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Math Help - [SOLVED] R-module isomorphisms, R[x]

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    [SOLVED] R-module isomorphisms, R[x]

    Let R be a ring. \prod_{i=0}^{\infty} R[tex].
    Last edited by Erdos32212; November 13th 2008 at 08:32 PM.
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    Quote Originally Posted by Erdos32212 View Post
    Let R be a ring. The set R[x] of polynomials in x with coefficients in R is an abelian group under addition, and can be made into an R-module in this way: if x \in R and f(x)= \sum a_ix^i, then c \cdot f(x)=\sum (ca_i)x^i. Using a similar construction, the set R[[x]] of power series in x with coefficients in R is also an R-module. Prove that there are R-module isomorphisms R[x] \cong \bigoplus_{i=0}^{\infty} R and R[[x]] \cong \prod_{i=0}^{\infty} R.
    the isomorphisms are defined very naturally: \Phi: R[x] \longrightarrow \bigoplus_{i=0}^{\infty} R is defined by \Phi(r_0 + r_1x + \cdots + r_nx^n)=(r_0,r_1, \cdots , r_n, 0, 0, \cdots), and \Psi:R[[x]] \longrightarrow \prod_{i=0}^{\infty} R is defined by:

    \Psi(r_0 + r_1x + r_2x^2 + \cdots)=(r_0,r_1, r_2, \cdots). for example to prove that \Psi is an R-module isomorphism: let t \in R and p(x)=\sum_{n=0}^{\infty}r_ix^i, \ q(x)=\sum_{i=0}^{\infty}s_ix^i be in R[[x]]. first see

    that \Psi(p(x))=(0,0,0, \cdots) if and only if r_j=0, \ \forall j, that is p(x)=0. this shows that \Psi is well-defined and also it's one-to-one. next we show that \Psi is a homomorphism:

    \Psi(p(x)+q(x))=\Psi(r_0+s_0 + (r_1 + s_1)x + (r_2 + s_2)x^2 + \cdots)=(r_0 + s_0, r_1 + s_1, r_2 + s_2, \cdots)

    =(r_0,r_1,r_2, \cdots) + (s_0,s_1, s_2, \cdots)=\Psi(p(x))+\Psi(q(x)).

    we also have: \Psi(tp(x))=\Psi(tr_0 + tr_1x + \cdots)=(tr_0,tr_1, \cdots)=t(r_0,r_1, \cdots)=t \Psi(p(x)). so \Psi is an R-module homomorphism. finally if \alpha=(a_0, a_1, a_2, \cdots) \in \prod_{i=0}^{\infty} R, then

    let f(x)=a_0 + a_1 x + a_2 x^2 + \cdots \in R[[x]]. then obvioulsy \Psi(f(x))=\alpha. thus \Psi is onto and we're done. an identical proof will work for \Phi.
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