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I am trying to prove the following question: Let $K$ be a field with $E$ finite algebraic extension of $K$, $\tilde{E}=E(x_1,...,x_n)$ and $\tilde{K}=K(x_1,...,x_n)$ (fields of quotients of polynomials). Prove that $\tilde{E}$ is a finite algebraic extension of $\tilde{K}$. I'm trying to proceed by induction, so I need to answer the question on the title.

I am having trouble proving this because the field of quotients of polynomials really confuses me, and I'm having a hard time trying to find a basis of E(x).

I found the question here If $E$ is algebraic extension field of $F$, then also $E(x)$ is algebraic extension of $F(x)$? however, the answer there uses "trascendence degree", a topic not covered in class.

Any hints are appreciated

Luis Villagómez
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    Hint. Consider a special case when $E = K(a)$. Prove that in such case $\tilde{E} = \tilde{K}(a)$. By induction on the number of generators of $E$ over $K$ reduce the problem to the special case. – pmp May 20 '22 at 02:41

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