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I can prove the converse but not this direction. In fact, the converse is proved in Gallian. But I have yet to see a proof in this direction. Is this certainly true? Necessarily true?

I am looking for a proof by contradiction. If a is algebraic then we have some polynomial that a is the root of with degree n, so we have an n dimensional vector space over Q. So perhaps a cardinality argument could be used. But I still find the field of fractions a little hard to grasp as a splitting field. I also find it difficult to work with isomorphism and concepts of dimension at the same time. Perhaps I can find an element in Q(x) that can't be expressed in terms of the assumed finite basis for Q(a) for a algebraic...

This isn't homework but just practice for a final. I really just want to know that this is indeed a fact and see why. If anyone could prove it or link me to a proof, that would be great.

OLP
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You're there:

The dimension of $\Bbb Q(x)$ over $\Bbb Q$ is infinite, and the field isomorphism $\Bbb Q(x)\to\Bbb Q(a)$ is particularly also a vector space isomorphism over $\Bbb Q$.

Berci
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  • Is it really that simple? I've been looking for theorems in Gallian to find a punchier way to justify this intuitively plausible statement. – OLP Apr 11 '15 at 19:52
  • So the isomorphism assures the dimension of Q(a) is infinite, contradicting what would obtain if a were algebraic? – OLP Apr 11 '15 at 19:54
  • yes yes $,!,!$ – Berci Apr 11 '15 at 19:56
  • Thanks for the hint. I'm having trouble seeing how to formalize this insight. I'm not used to relating isomorphisms and linear algebra concepts like dimension. I've seen no examples, so it's a bit difficult. – OLP Apr 11 '15 at 20:10
  • How do we know we can build the isomorphism without assuming that a is transcendental? Do we not need nonzero denominators for isomorphism? – OLP Apr 11 '15 at 20:23
  • What is the hypothesis? Isn't it that $\Bbb Q(a)\cong\Bbb Q(x)$? And, all rationals must stay fixed under any field isomorphism (as $0,1$ stay fixed), and thus it will be also a linear map. And is bijective. – Berci Apr 11 '15 at 20:24
  • Yes, but my confusion is that doesn't suggest which isomorphism exist. – OLP Apr 11 '15 at 20:24
  • I believe in the truth of what I'm trying to proof, but I can't see how to formalize it in an undeniable way. – OLP Apr 11 '15 at 20:25
  • What are you uncertain of? – Berci Apr 11 '15 at 20:26
  • Ok I am processing that last statement about ANY isomorphism. That's probably what I need . I know this is trivially easy for you, so I appreciate the trouble you are taking. How would you clinch the argument? I see the general approach, but not contradiction itself in terms of theorems to invoke. – OLP Apr 11 '15 at 20:26