HW5

1. Let $D$ be a division ring and $n$ a positive integer.

• (a) Show that the natural left $M_n(D)$-module, consisting of column vectors of length $n$ with entries in $D$, is a simple module.

• (b) Show that $M_n(D)$ is semisimple and has up to isomorphism only one simple module.

• (c) Show that every algebra of the form

  $$M_{n_1}\left(D_1\right)\oplus \cdots \oplus M_{n_r}\left(D_r\right)$$

  is semisimple, where $D_i$ are division rings.

• (d) Show that $M_n(D)$ is a simple ring, namely, one in which the only $2$-sided ideals are the zero ideal and the whole ring.

\begin{proof} a) Define $M$ is the set of all column vectors of length $n$. Then $(1,0,\cdots ,0{)}^T\in M$. For any $(d_1,\cdots ,d_n{)}^T\in M$ with $d_i\in D$, notice that

$$\left[\begin{array}{cccc}{d}_1& \ast & \cdots & \ast \\ d_2& \ast & \cdots & \ast \\ ⋮& ⋮& & ⋮\\ d_n& \ast & \cdots & \ast \end{array}\right]\left[\begin{array}{c}1\\ 0\\ ⋮\\ 0\end{array}\right]=\left[\begin{array}{c}{d}_1\\ d_2\\ ⋮\\ d_n\end{array}\right]$$

and then $⟨(1,0,\cdots ,0{)}^T⟩=M$. Therefore, $M$ is a simple module.

b) It suffices to show ${}_{M_n(D)}{M}_n(D)$ is a semisimple left $M_n(D)$-module. Let $e_1=(1,0,\cdots ,0{)}^T$ be a column vector, and let $m_1=[e_1,0,\cdots ,0]$, $m_2=[0,e_1,0,\cdots ,0]$, $\cdots$, $m_n=[0,\cdots ,0,e_1]$ be elements of ${}_{M_n(D)}{M}_n(D)$. By a) we know $N_i=⟨m_i⟩\simeq M$ is a simple module. Also note that ${}_{M_n(D)}{M}_n(D)=N_1\oplus \cdots \oplus N_n$, hence $M_n(D)$ is semisimple.

c) By b), each $M_{n_j}(D_j)$ is semisimple and so is a sum of simple submodules. Then $M_{n_1}\left(D_1\right)\oplus \cdots \oplus M_{n_r}\left(D_r\right)$ is also a sum of simple modules. Therefore, $M_{n_1}\left(D_1\right)\oplus \cdots \oplus M_{n_r}\left(D_r\right)$ is semisimple.

d) Let $I\subseteq M_n(D)$ be a $2$-sided ideal. If $I\ne 0$, then there exists non-zero element $i\in I$. By linear algebra, we know there exist $L,R\in M_n(D)$ such that $LiR=[e_1,0,\cdots ,0]\in I$. Then for any $m_0\in M_n(D)$, by linear algebra there exist $L^{\prime }$ and $R^{\prime }$ such that $L^{\prime }i{R}^{\prime }=m_0$ and so $m_0\in I$. Therefore, $I=M_n(D)$ and so $M_n(D)$ is a simple ring. \end{proof}

2. Let $R$ be an arbitrary ring. Then every $2$-sided ideal of the matrix ring $M_n(R)$ can be expressed in the form $M_n(I)$, where $I$ is a 2-sided ideal of $R$. In particular, $R$ is simple if and only if $M_n(R)$ is simple. (Hint: For an ideal $L$ of $M_n(R)$, let $I$ be the set of $(1,1)$-entries of $L$. Show that $I$ is an ideal of $R$ and $L=M_n(I)$. )

\begin{proof} Let $L$ be an ideal of $M_n(R)$, and let $I$ be the set of $(1,1)$-entries of $L$. Note that for any $r\in R$ and $i\in I$, there exists $\ell \in L$ whose $(1,1)$-entry is $i$ and ${\ell }^{\prime }:=\ell \cdot \mathrm{diag}(r,1,\cdots ,1)\in L$. Since the $(1,1)$-entry of ${\ell }^{\prime }$ is $ir$, we have $ir\in I$. Similarly, we can prove $ri\in I$ and so $I$ is an ideal of $R$. Let $I_2$ be the set of $(2,1)$-entries of $L$. Note that $m_L:=\left[\begin{array}{ccccc}0& 1& 0& \cdots & 0\\ 1& 0& 0& \cdots & 0\\ ⋮& ⋮& ⋮& & ⋮\\ \ast & \ast & \ast & \cdots & \ast \end{array}\right]$ exchange $(1,1)$-entry and $(2,1)$-entry by left multiplication, then $I_2=I$. Using the same argument, we can prove that the set of $(s,t)$-entries of $L$ is $I$ for any $1⩽s,t⩽n$. Therefore, every $2$-sided ideal of the matrix ring $M_n(R)$ can be expressed in the form $M_n(I)$, where $I$ is a 2-sided ideal of $R$.

