A question about ring homomorphisms

  • Thread starter Artusartos
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Homework Statement



If R is a domain with F=Frac(R), prove that Frac(R[x]) is isomorphic to F(x).

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The Attempt at a Solution



Let [tex]\phi : Frac(R[x]) \rightarrow F(x)[/tex] be a map sending (f(x),g(x)) to f(x)/g(x). We need to show that [tex]\phi[/tex] is a ring homomorphism. Let f,g,h,k be in R[x] such that f/h and g/k is in Frac(R[x]).

We know that

[tex]\phi(1,1) = 1/1= 1[/tex]

[tex]\phi (fg, hk) = \frac{fg}{hk} = \frac{f}{h}\frac{g}{k} = \phi(f,h)\phi(g,k)[/tex]

But I'm confused with the addition part...

[tex]\phi(f+g,h+k) = \frac{f+g}{h+k}[/tex]

[tex]\phi(f,h)+\phi(g,k) = \frac{f}{h}+\frac{g}{k} = \frac{kf+gh}{h+k}[/tex]

But now [tex]\phi(f+g,h+k) \not= \phi(f,h) + \phi(g,k)[/tex]

Can anybody help with this?

Thanks in advance
 

Answers and Replies

  • #2
HallsofIvy
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You appear to be interpreting "Frac[R(x)]" as pairs, (f(x), g(x)). How are you defining addition in Frac[R(x)]?
 
  • #3
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You appear to be interpreting "Frac[R(x)]" as pairs, (f(x), g(x)). How are you defining addition in Frac[R(x)]?

Oh...ok I think I get what you mean...because [tex]\phi(f,h + g,k) = \phi(fk+gh, hk)[/tex], right?
 
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