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Why is the fraction field of $F[x,y,z]/(x^4+y^4+zy)$ equal to $F(y,z)[x]/(x^4+y^4+zy)$? Is there a general formula to compute the fraction field of $F[x_1,...,x_n]/(f)$ or $F[x_1,...,x_n]/(f_1,...,f_m)$

roi_saumon
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  • $F[x_1,...,x_n]/(x^2+y^4+zy)$ makes no sense. The variables in the polynomial ring are $x_i$, but the polynomial you divide out by is generated by a polynomial in $x, y, z$. – Arthur Jan 29 '19 at 18:58
  • Thanks, it was a typo, I edited. – roi_saumon Jan 29 '19 at 19:12
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    Did you really mean to change from “$/(x^2+y^4+zy)$” to “$/(x^4+y^4+zy)$” ? – Lubin Jan 29 '19 at 19:40

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Hint:

This is because $\;F(y,z)\longrightarrow F(y,z)[x]/(x^4+y^4+zy)$ is an integral morphism and we have the following well-known result:

Let $B$ be an integral domain, $A$ a subring of $B$ such that $B$ is integral over $A$. Then $B$ is a field if and only if $A$ is a field.

Bernard
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  • Thank you, but what is $A$ and $B$? is $A=F(y,z)$ and $B=F[x,y,z]/(x^4+y^4+zy)$? When you say the morphism is integral, does it mean that $F[x,y,z]/(x^4+y^4+zy)$ is integral over $F(y,z)[x]/(x^4+y^4+zy)$? – roi_saumon Jan 29 '19 at 20:39
  • No, I mean $B=F(y,z)[x]/(x^4+y^4+zy)$ is integral over $A=F(y,z)$. – Bernard Jan 29 '19 at 20:42