Furthermore, $R$ is simple iff $M_n(I)=M_n(R)$ or $0$ for each $2$-sided ideal of $I\subseteq R$ iff the only $2$-sided ideals of $R$ are $0$ and $R$ iff $R$ is simple. \end{proof}

**3. If $W$ is a submodule of a semisimple module $V$, then $W$ and $V/W$ are also semisimple.

\begin{proof} If $W$ is a submodule of a semisimple module $V$, then $W$ is a direct summand of $V$. Hence $V=W\oplus U$ for some submodule $U$ and $V/W\simeq U$ is also semisimple. \end{proof}

4. Let $U$ be an $A$-module for a semisimple finite-dimensional algebra $A$ (over a field). Show that if ${\mathrm{End}}_A(U)$ is a division ring then $U$ is simple.

\begin{proof} Since $A$ is a semisimple finite-dimensional algebra, $A$ is an Artinian ring and so $U$ is semisimple. If $U$ is not simple, then $U=V\oplus W$ for some $A$-submodules $V$ and $W$. Define $f:U\to U$ induced from the projection map $f_0:U\to V$, and define $g:U\to U$ induced from the projection map $g_0:U\to W$. Then $f\circ g=0$ and $f,g\in {\mathrm{End}}_A(U)$ yield that ${\mathrm{End}}_A(U)$ is not a division ring. Therefore, $U$ is simple. \end{proof}

5. Let $k$ be a field of characteristic $0$. Suppose the group algebra $kG$ is semisimple, and the simple $kG$-modules are $S_1,\dots ,S_r$ with degrees $d_i=$ ${\dim }_k{S}_i$. Show that

$$\sum _{i=1}^r{d}_i^2⩾|G|$$

with equality if and only if

$${\mathrm{End}}_{kG}\left(S_i\right)=k$$

for all $i$.

\begin{proof} Since $kG$ is semisimple, $kG=S_1^{{\alpha }_1}\oplus \cdots \oplus S_r^{{\alpha }_r}$ for some ${\alpha }_i\in \mathbb{N}$. It deduces that

$$|G|=\sum _{i=1}^r{\alpha }_i{d}_i.$$

Note that a $kG$-homomorphism $f:kG\to S_i$ is determined by $f(1)$, then ${\mathrm{Hom}}_{kG}(kG,S_i)\simeq S_i$ and so ${\dim }_k\mathrm{Hom}(kG,S_i)=d_i$.

On the other hand, ${\dim }_k\mathrm{Hom}(S_i^{{\alpha }_i},S_i)={\alpha }_i\dim ({\mathrm{End}}_{kG}(S_i))=d_i$ deduces that ${\alpha }_i=d_i/\dim ({\mathrm{End}}_{kG}(S_i))$. Therefore,

$$|G|=\sum _{i=1}^r{\alpha }_i{d}_i=\sum _{i=1}^r\frac{1}{\dim ({\mathrm{End}}_{kG}(S_i))}{d}_i^2⩽\sum _{i=1}^r{d}_i^2,$$

where the equality holds iff $\dim ({\mathrm{End}}_{kG}(S_i))=1$ iff ${\mathrm{End}}_{kG}(S_i)=k$. \end{proof}

6. Suppose that we have $R$-module homomorphisms $U\stackrel{f}{\to }V\stackrel{g}{\to }W$. Show that: if $gf$ is an essential epimorphism and $f$ is an epimorphism, then both $f$ and $g$ are essential epimorphisms.

\begin{proof} If $f$ is not an essential epimorphism, then there exists $U_0\subset U$ such that $f{|}_{U_0}:U_0\to V$ is surjective and so for $gf{|}_{U_0}:U_0\to W$, which contradicts with $gf$ being an essential epimorphism. Therefore, $f$ is essential.

Since $gf$ is an essential epimorphism, $g$ is surjective. Assume that $g$ is not essential, then there exists $V_0\subset V$ such that $g{|}_{V_0}:V_0\to V$ surjective. Define $U_1=f^{-1}(V_0)$. If $U_1=U$, then $f$ is not surjective, contradiction. So $U_1\subset U$ and then $gf{|}_{U_1}:U_1\to W$ is surjective, which is impossible as $gf$ is essential. Therefore, $g$ is an essential epimorphism. \end{proof}