HW6

Let $A$ be a ring with a $1$.

1. Let $V$ be an $A$-module.

• (a) Show that an endomorphism $e:V\to V$ is a projection onto a $A$-submodule $W$ if and only if $e$ is idempotent as an element of ${\mathrm{End}}_A(V)$.
• (b) Show that direct sum decompositions $V=W_1\oplus W_2$ as $A$-modules are in bijection with primitive idempotent decompositions $1=$ $e+f$ in ${\mathrm{End}}_A(V)$.
• (c) Show that $e\in {\mathrm{End}}_A(V)$ is primitive if and only if $e(V)$ is indecomposable.
• (d) Suppose that $V$ is semisimple with finitely many simple summands and let $e_1,e_2\in {\mathrm{End}}_A(V)$ be idempotents. Show that $e_1(V)\cong$ $e_2(V)$ as $A$-modules if and only if $e_1$ and $e_2$ are conjugate by an invertible element of ${\mathrm{End}}_A(V)$.
• (e) Let $k$ be a field. Show that all primitive idempotent elements in $M_n(k)$ are conjugate under the action of the unit group $G{L}_n(k)$, which consists of all the invertible matrices in $M_n(k)$.

\begin{proof} a) If $e$ is a projection onto $W$, then $e{|}_W:W\to W$ is an identity map. Since $e(v)\in W$, one have $e^2(v)=e(e(v))=e(v)$ for any $v\in V$ and $e^2=e$. Thus $e$ is idempotent.

If $e$ is idempotent map, then $e^2=e\in \mathrm{End}(V)$. Define $W=\mathrm{im}e$, then $e{|}_W:W\to W$ is an identity map, because each $w\in W$ can be written as $w=e(v)$ with $e(e(v))=e(v)$. Thus $e$ is a projection onto $W$.

b) For any direct sum decomposition $V=W_1\oplus W_2$, define projections $e:V\to W_1$ and $f:V\to W_2$. By a), $e,f$ are idempotent, and $\mathrm{im}(ef)\subseteq W_1\cap W_2=\{0\}$. So $e+f={\mathrm{id}}_V$ is a idempotent decomposition.

Conversely, if $1=e+f$ in ${\mathrm{End}}_A(V)$ is a idempotent decompositions, then $e,f$ are projection by a). Define $W_1=\mathrm{im}e$ and $W_2=\mathrm{im}f$. It remains to show $V=W_1\oplus W_2$. Since $1=e+f$, we have $V=W_1+W_2$. If $w\in W_1\cap W_2$, then $e(w)=f(w)=w$ and then $1(w)=(e+f)(w)=2w$. It deduces that $w=0$ and so $W_1\cap W_2=\{0\}$. Therefore, $V=W_1\oplus W_2$.

c) If $e\in {\mathrm{End}}_A(V)$ is primitive and $e(V)=W_1\oplus W_2$, then define $e_1,e_2$ as projections onto $W_1$ and $W_2$, respectively. Then by a) $e_1,e_2$ are idempotent. Since $\mathrm{im}(e_1{e}_2)=W_1\cap W_2=\{0\}$, $e=e_1+e_2$ is a idempotent decomposition, which is a contradiction.

If $e(V)$ is indecomposable and $e$ is not imprimitive, then there is a idempotent decomposition $e=e_1+e_2$. By a), $\mathrm{im}{e}_1\oplus \mathrm{im}{e}_2=e$ is a decomposition of $e(V)$.

