## Posts Tagged ‘semisimple’

Throughout $k$ is a field, $K$ is the algebraic closure of $k$ and $A$ is a finite dimensional central simple $k$-algebra.

Lemma. $A \otimes_k K \cong M_n(K),$ for some integer $n.$

Proof. Let $S:=A \otimes_k K.$ By the first part of the corollary in this post we know that $S$ is simple. We also have

$Z(S) = Z(A) \otimes_k K = k \otimes_k K \cong K.$

It is easy to see that if $\{a_i \}$ is a $k$-basis for $A,$ then $\{a_i \otimes_k 1 \}$ is an $K$-basis for $S.$ Thus $\dim_K S = \dim_k A.$ So $S$ is a finite dimensional central simple $K$-algebra and hence, since $K$ is algebraically closed, $S \cong M_n(K),$ for some $n,$ by Remark 2 in this post. $\Box$

Theorem. If $A$ is a finite dimensional central simple $k$-algebra, then $\dim_k A$ is a perfect square.

Proof.  By the lemma, there exists an integer $n$ such that $A \otimes_k K \cong M_n(K).$ Thus

$\dim_k A = \dim_K A \otimes_k K = \dim_K M_n(K) = n^2. \ \Box$

Definition. The degree of $A$ is defined by $\deg A = \sqrt{\dim_k A}.$

Remark. Let $R$ be a finite dimensional $k$-algebra. Then $R$ is reduced if and only if $R$ is a finite direct product of finite dimensional division $k$-algebras. In this case, $\dim_k R=n^2\dim_k Z(R)$ for some integer $n \geq 1.$

Proof. obviously $R$ is (left) Artinian because $\dim_k R < \infty$ and so $J(R)$ is nilpotent. Thus $J(R)=(0)$ because $R$ is reduced and so $R$ is semisimple. The result now follows from the Artin-Wedderburn theorem. The converse is trivial. Finally, the fact that, by the above theorem, $\dim_{Z(D)}D$ is a perfect square for any finite dimensional division algebra $D,$ proves the last part of the remark.  $\Box$

## von Neumann Regular rings (1)

Posted: October 22, 2010 in Noncommutative Ring Theory Notes, von Neumann Regular rings
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Definition. A ring $R$ is called von Nemann regular, or just regular, if for every $a \in R$ there exists $x \in R$ such that $a=axa.$

Remark 1. Regular rings are semiprimitive. To see this, let $R$ be a regular ring. Let $a \in J(R),$ the Jacobson radical of $R,$ and choose $x \in R$ such that $a=axa.$ Then $a(1-xa)=0$ and, since $1-xa$ is invertible because $a$ is in the Jacobson radical of $R,$ we get $a=0.$

Examples 1. Every division ring is obviously regular because if $a = 0,$ then $a=axa$ for all $x$ and if $a \neq 0,$ then $a=axa$ for $x = a^{-1}.$

Example 2. Every direct product of regular rings is clearly a regular ring.

Example 3. If $V$ is a vector space over a division ring $D,$ then ${\rm End}_D V$ is regular.

Proof. Let $R={\rm End}_D V$ and $f \in R.$ There exist vector subspaces $V_1, V_2$ of $V$ such that $\ker f \oplus V_1 = {\rm im}(f) \oplus V_2 = V.$  So if $u \in V,$ then $u=u_1+u_2$ for some unique elements $u_1 \in {\rm im}(f)$ and $u_2 \in V_2.$ We also have $u_1 = v_1 + v$ for some unique elements $v_1 \in \ker f$ and $v \in V_1.$ Now define $g: V \longrightarrow V$ by $g(u)=v.$ It is obvious that $g$ is well-defined and easy to see that $g \in R$ and $fgf=f. \ \Box$

Example 4. Every semisimple ring is regular.

Proof. For a division ring $D$ the ring $M_n(D) \cong End_D D^n$ is regular by Example 3. Now apply Example 2 and the Wedderburn-Artin theorem.

Theorem. A ring $R$ is regular if and only if every finitely generated left ideal of $R$ is generated by an idempotent.

