Is the field of rational numbers Q isomorphic to the fraction field Frac ⁡ ( Q [ x ] )?Both are fields, I can't disprove it by some algebraic properties that hold for one but not for another.I know that Frac ⁡ ( Z [ x ] ) ≅ Frac ⁡ ( Q [ x ] ) so I have tried to find out whether Frac ⁡ ( Z [ x ] ) ≅ Q , but still have no ideas;I have also tried to define ϕ : Q → Frac ⁡ ( Q [ x ] ) by ϕ ( p / q ) = ( x p − 1 ) / ( x q − 1 )but it is not a ring homomorphism.I guess that if ϕ is a ring homomorphism, then any p / q should be mapped to some form of fraction of polynomials p ( x ) / q ( x ) such that the product and sum of two polynomials of those form should have the same form, but I don't know how to do.

Question

Is the field of rational numbers       Q   isomorphic to the fraction field   Frac  ⁡  (      Q    [  x  ]  )?Both are fields, I can&#039;t disprove it by some algebraic properties that hold for one but not for another.I know that   Frac  ⁡  (      Z    [  x  ]  )  ≅  Frac  ⁡  (      Q    [  x  ]  ) so I have tried to find out whether   Frac  ⁡  (      Z    [  x  ]  )  ≅      Q  , but still have no ideas;I have also tried to define   ϕ  :      Q    →  Frac  ⁡  (      Q    [  x  ]  ) by  ϕ  (  p      /    q  )  =  (      x    p    −  1  )      /    (      x    q    −  1  )but it is not a ring homomorphism.I guess that if   ϕ is a ring homomorphism, then any     p      /    q   should be mapped to some form of fraction of polynomials     p  (  x  )      /    q  (  x  )   such that the product and sum of two polynomials of those form should have the same form, but I don&#039;t know how to do.

Brennan Frye · Accepted Answer

No, they are not isomorphic.There is only one possible morphism   f  :      Q    →  Frac  ⁡  (      Q    [  x  ]  )  =      Q    (  x  ): indeed, since it is a ring morphism it must send   1 to   1. Then for all   n  ∈      N    f  (  n  )  =  f  (  1  +  ⋯  +  1  )  =  f  (  1  )  +  ⋯  +  f  (  1  )  =  n  ⋅  f  (  1  )  =  n, and so   f  (  −  n  )  =  −  f  (  n  )  =  −  n. Finally, if   p      /    q is some fraction, then   q  f  (  p      /    q  )  =  f  (  q  ⋅  p      /    q  )  =  f  (  p  )  =  p    ⟹    f  (  p      /    q  )  =  p      /    q. So in conclusion,   f is necessarily the map that sends a rational   p      /    q to the constant fraction   p      /    q  ∈      Q    (  x  ).But this   f is not an isomorphism (it&#039;s not surjective), and since it&#039;s the only morphism       Q    →  Frac  ⁡  (      Q    [  x  ]  ), the two rings are not isomorphic.

lnwlf1728112xo85f · Accepted Answer

Q   is a prime field and       Q    (  x  ) - no, since       Q    ⊊      Q    (  x  ). Hence they are non-isomorphic.

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