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A question about homomorphisms...

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


Let ϕ:Frac(R[x])→F(x)\phi : Frac(R[x]) \rightarrow F(x) be a map sending (f(x),g(x)) to f(x)/g(x). We need to show that ϕ\phi 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

ϕ(1,1)=1/1=1\phi(1,1) = 1/1= 1

ϕ(fg,hk)=fghk=fhgk=ϕ(f,h)ϕ(g,k)\phi (fg, hk) = \frac{fg}{hk} = \frac{f}{h}\frac{g}{k} = \phi(f,h)\phi(g,k)

ϕ(f,h+g,k)=ϕ(fk+gh,hk)=kf+ghhk\phi(f,h + g,k) = \phi(fk+gh, hk) = \frac{kf+gh}{hk}

ϕ(f,h)+ϕ(g,k)=fh+gk=kf+ghhk\phi(f,h)+\phi(g,k) = \frac{f}{h}+\frac{g}{k} = \frac{kf+gh}{hk}



Am I right so far?

Now I have to show that this is a bijection...but I'm a bit confused...

For injection: Since (f(x),g(x)) maps to f(x)/g(x)...if f(x) and g(x) are equal, then we will that f(x)/g(x) = 1. For example, f(x)/f(x) = 1 = g(x)/g(x)...but we know that f(x) is not the same as g(x) in R[x].

For surjection: Isn't it obvious that for every f(x)/g(x), there exists an f(x) and g(x) in R[x]?

Thanks in advance

Thanks in advance
(edited 11 years ago)
Reply 1
Avatar for Mark13
Mark13
Original post by Artus
If R is a domain with F=Frac(R), prove that Frac(R[x]) is isomorphic to F(x).


Let ϕ:Frac(R[x])→F(x)\phi : Frac(R[x]) \rightarrow F(x) be a map sending (f(x),g(x)) to f(x)/g(x). We need to show that ϕ\phi is a ring homomorphism.


If you note that F(x) is the field of fractions of F[X], the question becomes more straightforward. EDIT: Just to make it a bit clearer - you can prove the statement without explicitly constructing a homomorphism if you use this method.

But referring to what you've written already, presumably you're writing (f(x),g(x)) for the equivalence class of pairs of elements (a(x),b(x)) (with a(x) and b(x) elements of R[X]) such that f(x)b(x)-g(x)a(x)=0? You need to check that your map phi is well-defined - if (f(x),g(x))=(h(x),k(x)), then does f(x)/g(x)=h(x)/k(x) in F(x)?
(edited 11 years ago)
Reply 2
Avatar for Artus
Artus
OP
Original post by Mark13
If you note that F(x) is the field of fractions of F[X], the question becomes more straightforward. EDIT: Just to make it a bit clearer - you can prove the statement without explicitly constructing a homomorphism if you use this method.

But referring to what you've written already, presumably you're writing (f(x),g(x)) for the equivalence class of pairs of elements (a(x),b(x)) (with a(x) and b(x) elements of R[X]) such that f(x)b(x)-g(x)a(x)=0? You need to check that your map phi is well-defined - if (f(x),g(x))=(h(x),k(x)), then does f(x)/g(x)=h(x)/k(x) in F(x)?


Actually, I did mean that F(x) was teh field of fractions of F[x]. In our textbook, they say that [a,b] is a/b...so I said that (f(x), g(x)) is f(x)/g(x)...do you think that's wrong?
Reply 3
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Mark13
Original post by Artus
Actually, I did mean that F(x) was teh field of fractions of F[x]. In our textbook, they say that [a,b] is a/b...so I said that (f(x), g(x)) is f(x)/g(x)...do you think that's wrong?


That's fine, you just need to make sure you check your map is well-defined.

If you can show that there's a subfield of Frac(R[X]) isomorphic to F, and hence a subfield isomorphic to F[X], then can you see why it must be the case that Frac(R[X]) is isomorphic to Frac(F[X])?
Reply 4
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Artus
OP
Original post by Mark13
That's fine, you just need to make sure you check your map is well-defined.

If you can show that there's a subfield of Frac(R[X]) isomorphic to F, and hence a subfield isomorphic to F[X], then can you see why it must be the case that Frac(R[X]) is isomorphic to Frac(F[X])?


I'm not sure, but I'm not trying to prove that Frac(R[x]) is isomorphic to Frac(F[x]), I'm trying to prove that it is isomorphic to F(x)...
Reply 5
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Mark13
Original post by Artus
I'm not sure, but I'm not trying to prove that Frac(R[x]) is isomorphic to Frac(F[x]), I'm trying to prove that it is isomorphic to F(x)...


But F(X) is Frac(F[X]).

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