Proof. Suppose first that every finitely generated left ideal of $R$ can be generated by an idempotent. Let $x \in R.$ Then $I=Rx = Re$ for some idempotent $e.$ That is $x = re$ and $e=sx$ for some $r,s \in R.$ But then $xsx=xe=re^2=re=x.$ Conversely, suppose that $R$ is regular. We first show that every cyclic left ideal $I=Rx$ can be generated by an idempotent. This is quite easy to see: let $y \in R$ be such that $xyx=x$ and let $yx=e.$ Clearly $e$ is an idempotent and $xe=x.$ Thus $x \in Re$ and so $I \subseteq Re.$ Also $e=yx \in I$ and hence $Re \subseteq I.$ So $I=Re$ and we’re done for this part. To complete the proof of the theorem we only need to show that if $J=Rx_1 + Rx_2,$ then there exists some idempotent $e \in R$ such that $J=Re.$ To see this, choose an idempotent $e_1$ such that $Rx_1=Re_1.$ Thus $J=Re_1 + Rx_2(1-e_1).$  Now choose an idempotent $e_2$ such that $Rx_2(1-e_1)=Re_2$ and put $e_3=(1-e_1)e_2.$ See that $e_3$ is an idempotent, $e_1e_3=e_3e_1=0$ and $Re_2=Re_3.$ Thus $J=Re_1 + Re_3.$ Let $e=e_1+e_3.$ Then $e$ is an idempotent and $J=Re. \Box$

Corollary. If the number of idempotents of a regular ring $R$ is finite, then $R$ is semisimple.

Proof. By the theorem, $R$ has only a finite number of left principal ideals. Since every left ideal is a sum of left principal ideals, it follows that $R$ has only a finite number of left ideals and hence it is left Artinian. Thus $R$ is semisimple because $R$ is semiprimitive by Remark 1. $\Box$

Remark 2. The theorem is also true for finitely generated right ideals. The proof is similar.

Remark 3. Since, by the Wedderburn-Artin theorem, a commutative ring is semisimple if and only if it is a finite direct product of fields, it follows from the Corollary that if the number of idempotents of a commutative von Neumann regular ring $R$ is finite, then $R$ is a finite direct product of fields.

## Ring of endomorphisms (3)

Posted: June 9, 2010 in Noncommutative Ring Theory Notes, Ring of Endomorphisms
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Schur’s lemma states that if $A$ is a simple $R$ module, then $\text{End}_R(A)$ is a division ring. A similar easy argument shows that:

Example 6. For simple $R$-modules $A \ncong B$ we have $\text{Hom}_R(A,B)=\{0\}.$

Let’s generalize Schur’s lemma: let $M$ be a finite direct product of simple $R$-submodules. So $M \cong \bigoplus_{i=1}^k M_i^{n_i},$ where each $M_i$ is a simple $R$-module and $M_i \ncong M_j$ for all $i \neq j.$ Therefore, by Example 6 and Theorem 1, $\text{End}_R(M) \cong \bigoplus_{i=1}^k \mathbb{M}_{n_i}(D_i),$ where $D_i = \text{End}_R(M_i)$ is a division ring by Schur’s lemma. An important special case is when $R$ is a semisimple ring. (Note that simple submodules of a ring are exactly minimal left ideals of that ring.)

Theorem 2. (Artin-Wedderburn) Let $R$ be a semisimple ring. There exist a positive integer $k$ and division rings $D_i, \ 1 \leq i \leq ,$ such that $R \cong \bigoplus_{i=1}^k \mathbb{M}_{n_i}(D_i)$.

Proof. Obvious, by Example 1 and the above discussion. $\Box$

Some applications of Theorem 2.

1. A commutative semisimple ring is a finite direct product of fields.

2. A reduced semisimple ring is a finite direct product of division rings.

3. A finite reduced ring is a finite direct product of finite fields.

## Quotient rings; some facts (2)

Posted: January 2, 2010 in Noncommutative Ring Theory Notes, Quotient Rings
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For the first part see here.

