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Let $L/F$ be a field extension and let $a_1,...,a_n$ elements in $L$ that are algebraically independent over $F$ (i.e. there is no non-zero polynomial $f\in F[x_1,...,x_n]$ s.t. $f(a_1,...,a_n)=0$). Prove that $F(a_1,...,a_n)$ is isomorphic to the fraction field of $F[x_1,...,x_n]$.

Proof : Let $\varphi:F[x_1,...,x_n]\to L$ s.t. $\varphi:x_i\longmapsto a_i$. This is clearly injective.

This injectivity along with the existence of inverse in $L$, means that we can extend $\varphi$ to an injective homomorphism $\varphi:F(x_1,...,x_n)\to L$.

Question : I don't understand this last sentence. How $\varphi$ can be prolonged to $F[x_1,...,x_n]$ to $F(x_1,...,x_n)$ ?

After, they say that this is surjective because $F$ and each of the $a_i$ is in the image, and these generate $L$ over $F$.

Question : Why it generate $L$ over $F$ ?

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    For your first question: You already have a definition for $\varphi(f)$ for $f \in F[x_1, \ldots, x_n]$. Now define $\varphi(f/g) = \varphi(f)/\varphi(g)$ and you've extended the map to $F(x_1, \ldots, x_n)$. – 2017-01-21
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    @SpamIAm: Why can I do that ? – 2017-01-21
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    What are you concerned about? The one thing to note is that since $a_1, \ldots, a_n$ are algebraically independent, then $\varphi(g) = g(a_1, \ldots, a_n) \neq 0$ for all $g \in F[x_1, \ldots, x_n]$ with $g \neq 0$. Thus we are never dividing by $0$ using the definition I gave above. – 2017-01-21
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    I'd say that if $\varphi : A \to B$ is an injective morphism of integral domains, then $\tilde{\varphi}(a/b) = \varphi(a)/ \varphi(b)$ is an injective morphism of the fraction fields $Frac(A) \to Frac(B)$ ($\varphi$ is injective means that $\varphi(a) = 0 \implies a= 0$) – 2017-01-21
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    Here $B= L$ is a field, so it is an integral domain being its own fraction field. – 2017-01-21
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    And no $\varphi$ isn't surjective $F(x_1,\ldots,x_n) \to L $ unless $L = F(a_1,\ldots,a_n)$ – 2017-01-21

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