Exercise 4.1.6

Proof.

Let $f\in F(x_1,\dots ,x_n)$, $f=p∕q,p,q\in F[x_1,\dots ,x_n],q\ne 0$. Since ${\alpha }_1,\dots ,{\alpha }_n$ are algebraically independent over $F$, $q({\alpha }_1,\dots ,{\alpha }_n)\ne 0$. We can so define

(this quotient doesn’t depend on the choice of the representative $p∕q$ of $f$).

$\phi$ is a ring homomorphism.

By definition of $F({\alpha }_1,\dots ,{\alpha }_n)$, $\phi$ is surjective.

Let $f=p∕q\in F(x_1,\dots ,x_n)$, with $p,q\in F[x_1,\dots ,x_n],q\ne 0$. If $f\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$, then $p({\alpha }_1,\dots ,{\alpha }_n)∕q({\alpha }_1,\dots ,{\alpha }_n)=0$, thus $p({\alpha }_1,\dots ,{\alpha }_n)=0$. Since ${\alpha }_1,\dots ,{\alpha }_n$ are algebraically independent, $p=0$. Consequently $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\{0\}$, and so $\phi$ is a ring isomorphism between two fields: it is a field isomorphism.

Conclusion: If ${\alpha }_1,\dots ,{\alpha }_n\in L$ are algebraically independent over $F$, then

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