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Math Help - Ring of Endomorphism =3

  1. #1
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    Wink Ring of Endomorphism =3

    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
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  2. #2
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    Quote Originally Posted by GTK X Hunter View Post
    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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