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Field question Divisibility

  • Thread starter kathrynag
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  • #1
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Let f(x), g(x) be in F[x]. Show that if g(x)|f(x) and f(x)|g(x), then f(x)=kg(x) for some k in F.

Since g(x)|f(x), then f(x)=g(x)r(x) for some r(x) in F[x].
Similarily, since f(x)|g(x), then g(x)=f(x)s(x)
So f(x)=f(x)s(x)r(x)
 
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Answers and Replies

  • #2
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I tried doing something with degrees.
deg(fsr)=deg(f)+deg(s)+deg(r)
So deg(f)=deg(f)+deg(s)+deg(r)
0=deg(s)+deg(r)
deg(s)=-deg(r)
 
  • #3
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What do you mean with F(x)? You mean polynomials or rational functions? The standard notation for polynomials is F[x]...

Now, you got that f(x)=f(x)s(x)r(x). Now apply that F[x] is an integral domain...
 
  • #4
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I mean polynomials F[x]
 
  • #5
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I haven't learned about integral domains yet
 
  • #6
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An integral domain is just a ring such that ab=0 implies a=0 and b=0.
Now, since F[x] is clearly an integral domain, what does this imply for the equation f(x)=f(x)s(x)r(x)??
 
  • #7
Delta2
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I tried doing something with degrees.
deg(fsr)=deg(f)+deg(s)+deg(r)
So deg(f)=deg(f)+deg(s)+deg(r)
0=deg(s)+deg(r)
deg(s)=-deg(r)
you almost got it, you know that deg(s)>=0 and deg(r)>=0 and deg(s)=-deg(r) hence it should be deg(s)=deg(r)=0.
 
  • #8
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Guess I don't know what to do with that that information
 
  • #9
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I don't how to go from there to f(x)=kg(x)
 
  • #10
Delta2
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er deg(s)=0 means there is a constant c in F such that s(x)=c.
 
  • #11
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ok so s(x)=c
g(x)=cf(x)
but we had r(x)=s(x)=0 so there is r(x)=k
f(x)=kg(x)
 

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