array}for\; l\in \mathrm{ann}(x_1,...,x_{n-1}).$$ If $l{x}_n=\kappa x_n$ then $l-\kappa \in \mathrm{ann}(x_1,...,x_{n-1})\cap \mathrm{ann}\left(x_n\right)=\mathrm{ann}(x_1,...,x_n).$ So $(l-\kappa )u=0$ and $lu=\kappa u$ which shows that $\alpha$ is well-defined. So $\alpha \in 𝔻={\mathrm{End}}_R\left(U\right).$ Now $\mathrm{ann}(x_1,...,x_{n-1})(u-\alpha x_n)=0$ and so, by the induction hypothesis, $u-\alpha x_n\in \mathrm{span}\{x_1,...,x_{n-1}\}.$ So $u\in \mathrm{span}\{x_1,...,x_n\}.$ $\square$

Radicals of algebras

Let $A$ be an algebra over a field $𝔽.$ The radical of $A$ is the intersection of the maximal left ideals of $A,$ $$\mathrm{Rad}\left(A\right)=\bigcap _{L_{\max }}{L}_{\max }.$$

Assume $A$ satisfies the descending chain condition on left ideals. Then $A$ is completely reducible if and only if $\mathrm{Rad}\left(A\right)=0.$

A nilpotent ideal is an ideal $I$ such that $I^k=0$ for some $k\in {\mathbb{Z}}_{>0}.$ A nilpotent element is an element $x\in A$ such that $x^k=0$ for some $k\in {\mathbb{Z}}_{>0}.$

If $\stackrel{\_}{t}:A\to ℂ$ is a trace on $A$ then $$\mathrm{Rad}\left(\stackrel{\_}{t}\right)=\{a\in A|\stackrel{\_}{t}\left(ab\right)=0for\; allb\in A\}.$$

$\mathrm{Rad}\left(A\right)=\mathrm{Rad}\left(\stackrel{\_}{t}\right),$ if $\stackrel{\_}{t}$ is the trace of a faithful representation of $A.$

Notes and References

These notes are originally from http://researchers.ms.unimelb.edu.au/~aram@unimelb/notespre2005.html the file http://researchers.ms.unimelb.edu.au/~aram@unimelb/Notespre2005/radicals12.25.03.pdf

References

[BouA] N. Bourbaki, Algebra I, Chapters 1-3, Elements of Mathematics, Springer-Verlag, Berlin, 1990.

[BouL] N. Bourbaki, Groupes et Algèbres de Lie, Chapitre IV, V, VI, Eléments de Mathématique, Hermann, Paris, 1968.

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