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Tristan Meyers

Tristan Meyers

Answered question

2022-04-12

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.

Answer & Explanation

Brennan Frye

Brennan Frye

Beginner2022-04-13Added 20 answers

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's not surjective), and since it's the only morphism Q Frac ( Q [ x ] ), the two rings are not isomorphic.
lnwlf1728112xo85f

lnwlf1728112xo85f

Beginner2022-04-14Added 1 answers

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

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