6) If $R$ is simple, then $Z(R)=Z(Q).$

Proof. Let $x=s^{-1}a \in Z(Q).$ Then from $xs=sx$ we get $sa=as$ and thus $s^{-1}a=as^{-1}.$ Hence for every $b \in R$ we’ll have $s^{-1}ab=bs^{-1}a=bas^{-1},$ which gives us $abs=sba.$ Also, since $R$ is simple, $RsR=R,$ which means $\sum_{i=1}^n b_isc_i = 1,$ for some $b_i, \ c_i \in R.$ Thus $\sum_{i=1}^n sb_iac_i = \sum_{i=1}^n ab_isc_i = a=sx$ and so $x=\sum_{i=1}^n b_iac_i \in R.$ Therefore $x \in Z(R)$ which proves $Z(Q) \subseteq Z(R).$ Conversely, let $b \in Z(R)$ and $x=s^{-1}a \in Q$. Since $bs=sb,$ we have $s^{-1}b=bs^{-1}$ and thus $bx=bs^{-1}a=s^{-1}ba=s^{-1}ab=xb$ and so $b \in Z(Q). \Box$

7) The left uniform dimension of $R$ and $Q$ are equal.

Proof. We saw in the previous section that the left ideals of $Q$ are exactly in the form $QI,$ where $I$ is a left ideal of $R.$ Clearly $\sum QI_i$ is direct iff $\sum I_i$ is direct.

8) Let $N$ be a nilpotent ideal of $R$ and let $I$ be the right annihilator of $N$ in $R.$ Then $I$ is an essential left ideal of $R$ and hence $QI$ is an essential left ideal of $Q.$

Proof. Let $I$ be the right annihilator of $N$ in $R.$ For an essential left ideal $J$ of $R$ the left ideal $QJ$ of $Q$ is essential in $Q$ because for every non-zero left ideal $K$ of $R : (0) \neq \ Q(J \cap K) \subseteq QJ \cap QK.$ So we only need to prove the first part of the claim. Let $J$ be any non-zero left ideal of $R$ and put $n=\min \{k \geq 0 : \ N^k J \neq (0) \}.$ Then $(0) \neq N^n J \subseteq I \cap J.$

9) If $Q$ is semisimple, then $R$ is semiprime.

Proof. So we need to prove that $R$ has no non-zero nilpotent ideal. Suppose that $N$ is a nilpotent ideal of $R$ and let $I$ be the right annihilator of $N$ in $R.$ Since $Q$ is semisimple, $QI \oplus A = Q,$ for some left ideal $A$ of $Q.$ But, from the previous fact, we know that $QI$ is essential in $Q$ and thus $A=(0),$ i.e. $QI=Q.$ Thus $s^{-1}a=1,$ for some $a \in I=\text{r.ann}_R N.$ So $s=a$ and $Ns=Na=(0).$ Thus $N=Nss^{-1}=(0).$

We proved, in the previous section, that if $R$ is prime, then $Q$ is prime too.

10) If $Q$ is simple, then $R$ is prime.

Proof. Let $I,J$ be two non-zero ideals of $R.$ We need to show that $IJ \neq (0).$ We have $QIQ=Q,$ because $I \neq (0)$ and $Q$ is simple. Therefore $1=\sum_{i=1}^n x_ia_iy_i,$ for some $x_i,y_i \in Q$ and $a_i \in I.$ We can write $x_i = s^{-1}b_i,$ for some $b_i \in R.$ Then $s=\sum_{i=1}^n b_ia_iy_i \in IQ.$ So $IQ$ is a right ideal of $Q$ which contains a unit. Thus $IQ=Q.$ Similarly $JQ=Q$ and hence $IJQ=Q.$ As a result, $IJ \neq (0). \Box$

## Semiprimitivity of C[G]

Posted: December 2, 2009 in Group Algebras, Noncommutative Ring Theory Notes
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Notation. For a ring $R$ let $J(R)$ be the Jacobson radical of $R.$

