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Is the field of rational numbers $\mathbb{Q}$ isomorphic to the fraction field $\operatorname{Frac}(\mathbb{Q}[x])$?

Both are fields, I can't disprove it by some algebraic properties that hold for one but not for another.

  • I know that $\operatorname{Frac}(\mathbb{Z}[x]) \cong \operatorname{Frac}(\mathbb{Q}[x])$ so I have tried to find out whether $\operatorname{Frac}(\mathbb{Z}[x]) \cong \mathbb{Q}$, but still have no ideas;

  • I have also tried to define $\phi : \mathbb{Q} \to \operatorname{Frac}(\mathbb{Q}[x])$ by $$\phi(p/q)=(x^p-1)/(x^q-1)$$ but it is not a ring homomorphism.

I guess that if $\phi$ 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.

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  • $\begingroup$ @DietrichBurde I don't really follow, are you sure the two questions are duplicate? $\endgroup$ – Najib Idrissi Mar 8 '16 at 14:09
  • $\begingroup$ I thought, that they show that the fraction field of $\mathbb{Q}[x]$ is $\mathbb{Q}(x)$, which is not isomorphic to $\mathbb{Q}$ ? But I will look for a better duplicate ( I have retracted it). $\endgroup$ – Dietrich Burde Mar 8 '16 at 14:11
  • $\begingroup$ @DietrichBurde Where is it shown that $\mathbb{Q}(x)$ is not isomorphic to $\mathbb{Q}$ in the linked question? I mean, that's pretty much the whole question... $\endgroup$ – Najib Idrissi Mar 8 '16 at 14:12
  • $\begingroup$ It's a duplicate of this MSE-question. $\endgroup$ – Dietrich Burde Mar 8 '16 at 14:25
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    $\begingroup$ I haven't tried to search this problem by text-without-symbols description, so I didn't know that it is a duplicate of this $\endgroup$ – Longitude Mar 8 '16 at 14:39
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No, they are not isomorphic.

There is only one possible morphism $f : \mathbb{Q} \to \operatorname{Frac}(\mathbb{Q}[x]) = \mathbb{Q}(x)$: indeed, since it is a ring morphism it must send $1$ to $1$. Then for all $n \in \mathbb{N}$, $f(n) = f(1+\dots+1) = f(1)+\dots+f(1) = n \cdot f(1) = n$, and so $f(-n) = -f(n) = -n$. Finally, if $p/q$ is some fraction, then $q f(p/q) = f(q \cdot p/q) = f(p) = p \implies f(p/q) = p/q$. So in conclusion, $f$ is necessarily the map that sends a rational $p/q$ to the constant fraction $p/q \in \mathbb{Q}(x)$.

But this $f$ is not an isomorphism (it's not surjective), and since it's the only morphism $\mathbb{Q} \to \operatorname{Frac}(\mathbb{Q}[x])$, the two rings are not isomorphic.

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  • $\begingroup$ thank for you help. I always forget to try the most basic things (do sth by definition first) while doing unfamiliar problem $\endgroup$ – Longitude Mar 8 '16 at 14:42
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$\mathbb{Q}$ is a prime field and $\mathbb{Q}(x)$ - no, since $\mathbb{Q}\varsubsetneq \mathbb{Q}(x)$. Hence they are non-isomorphic.

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