d) Assume that $V=V_1\oplus \cdots \oplus V_n$, where $V_i$ are simple $A$-modules. Since $e_1$ and $e_2$ are idempotents, $e_1(V)$ and $e_2(V)$ are direct sum of $V_i$. If $e_1(V)\simeq e_2(V)$, then there is an isomorphism $\phi :e_1(V)\to e_2(V)$. Since $V$ is semisimple, we can suppose that $V=e_1(V)\oplus W_1=e_2(V)\oplus W_2$ and $W_1\simeq W_2$. There is an isomorphism ${\phi }_2:W_1\to W_2$. Define $\Phi :={\phi }_1\oplus {\phi }_2$, and consider $e_1^{\prime }:=\Phi \circ e_1\circ {\Phi }^{-1}$. Notice that $e_1^{\prime }{|}_{e_2(V)}=\mathrm{id}{|}_{e_2(V)},e_1^{\prime }{|}_{W_2}=0$, then $e_1^{\prime }=e_2$, that is, $e_1$ and $e_2$ are conjugate by an invertible element $\Phi$ of ${\mathrm{End}}_A(V)$.

Conversely, assume that $e_1$ and $e_2$ are conjugate by an invertible element $\Psi$ of ${\mathrm{End}}_A(V)$, and $e_2=\Psi \circ e_1\circ {\Psi }^{-1}$. Since

$$e_2{|}_{e_2(V)}=\mathrm{id}{|}_{e_2(V)}=\Psi \circ e_1\circ {\Psi }^{-1}{|}_{e_2(V)},$$

one have $e_1{|}_{{\Psi }^{-1}(e_2(V))}=\mathrm{id}{|}_{{\Psi }^{-1}(e_2(V))}$. Similarly, since $e_2{|}_{W_2}=0$, $e_1{|}_{{\Psi }^{-1}(W_2)}=0$. Furthermore, notice that $V={\Psi }^{-1}(e_2(V))\oplus {\Psi }^{-1}(W_2)$. So $e_1(V)={\Psi }^{-1}(e_2(V))$ and $W_1={\Psi }^{-1}(W_2)$. Thus $e_1(V)\stackrel{\Psi {|}_{e_1(V)}}{\simeq }{e}_2(V)$.

e) If $e\in M_n(k)$ is a primitive idempotent element, then $e^2=e$ and $e(k^n)$ is indecomposable. Note that $e^2=e$ yields that all eigenvalues of $e$ are $0$ and $1$. Since $e(k^n)$ is indecomposable, the Jordan canonical form of $e$ is $M:=\mathrm{diag}(1,0,\cdots ,0)$. Since all primitive idempotent elements in $M_n(k)$ are similar to $M$, they are conjugate under the action of the unit group ${\mathrm{GL}}_n(k)$. \end{proof}

2. Prove: No idempotent $e$ of $A$ is contained in $J(A)$. (Hint: $e(1-e)=0$.)

\begin{proof} If idempotent $e$ is contained in $J(A)$, then $1-e$ is a unit. However, $e=e^2$ and $e(1-e)=0$, which is a contradiction. \end{proof}

3. Prove: Let $A$ and $Z(A)$ be Artinian. Then $J(Z(A))=J(A)\cap Z(A)$.

\begin{proof} Since $Z(A)$ is Artinian, each $z\in J(Z(A))$ is nilpotent. Then $z\in J(A)$ as $J(A)$ is the the maximal nilpotent left ideal of $A$. In addition, note that $J(Z(A))\subseteq Z(A)$. Thus $J(Z(A))\subseteq J(A)\cap Z(A)$.

For any $a\in J(A)\cap Z(A)$, $1-a$ is a unit and $1-a\in Z(A)$. Thus, $a\in J(Z(A))$ and so $J(Z(A))\supseteq J(A)\cap Z(A)$. Now we finish the proof. \end{proof}

4. Prove: Let $A=A_1\oplus \cdots \oplus A_n$ be a direct sum of 2-sided ideals. Then

• (i) $Z(A)=Z\left(A_1\right)\oplus Z\left(A_2\right)\oplus \cdots \oplus Z\left(A_n\right)$.
• (ii) $J(A)=J\left(A_1\right)\oplus J\left(A_2\right)\oplus \cdots \oplus J\left(A_n\right)$.

