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 $R=ℤ.$ Then every submodule of ${}_{R}R$ is generated by one element. The ring $ℤ$ is noetherian but not artinian: $ℤ\supseteq pℤ\supseteq {p}^{2}ℤ\supseteq \cdots .$

If

1. The annihilator of $M$ is $\mathrm{ann}\left(M\right)=\left\{r\in R\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}rM=0\right\}.$
2. The radical of $M$ is $Rad(M) = ⋂Pmax Pmax,$ the intersection of the maximal proper submodules of $M.$
3. The socle of $M$ is $soc(M) = ∑Pmin Pmin,$ 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 $0 = soc0(M) ⊆ soc1(M) ⊆ ⋯$ where ${\mathrm{soc}}^{1}\left(M\right)=M$ and ${\mathrm{soc}}^{i}\left(M\right)$ is determined by $soci(M) soci-1(M) = soc( M soci(M) ).$
6. The radical series of $M$ is $0 = Rad0(M) ⊇ Rad1(M) ⊇ ⋯$ 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 $socj(M) ⊇ Radn-j(M), 0≤j≤n.$

Let $R$ be a ring.

1. $\mathrm{Rad}\left(R\right)=\bigcap _{{L}_{\mathrm{max}}}{L}_{\mathrm{max}},$ 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. $\mathrm{Rad}\left(R\right)=\left\{x\in R\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}1-axb\phantom{\rule{.5em}{0ex}}is invertible for all\phantom{\rule{.5em}{0ex}}a,b\in R\right\}.$
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 $\mathrm{ann}\left(m\right)=\left\{r\in R\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}rm=0\right\}$ is a maximal left ideal of $R$ because $R/\mathrm{ann}\left(m\right)\cong M.$ The primitive ideal $ann(M) = {r∈R | rM=0} = ⋂m∈M ann(m).$ Let $s\in \mathrm{Rad}\left(R\right).$ Then $R\left(1-x\right)=R$ since $1-x$ is not in any maximal left ideal. So $t\left(1-x\right)=1$ for some $t\in R.$ So $1-t=-tx\in \mathrm{Rad}\left(R\right).$ So $1-\left(1-t\right)=t$ 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 $\mathrm{Rad}\left(R\right)=\left\{x\in R\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}1-axb\phantom{\rule{.5em}{0ex}}is invertible for all\phantom{\rule{.5em}{0ex}}a,b\in R\right\}.$ Assume $1-axb$ is invertible for all $a,b\in R.$ Let ${L}_{\mathrm{max}}$ be a maximal left ideal not containing $x.$ Then $1=ax+l$ for some So $1-ax\in {L}_{\mathrm{max}}.$ So ${L}_{\mathrm{max}}=R$ which is a contradiction. So $x$ is an element of every maximal left ideal. So $\left\{x\in R\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}1-axb\phantom{\rule{.5em}{0ex}}is invertible for all\phantom{\rule{.5em}{0ex}}a,b\in R\right\}\subseteq \mathrm{Rad}\left(R\right).$ 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 $\left(1+x+{x}^{2}+\cdots +{x}^{k-1}\right)\left(1-x\right)=1$ 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 $\mathrm{Rad}\left(R\right)M=M,$ $mk = ∑i=1k aimi, with ai∈Rad(R).$ So $\left(1-{a}_{k}\right){m}_{k}=\sum _{i=1}^{k-1}{a}_{i}{m}_{i}.$ But $1-{a}_{k}$ has a left inverse in $R.$ So ${m}_{k}=\sum _{i=1}^{k-1}{\left(1-{a}_{k}\right)}^{-1}{a}_{i}{m}_{i},$ 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) $\mathrm{soc}\left(M\right)=\left\{m\in M\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}\mathrm{Rad}\left(R\right)m=0\right\}.$
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 $\left(1-a\right)x=0.$ 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 $R⊕⋯⊕R → M.$ 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, $\mathrm{Rad}\left(R\right)\left(M/{N}_{\mathrm{max}}\right)=0$ for every maximal proper submodule ${N}_{\mathrm{max}}\subseteq M.$ So $\mathrm{Rad}\left(R\right)M\subseteq {N}_{\mathrm{max}}$ for every ${N}_{\mathrm{max}}.$ 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, $\mathrm{Rad}\left(M/{M}_{1}\right)=0.$ So $\mathrm{Rad}\left(M\right)\subseteq {M}_{1}=\mathrm{Rad}\left(R\right)M.$ (bc) The set $N = {m∈M | Rad(R)m=0}$ 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
2. If $M=\underset{\lambda \in \stackrel{^}{A}}{⨁}{\left({R}^{\lambda }\right)}^{\oplus {m}_{\lambda }}$ is a finite direct sum of simple modules then $EndR(M) = ⨁λ∈A^ Mmλ (𝔻λ),$ 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, $\mathrm{ker}\phi$ 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 $M=\underset{\lambda \in \stackrel{^}{A}}{⨁}\underset{i=1}{\overset{{m}_{\lambda }}{⨁}}{R}^{\lambda ,i},$ with ${R}^{\lambda ,i}\cong {R}^{\lambda },$ for $1\le i\le {m}_{\lambda },$ then $EndR(M) = ⨁λ∈A^ ⨁ i,j=1 mλ EndR( Rλ,i,Rλ,j ) = ⨁λ∈A^ Mmλ (𝔻λ).$ $\square$

