Ring of Endomorphism =3

**GTK X Hunter** · December 23rd 2009, 09:28 AM

Let $R$ be a ring with unity and let $End(\langle R,+\rangle)$ be the ring of endomorphisms of $\langle R,+\rangle$ . Let $a\in R$ , and let $\lambda_a : R \rightarrow R$ be given by $\lambda_a(x)=ax,\forall x\in R$ .
a) Show that $\lambda_a$ is an endomorphism of $\langle R,+\rangle$
b) Show that $R'=\{\lambda_a|a\in R\}$ is a subring of $End(\langle R,+\rangle)$
c) Prove the analogue of Cayley's theorem for $R$ by showing that $R'$ of (b) is isomorphic to $R$

**NonCommAlg** · December 23rd 2009, 05:12 PM

Originally Posted by GTK X Hunter

Let $R$ be a ring with unity and let $End(\langle R,+\rangle)$ be the ring of endomorphisms of $\langle R,+\rangle$ . Let $a\in R$ , and let $\lambda_a : R \rightarrow R$ be given by $\lambda_a(x)=ax,\forall x\in R$ .
a) Show that $\lambda_a$ is an endomorphism of $\langle R,+\rangle$

(R,+) is an abelian group. so an endomorphism of (R,+) is just a group homomorphism: $\lambda_a(x+y)=a(x+y)=ax + ay=\lambda_a(x)+\lambda_a(y).$

b) Show that $R'=\{\lambda_a|a\in R\}$ is a subring of $End(\langle R,+\rangle)$

$\lambda_a(x)+\lambda_b(x)=ax+bx=(a+b)x=\lambda_{a+ b}(x).$ so $\lambda_a + \lambda_b = \lambda_{a+b} \in R',$ which means $R'$ is closed under addition.

$\lambda_a \lambda_b (x)=\lambda_a(bx)=(ab)x=\lambda_{ab}(x).$ so $\lambda_a \lambda_b = \lambda_{ab} \in R',$ which means $R'$ is closed under multiplication.

if $R$ has an identity element $1,$ then $\lambda_1(x)=x,$ for all $x \in R$ and $\lambda_1 \lambda_a = \lambda_a$ and thus $\lambda_1 = 1_{\text{End}(R)}.$

c) Prove the analogue of Cayley's theorem for $R$ by showing that $R'$ of (b) is isomorphic to $R$

for this part you do need $R$ to have $1.$ define the map $\varphi : R \longrightarrow R'$ by $\varphi(a)=\lambda_a.$ then $\varphi (a+b)= \lambda_{a+b}=\lambda_{a} + \lambda_{b}=\varphi (a) + \varphi (b).$ we also have $\varphi (ab)=\lambda_{ab}=\lambda_a \lambda_b = \varphi (a) \varphi (b).$

so $\varphi$ is a ring homomorphism. to show that it's 1-1, let $\varphi (a)=0.$ then $ax=\lambda_a(x)=0$ for all $x \in R.$ put $x=1_R$ to get $a=0.$ finally, it's trivial that $\varphi$ is onto.

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