\begin{proof} i) Since $Z(A_i)\cap Z(A_j)\subseteq A_i\cap A_j=\{0\}$, $Z(A_1)+\cdots +Z(A_n)=Z(A_1)\oplus \cdots Z(A_n)$. Since $Z(A_i)\subseteq Z(A)$ for all $i$, $Z(A)\supseteq Z(A_1)\oplus \cdots \oplus Z(A_n)$. For any $a\in Z(A)\subseteq A$, $a$ can be written as $a=a_1+\cdots +a_n$ with $a_i\in A_i$. For any $r\in R$, one have

$$r{a}_1+\cdots +r{a}_n=ra=ar=a_{1}r+\cdots +a_{n}r$$

and $r{a}_i,a_{i}r\in A_i$. Since $A=A_1\oplus \cdots \oplus A_n$, the expression of $ra$ is unique and $r{a}_i=a_{i}r$ for all $i$ and $r\in R$. Hence $a_i\in Z(A_i)$ and so $Z(A)\subseteq Z(A_1)\oplus \cdots \oplus Z(A_n)$. Hence $Z(A)=Z(A_1)\oplus \cdots \oplus Z(A_n)$.

ii) Assume that $I\subseteq A$ is a maximal ideal of $A$, then $I\cap A_i\subseteq A_i$ is an ideal of $A_i$. If $I⊉A_i$ and $I\cap A_i⊊A_i$, then $I\cap A_i$ is a maximal ideal of $A_i$. Otherwise, there exists an ideal $I_i\subset A_i$ such that $I\cap A_i\subset I_i\subset A_i$ and $I\cup I_i$ is a proper ideal containing $I$, which contradicts with $I$ maximal.

Let $\mathcal{I}$ be the set of maximal ideals of $A$, and let ${\mathcal{I}}_i$ be the set of maximal ideals of $A_i$. For any $i\in \{1,\cdots ,n\}$, if $A_i$ has a maximal ideal $M_i$, then $A_1\oplus \cdots \oplus A_{i-1}\oplus M_i\oplus A_{i+1}\oplus \cdots \oplus A_n\in \mathcal{I}$; if $A_i$ is a simple ring, then $A\setminus A_i$ is a maximal ideal of $A$. Hence $\{I\cap A_i:I\in \mathcal{I}\}=\{A_i\}\cup {\mathcal{I}}_i$, which deduces that $J(A)\cap A_i=J(A_i)$. Therefore,

$$J(A)=J(A)\cap (A_1\oplus \cdots \oplus A_n)=(J(A)\cap A_1)\oplus \cdots \oplus (J(A)\cap A_n)=J(A_1)\oplus \cdots \oplus J(A_n).$$

Now we finish the proof. \end{proof}

5. Let $\varphi :U\to V$ be a homomorphism of $A$-modules. Show that $\varphi (\mathrm{Soc}U)\subseteq \mathrm{Soc}(V)$, and that if $\varphi$ is an isomorphism then $\varphi$ restricts to an isomorphism $\mathrm{Soc}U\to \mathrm{Soc}V$.

\begin{proof} Since $\mathrm{soc}U$ is the largest semisimple submodule of $U$, one can assume that $\mathrm{soc}U=U_1\oplus \cdots U_n$ where $U_i$ are simple $A$-modules. Since $\varphi (U_i)\simeq U_i/\ker \varphi {|}_{U_i}$, $\varphi (U_i)\in \{0,V_i\}$ with $V_i\simeq U_i$. Thus

$$\varphi (\mathrm{soc}U)=\varphi (U_1\oplus \cdots \oplus U_n)=\varphi (U_1)\oplus \cdots \oplus \varphi (U_n)$$

is semisimple and so $\varphi (\mathrm{soc}U)\subseteq \mathrm{soc}V$. Furthermore, if $\varphi$ is an isomorphism, $\ker \varphi {|}_{U_i}=\{0\}$ and so $\varphi (U_i)=V_i\simeq U_i$ for all $i$. It deduces that $\mathrm{soc}U\stackrel{\varphi }{\simeq }\mathrm{soc}V$. \end{proof}