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 $\mathrm{soc}\left(M/N\right)=M/N.$

 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 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 $Rad(M) = ⋂Ni = 0.$ Then $φ: M → M/N1 ⊕⋯⊕ M/Nk m ↦ ( m+N1 ,..., m+Nk )$ has $\mathrm{ker}\phi =0.$ 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. $R\cong \underset{\lambda \in \stackrel{^}{A}}{⨁}{M}_{{d}_{\lambda }}\left({𝔻}_{\lambda }\right),$ where $\stackrel{^}{A}$ 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 is a ring isomorphism. Thus, by Schur's lemma, $Rop ≅ EndR( RR ) ≅ ⨁λ∈A^ Mdλ (𝔻λ), and thus R ≅ ⨁λ∈A^ Mdλ (𝔻λ) op ≅ ⨁λ∈A^ Mdλ (𝔻λ).$ $\square$

## 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 $n\in {ℤ}_{\ge 1},$ $𝔻$ 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 $U(α,V) = {β∈End𝔻(U) | ResVU(β) = ResVU(α)} open$ 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 $Endℂ(U) ≅ M∞(ℂ) = { infinite matrices with a finite number of nonzero entries in each column }.$ Let $I = {finite rank elements of M∞(ℂ)} = { α∈Endℂ(U) | im(α) is finite dimensional }$ and let $R = {n⋅1+l | n∈ℤ, l∈I}.$ Then $R$ is a dense subring of ${\mathrm{End}}_{ℂ}\left(U\right),$ $ℂ=EndR(U) and R≠Endℂ(U).$

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 $ann(x1,...,xn)u = 0 ⇔ u∈𝔻-span { x1,...,xn }.$

 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 If ${x}_{n}\notin 𝔻-span\left\{{x}_{1},...,{x}_{n}\right\}$ then, by the lemma, $\mathrm{ann}\left({x}_{1},...,{x}_{n-1}\right){x}_{n}\ne 0.$ Since $\mathrm{ann}\left({x}_{1},...,{x}_{n-1}\right){x}_{n}$ is a nonzero $R-$submodule of $U$ and $U$ is simple $\mathrm{ann}\left({x}_{1},...,{x}_{n-1}\right){x}_{n}=U.$ So Then (b) Let $R$ be artinian and let $U$ be a simple module. Let $I$ be a minimal element of ${ann(x1,...,xk) | x1,...,xk∈U}$ and let ${x}_{1},...,{x}_{k}$ be the finite subset of $U$ such that $I=\mathrm{ann}\left({x}_{1},...,{x}_{k}\right).$ Let $u\in I.$ If $\mathrm{ann}\left({x}_{1},...,{x}_{k}\right)u\ne 0$ then $\mathrm{ann}\left({x}_{1},...,{x}_{k},u\right)\subseteq I$ and $\mathrm{ann}\left({x}_{1},...,{x}_{k},u\right)\ne I,$ a contradiction to the minimality of $I.$ So $\mathrm{ann}\left({x}_{1},...,{x}_{n}\right)u=0.$ So $u\in 𝔻-span\left\{{x}_{1},...,{x}_{n}\right\}.$ So $U$ is finite dimensional. Now (c) follows from (b). $\square$

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

Let $A$ be an algebra over a field $𝔽.$ The radical of $A$ is the intersection of the maximal left ideals of $A,$ $Rad(A) = ⋂Lmax Lmax.$

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

If $\stackrel{_}{t}:A\to ℂ$ is a trace on $A$ then $Rad(t_) = {a∈A | t_(ab)=0 for all b∈A}.$

$\mathrm{Rad}\left(A\right)=\mathrm{Rad}\left(\stackrel{_}{t}\right),$ 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.