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A rational form is defined like this:

Let $F$ be any field. Write $F[t]$ for the set of all polynomials $p(t)=a_0+a_1t+\cdots+a_nt^n$ in an indeterminate $t$, with coefficients $a_k$ in $F$. Write $F(t)$ for the set of all ratios $p(t)/q(t)$ with $p(t)$, $q(t) \in F[t]$ and $q(t)$ is not the zero polynomial. $F(t)$ is a field; it is called the field of rational forms over $F$.

I need to show that the field $\Bbb Q$ is NOT isomorphic to the field $\Bbb Q(t)$ of rational forms over $\Bbb Q$.

Here's what I have:
If $\phi:\Bbb Q \to \Bbb Q(t)$ is an isomorphism between $\Bbb Q$ and $\Bbb Q(t)$, then $\phi(r)=\phi(1+1+ \cdots + 1)=\phi(1)+\phi(1)+\cdots+\phi(1)=r$ for some arbitrary $r \in \Bbb Q$. Likewise $\phi(p/q)=\phi(p)\ \phi(q^{-1})=p\ (\phi(q))^{-1}=p/q$ for some arbitrary $p$, $q \in \Bbb Q$. Thus $\phi$ takes $\Bbb Q$ to $\Bbb Q$. But $\Bbb Q \subset \Bbb Q(t)\ $ & $\ \Bbb Q \neq \Bbb Q(t)$. Thus $\phi$ is not a surjective mapping, thus $\phi$ is not an isomorphism, and thus $\Bbb Q$ is not isomorphic to $\Bbb Q(t)$.

This seems too simple. Does this adequately prove the statement?

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  • $\begingroup$ Go the other direction: assume an isomorphism from $\Bbb Q(t)$ to $\Bbb Q$. Note that a proper subfield of $\Bbb Q(t)$ surjects onto $\Bbb Q$. Where can you map $t$? $\endgroup$ – user452 Jun 16 '14 at 18:42
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I like to have the map going the other way. Say $\psi:\Bbb Q(t)\to \Bbb Q$. Let $p=\psi(t)$. Now what about $\frac 1p t$?

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  • $\begingroup$ Could you expand on your suggestion? All I get is: let $p= \psi(t) \in \Bbb Q$. Then $\psi( \frac t p) = \frac {\psi(t)} {\psi(p)} = \frac p {\psi(p)}$. We know that $p \in \Bbb Q$, but I don't know what $\psi(p)$ would be. Isn't it enough to show that $\Bbb Q \subsetneq \Bbb Q(t)$ to show that there is no isomorphism between them? $\endgroup$ – user137731 Jun 17 '14 at 19:48
  • $\begingroup$ A variant is writing $p=m/n$ for integers $m,n$, and noting that neccessarily $\psi(nt)=\psi(m) =\psi(1+1+\cdots+1)= m$. $\endgroup$ – Arthur Jun 17 '14 at 21:37
  • $\begingroup$ So your argument is that $\psi(t/p)=1 \Rightarrow \psi(t)= \psi( \psi(t)) \Rightarrow \psi(t)=t$. But $t \in \Bbb Q(t)$ and $t \notin \Bbb Q$, therefore there is no isomorphism between them? $\endgroup$ – user137731 Jun 17 '14 at 22:08
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    $\begingroup$ My argument is that $\phi(t/p) = 1 = \phi(1)$, thus the homomorphism cannot be injective. It is easy to show that for any ring homomorphism $\phi:R_1 \to R_2$, any element $x$ and any integer $n$, we have $\phi(nx) = n\phi(x)$, as long as $1$ exists. That is why I gave you a second example in the comments. It is a bit more work to show the same result for any fraction, which basically means that if $\phi:\Bbb Q \to \Bbb Q(t)$ is a non-trivial homomorphism, then it has to be the identity on $\Bbb Q$. $\endgroup$ – Arthur Jun 17 '14 at 22:36
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    $\begingroup$ Strict inclusion is usually not enough to prove the non-existance of isomorphisms between rings. Take for instance the polynomial ring $R[x]$ and the subring consisting only of even polynomials. That's a strict inclusion, yet there is a clear isomorphism between them. You need more to show that your rings are non-isomorphic. $\endgroup$ – Arthur Jun 17 '14 at 22:39

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