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Thread: Ring of Endomorphism =3

  1. #1
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    Wink Ring of Endomorphism =3

    Let $\displaystyle R$ be a ring with unity and let $\displaystyle End(\langle R,+\rangle)$ be the ring of endomorphisms of $\displaystyle \langle R,+\rangle$. Let $\displaystyle a\in R$, and let $\displaystyle \lambda_a : R \rightarrow R$ be given by $\displaystyle \lambda_a(x)=ax,\forall x\in R$.
    a) Show that $\displaystyle \lambda_a$ is an endomorphism of $\displaystyle \langle R,+\rangle$
    b) Show that $\displaystyle R'=\{\lambda_a|a\in R\}$ is a subring of $\displaystyle End(\langle R,+\rangle)$
    c) Prove the analogue of Cayley's theorem for $\displaystyle R$ by showing that $\displaystyle R'$ of (b) is isomorphic to $\displaystyle R$
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  2. #2
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    Quote Originally Posted by GTK X Hunter View Post
    Let $\displaystyle R$ be a ring with unity and let $\displaystyle End(\langle R,+\rangle)$ be the ring of endomorphisms of $\displaystyle \langle R,+\rangle$. Let $\displaystyle a\in R$, and let $\displaystyle \lambda_a : R \rightarrow R$ be given by $\displaystyle \lambda_a(x)=ax,\forall x\in R$.
    a) Show that $\displaystyle \lambda_a$ is an endomorphism of $\displaystyle \langle R,+\rangle$
    (R,+) is an abelian group. so an endomorphism of (R,+) is just a group homomorphism: $\displaystyle \lambda_a(x+y)=a(x+y)=ax + ay=\lambda_a(x)+\lambda_a(y).$

    b) Show that $\displaystyle R'=\{\lambda_a|a\in R\}$ is a subring of $\displaystyle End(\langle R,+\rangle)$
    $\displaystyle \lambda_a(x)+\lambda_b(x)=ax+bx=(a+b)x=\lambda_{a+ b}(x).$ so $\displaystyle \lambda_a + \lambda_b = \lambda_{a+b} \in R',$ which means $\displaystyle R'$ is closed under addition.

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

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

    c) Prove the analogue of Cayley's theorem for $\displaystyle R$ by showing that $\displaystyle R'$ of (b) is isomorphic to $\displaystyle R$
    for this part you do need $\displaystyle R$ to have $\displaystyle 1.$ define the map $\displaystyle \varphi : R \longrightarrow R'$ by $\displaystyle \varphi(a)=\lambda_a.$ then $\displaystyle \varphi (a+b)= \lambda_{a+b}=\lambda_{a} + \lambda_{b}=\varphi (a) + \varphi (b).$ we also have $\displaystyle \varphi (ab)=\lambda_{ab}=\lambda_a \lambda_b = \varphi (a) \varphi (b).$

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