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

**Erdos32212** · November 13th 2008, 02:07 PM

Let R be a ring. \prod_{i=0}^{\infty} R[tex].

**NonCommAlg** · November 13th 2008, 06:41 PM

Originally Posted by Erdos32212

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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