First exercise solutions

Antiautomorphism

Antiautomorphism

Assume $R\cong R^{op}$. Since $R\cong R^{op}$, there is a ring isomorphism $\varphi :R\to R^{op}$ such that $\varphi (r)=r$. We just need to check the reversing the multiplication. So,

$$\varphi (r\cdot r^{\prime })=r\cdot r^{\prime }=r^{\prime }{\cdot }_{op}r=\varphi (r^{\prime }){\cdot }_{op}\varphi (r).$$

It is an anti-automorphism between $R$ and $R^{op}$. On the other hand, assume there is an anti-automorphism $\alpha$. Let $i:R\to R^{op}$ as an identity map. Similarly as before,

$$i\circ \alpha (r\cdot r^{\prime })=i(\alpha (r^{\prime })\cdot \alpha (r))=\alpha (r^{\prime })\cdot \alpha (r)=\alpha (r){\cdot }_{op}\alpha (r^{\prime })=r{\cdot }_{op}{r}^{\prime }.$$

Therefore $R\cong R^{op}$.

Matrix ring with commutative ring

Using the previous problem, we just need to find the antiautomorphism $\alpha$. In this case, $\alpha (A)=A^T$ for $A\in M_n(K)$. Transpose map is clearly bijective and preserves $0,I,\text{ and }+$. About multiplication, $\alpha (AB)={\left(AB\right)}^T=B^T{A}^T=\alpha (B)\alpha (A),$ so that the transpose map is an antiautomorphism.

Real Quaternion

$\varphi :\mathbb{H}\to {\mathbb{H}}^{op}$ such as $a+bi+cj+dk\mapsto a-bi-cj-dk$ is an antiautomorphism. Do it! You should make your hand dirty yourself.

Group ring

Property

Module

1. Using correspondence between quotient module and submodule containing the submodule.
2. $F=\mathbb{Q}$. Suppose $M$ is a proper maximal submodule of $\mathbb{Q}$. Then by 1, $\mathbb{Q}/M$ is a simple module. As an abelian group, simple module is a cyclic group which isomorphic to $\mathbb{Z}$ or $\mathbb{Z}/n$. Since it is a simple module, it is isomorphic to $\mathbb{Z}/p$ for some prime $p$. Let $0\ne q\in \mathbb{Q}\setminus M$, then there exists $q^{\prime }$ such that $q=p{q}^{\prime }$. Then $q+M=p{q}^{\prime }+M=p(q^{\prime }+M)=0$. So that $q\in M$. Contradicts to the assumption $q\notin M$. Therefore there is no proper maximal submodule of $\mathbb{Q}$.
3. As a routine. Using Zorn’s lemma.

Simple Module

1. Let $N$ be a nontrivial submodule of ${\mathbb{R}}^2$. Let $0\ne \left(a,b\right)\in N$ and $\left[\begin{array}{cc}{v}_1& v_2\end{array}\right]\in M_2\left(\mathbb{R}\right)$. The multiplication $a{v}_1+b{v}_2\in N⊲{\mathbb{R}}^2$.

   1. Assume $ab\ne 0$. Since $v_1$ and $v_2$ are arbitrary elements of ${\mathbb{R}}^2$, $M_2\left(\mathbb{R}\right)\cdot (a,b)={\mathbb{R}}^2$.
   2. Without loss of generality, assume $a\ne 0$ and $b=0$. Similarly, $M_2\left(\mathbb{R}\right)\cdot (a,0)={\mathbb{R}}^2$. Therefore, ${\mathbb{R}}^2$ is simple $M_2\left(\mathbb{R}\right)$-module.

2. Using 1. $M_2\left(\mathbb{R}\right)=\left[\begin{array}{cc}\ast & 0\\ \ast & 0\end{array}\right]\oplus \left[\begin{array}{cc}0& \ast \\ 0& \ast \end{array}\right]{\cong }_{mod}{\mathbb{R}}^2\oplus {\mathbb{R}}^2$, where each components are simple

Central Idempotent

1. Only if part. $(0,1{)}^2=(0,1)$ and $(a,b)\cdot (0,1)=(0,b)=(0,1)\cdot (a,b)$. Therefore, there is nontrivial central idempotent $(0,1)$.
2. If part. Let $e$ be an nontrivial central idempotent. Since $e$ is an idempotent, $e(e-1)=0$. Using Chinese remainder theorem, $R\twoheadrightarrow R/e\times R/(e-1)$ has a kernel generated by $e(e-1)=0$. Since $Re$ and $R(e-1)$ are comaximal, $R{\cong }_{Ring}R/e\times R/(e-1)$.

Burnside’s Theorem

1. Let $N$ be a submodule. Without loss generality, if $N=\left\{\left(\begin{array}{c}\ast \\ 0\end{array}\right)|\ast \in \mathbb{C}\right\}$ , since $s\cdot \left(\begin{array}{c}\ast \\ 0\end{array}\right)=\left(\begin{array}{c}0\\ \ast \end{array}\right)\notin N$. Therefore $N$ has a form $\left\{\left(\begin{array}{c}\ast \\ \ast \end{array}\right)\right\}$.

