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Sep 6, 2011 at 3:38 comment added JSpecter Both $\mathbb{C}(X)$ and $\mathbb{C}$ have the same transcendence degree over $\mathbb{Q},$ so there exists an isomorphism between their algebraic closures. Restricting this isomorphism to $\mathbb{C}(X)$ gives the desired injection.
Sep 6, 2011 at 3:22 comment added Bill Cook Thanks! I guess my "sketch" of a "proof" was flawed. What is this injection from rational functions into the complex numbers? I guess I'm not familiar with that example.
Sep 6, 2011 at 3:20 vote accept Bill Cook
Sep 6, 2011 at 1:28 history edited JSpecter CC BY-SA 3.0
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Sep 6, 2011 at 1:06 history answered JSpecter CC BY-SA 3.0