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I got a doubt with this problem:

Let $M=\{f:\mathbb{N}\to \mathbb{Z}|\text{$f$ is a function}\}$, defining the sum in $M$ as $(f+g)(n)=f(n)+g(n)$, $M$ is an abelian group. Let $R=\{\phi :M\to M : \phi\ \text{is a morphism}\}$. $R$ is a ring with the pointwise addition and the product as the function composition. Define $$\phi_1(f)(n)=f(2n+1),\ \phi_2(f)(n)=f(2n) $$ Show that $\{\phi_1,\phi_2\}$ is a basis of $R$ as a $R$-module.

Well, I know what to do, but my problem is that I don't even know how to handle the problem. For instance, let $\varphi \in R$, I have to show that exists $\alpha,\beta \in R$ such that: $$ \alpha \phi_1+\beta \phi_2=\varphi $$ and then I don't know how to go on.

Juan Pablo
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1 Answers1

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Considering elements of $M$ as sequences, we can define 'joint inverse' for $\phi_1,\phi_2$: Let $$\alpha_0:=f\mapsto (0,f(0),0,f(1),0,f(2),\dots) \\ \beta_0:=f\mapsto (f(0),0,f(1),0,f(2),0,\dots)$$ So that, $\alpha_0\phi_1+\beta_0\phi_2=id$, and hence we can choose $\alpha:=\varphi\alpha_0$ and $\beta:=\varphi\beta_0$.

You also have to prove that $\phi_1,\phi_2$ are ($R$-) independent.

Berci
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    For $\alpha_0$, I think you should start with $f(0)$, right? Since OP defined $\phi_1(f)(1)=f(3)$, while $\phi_1(f)(0)=f(1)$. I think this idea is similar to the homotopies, if I remember correctly? – awllower Jul 13 '13 at 23:46
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    Ah yes, right. Thank you. – Berci Jul 13 '13 at 23:47
  • @Berci, Could you help me with the last thing you said? – Juan Pablo Jul 14 '13 at 03:49
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    @JuanPablo If $a\phi_1+b\phi_2=0$, then $\forall f, \forall n$, we would have... what? And is this reasonable? – awllower Jul 14 '13 at 14:45
  • @awllower we would have $a(\phi_1(f))(n)+b(\phi_2(f))(n)=0,\ \forall f,n$. My idea is that the foregoing equation implies that $a((f(1),f(3),f(5),\ldots))=(0,\ldots),\ \forall f$. And I don't get what you mean by reasonable. – Juan Pablo Jul 14 '13 at 16:24
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    Sorry, i wanted to mean that the original equation implies that $a((f(1),f(2),\ldots))=(0,\ldots)$, for all $f$. This is because, given $f \in M$ I can take $g=(0,f(0),0,f(1),0,f(2),\ldots)$, and then I replace $g$ in the original equation. – Juan Pablo Jul 14 '13 at 17:41
  • @JuanPablo Well, $a\phi_1(f)(n)=af(2n+1), b\phi_2(n)=bf(2n)$. So we would obtain $f(2n+1)+bf(2n)=0$. But this holds for arbitrary $f$, which I think is impossible. So this implies the linear independence. – awllower Jul 15 '13 at 00:27