   Let $(w_1,w_2)\in N$, where $w_1{w}_2\ne 0$. Then $(\xi w_1,{\xi }^{-1}{w}_2)\in N$. As a $\mathbb{C}\subset \mathbb{C}{D}_{2n}$-vector space, $(w_1,w_2)$ and $(\xi w_1,{\xi }^{-1}{w}_2)$ are linearly independent. Therefore

   $$\mathbb{C}{D}_{2n}\cdot (w_1,w_2)={\mathbb{C}}^2.$$

   That ${\mathbb{C}}^2$ is a simple $\mathbb{C}{D}_{2n}$-module.

2. By 1, using Burnside’s Theorem, for $\mathbb{C}\subset \mathbb{C}{D}_{2n}\subset M_2(\mathbb{C})$, $\mathbb{C}{D}_{2n}{\cong }_{Ring}{M}_2\mathbb{C}$, so that $\pi$ is onto.

The idempotent of the trivial representation

1. Think about the group table, we know that $g_0{\sum }_{g\in G}g={\sum }_{g\in G}g$ for every $g_0\in G$. Therefore, $g{e}_T=e_T=e_{T}g$ for all $g\in G$.

2. Exactly same as 1.

   $$\frac{1}{n^2}\sum _{g\in G}g\sum _{h\in G}h=\frac{1}{n^2}n\sum _{g\in G}g=\frac{1}{n}\sum _{g\in G}g=e_T.$$

3. Let $r\in k$ and $g_0\in G$. Then

   $$r{g}_0{\cdot }_{kG}{e}_T=\frac{r}{n}\sum _{g\in G}g=\frac{1}{n}\sum _{g\in G}g\cdot r{g}_0=e_{T}r{g}_0.$$

   Therefore $e_T\in Center(kG)$.

4. From problem 1 and 3, we showed that $r{g}_0{e}_T=r{e}_T$ for $r\in k$. So we have a canonical isomorphism as $r{e}_T\mapsto r$.

5. $$(1-e_T{)}^2=(1-e_T){\cdot }_{kG}(1-e_T)=1-e_T-e_T+e_T^2=1-e_T.$$

6. Fact 1)By the Group Ring, $kG$ is a semi simple ring.

   Fact 2)$e_T(1-e_T)=e_T-e_T^2=e_T-e_T=0$. So $(1-e_T)$ and $e_T$ are orthogonal each other.

   Fact 3)$(1-e_T)$ is central.

   $$rg(1-e_T)=rg-rg{e}_T=rg-r{e}_T=(1-e_T)rg.$$

   Therefore, we have a Block Decomposition.

7. Using non trivial central idempotent argument, $kG{\cong }_{Ring}kG/e_T\times kG/(1-e_T)$. So, it’s enough to show that $kG/(1-e_T){\cong }_{Ring}k$.

   Since $kG{e}_T$ and $kG(1-e_T)$ are two-sided ideals, $kG/kG(1-e_T){\cong }_{ring}kG{e}_T{\cong }_{Ring}k$ where the last equivalent is problem 4.

Simple Ring and Artin-Wedderburn Theorem

4. To find the block decomposition is equivalent to find the orthonormal non trivial central idempotent. We can easily see that $\mathbb{Q}{C}_2$ is semi simple ring by Group Ring.

   Let $e=p+qg$ be a nontrivial central idempotent. Clearly, it is central. So we just need to show the following.

   So $q(2p-1)=0$ and $(p-1/2{)}^2+q^2=1/4$. If $q=0$, then $p=0$ which implies $e$ is trivial idempotent. Hence $p=1/2$ and $q=\pm 1/2$.

   About orthogonality,

   $$(1/2+1/2g)\cdot (1/2-1/2g)=1/4-1/4+(1/4-1/4)g=0.$$

   Therefore, a block decomposition is as following.

   $$\mathbb{Q}{C}_2{\cong }_{\mathbb{Q}{C}_2-mod}\mathbb{Q}{C}_2(1/2+1/2g)\oplus \mathbb{Q}{C}_2(1/2-1/2g).$$

5. By the Artin-Weddenburn Theorem, $\mathbb{Q}{C}_2{\cong }_{Ring}{R}_1\times \cdots R_n$ where $R_i$‘s are simple rings. But we can find the isomorphism explicitly through the problem 1.

   As part Central Idempotent, we have a nontrivial central idempotent $(1/2+1/2g)$. Therefore

   $$\mathbb{Q}{C}_2{\cong }_{Ring}\mathbb{Q}{C}_2/(1/2+1/2g)\times \mathbb{Q}{C}_2/(-1/2+1/2g).$$

   We have a ring homomorphism ${\varphi }_{\pm }:\mathbb{Q}{C}_2\to \mathbb{Q}$ as $r+sg\mapsto r\mp s$ with kernel $\mathbb{Q}{C}_2(1\pm g)$. Therefore,

   $$\mathbb{Q}{C}_2{\cong }_{Ring}{\mathbb{Q}}_{+}\times {\mathbb{Q}}_{-}.$$

6. Suppose $\mathbb{Z}{C}_2$ has a block decomposition, such that there exists non trivial central idempotent $m+ng$. In the argument in problem 1, $m=0$ and $n=0$ which is trivial idempotent. Therefore, there is no block decomposition and does not equal to product of the rings.