Definition. Recall that if $k$ is a field and $G$ is a group, then the group algebra $k[G]$ has two structures. Firstly, as a vector space over $k,$ it has $G$ as a basis, i.e. every element of $k[G]$ is uniquely written as $\sum_{g \in G} x_g g,$ where $x_g \in k.$ In particular, $\dim_k k[G]=|G|,$ as cardinal numbers. Secondly, multiplication is also defined in $k[G].$ If $x = \sum_{g \in G} x_g g$ and $y = \sum_{g \in G} y_g g$ are two elements of $k[G],$ then we just multiply $xy$ in the ordinary fashion using distribution law. To be more precise, we define $xy = \sum_{g \in G} z_g g,$ where $z_g = \sum_{rs=g} x_r y_s.$

We are going to prove that $J(\mathbb{C}[G])=0,$ for every group $G.$

Lemma. $J(\mathbb{C}[G])$ is nil, i.e. every element of $J(\mathbb{C}[G])$ is nilpotent.

Proof. If $G$ is countable, we are done by this theorem. For the general case, let $\alpha \in J(\mathbb{C}[G]).$ So $\alpha =\sum_{i=1}^n c_ig_i,$ for some $c_i \in \mathbb{C}, \ g_i \in G.$ Let $H=\langle g_1,g_2, \cdots , g_n \rangle.$ Clearly $\alpha \in H$ and $H$ is countable. So to complete the proof, we only need to show that $\alpha \in J(\mathbb{C}[H]).$ Write $G = \bigcup_i x_iH,$ where $x_iH$ are the distinct cosets of $H$ in $G.$ Then $\mathbb{C}[G]=\bigoplus_i x_i \mathbb{C}[H],$ which means $\mathbb{C}[G]=\mathbb{C}[H] \oplus K,$ for some right $\mathbb{C}[H]$ module $K.$ Now let $\beta \in \mathbb{C}[H].$ Since  $\alpha \in J(\mathbb{C}[G]),$ there exists $\gamma \in \mathbb{C}[G]$ such that $\gamma (1 - \beta \alpha ) = 1.$ We also have $\gamma = \gamma_1 + \gamma_2,$ for some $\gamma_1 \in \mathbb{C}[H], \ \gamma_2 \in K.$ That now gives us $\gamma_1(1 - \beta \alpha)=1. \ \Box$

Theorem. $J(\mathbb{C}[G])=0,$ for any group $G.$

Proof. For any $x =\sum_{i=1}^n c_i g_i\in \mathbb{C}[G]$ define

$x^* = \sum_{i=1}^n \overline{c_i} g_i^{-1}.$

It’s easy to see that $xx^*=0$ if and only if $x=0$ and for all $x,y \in \mathbb{C}[G]: \ (xy)^*=y^*x^*.$ Now suppose that $J(\mathbb{C}[G]) \neq 0$ and let $0 \neq \alpha \in J(\mathbb{C}[G]).$ Put $\beta = \alpha \alpha^* \in J(\mathbb{C}[G]).$ By what I just mentioned $\beta \neq 0$ and $(\beta^m)^* = (\beta^*)^m=\beta^m,$ for all positive integers $m.$ By the lemma, there exists $k \geq 2$ such that $\beta^k = 0$ and $\beta^{k-1} \neq 0.$ Thus $\beta^{k-1} (\beta^{k-1})^* = \beta^{2k-2} = 0,$ which implies that $\beta^{k-1} = 0.$ Contradiction! $\Box$

Corollary. If $G$ is finite, then $\mathbb{C}[G]$ is semisimple.

Proof. We just proved that $J(\mathbb{C}[G])=(0).$ So we just need to show that $\mathbb{C}[G]$ is Artinian. Let

$I_1 \supset I_2 \supset \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$

be a descending chain of left ideals of $\mathbb{C}[G].$ Obviously each $I_j$ is a $\mathbb{C}$-subspace of $\mathbb{C}[G].$ Thus each $I_j$ is finite dimensional because $\dim_{\mathbb{C}} \mathbb{C}[G]=|G| < \infty.$ Hence $(*)$ will stablize at some point because $\dim_{\mathbb{C}} I_1 < \infty$ and $\dim_{\mathbb{C}}I_1 > \dim_{\mathbb{C}} I_2 > \cdots .$ Thus $\mathbb{C}[G]$ is (left) Artinian and the proof is complete because we know a ring is semisimple if and only if it is (left) Artinian and its Jacobson radical is zero. $\Box$