Mixing
Fraction field of $F[x,y,z]/(x^4+y^4+zy)$ equal to $F(y,z)[x]/(x^4+y^4+zy)$?
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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 $F[x_1,...,x_n]/(f_1,...,f_m)$
commutative-algebra
asked Jan 29 at 18:44
roi_saumonroi_saumon
62938
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add a comment |
$begingroup$
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 $F[x_1,...,x_n]/(f_1,...,f_m)$
commutative-algebra
asked Jan 29 at 18:44
roi_saumonroi_saumon
62938
$endgroup$
$begingroup$
$F[x_1,...,x_n]/(x^2+y^4+zy)$ 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 $x, y, z$.
$endgroup$
– Arthur
Jan 29 at 18:58
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Thanks, it was a typo, I edited.
$endgroup$
– roi_saumon
Jan 29 at 19:12
1
$begingroup$
Did you really mean to change from “$/(x^2+y^4+zy)$” to “$/(x^4+y^4+zy)$” ?
$endgroup$
– Lubin
Jan 29 at 19:40
add a comment |
$begingroup$
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 $F[x_1,...,x_n]/(f_1,...,f_m)$
commutative-algebra
asked Jan 29 at 18:44
roi_saumonroi_saumon
62938
$endgroup$
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 $F[x_1,...,x_n]/(f_1,...,f_m)$
commutative-algebra
commutative-algebra
asked Jan 29 at 18:44
roi_saumonroi_saumon
62938
asked Jan 29 at 18:44
roi_saumonroi_saumon
62938
edited Jan 29 at 19:44
roi_saumon
asked Jan 29 at 18:44
roi_saumonroi_saumon
62938
asked Jan 29 at 18:44
roi_saumonroi_saumon
62938
asked Jan 29 at 18:44
roi_saumonroi_saumon
62938
62938
$begingroup$
$F[x_1,...,x_n]/(x^2+y^4+zy)$ 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 $x, y, z$.
$endgroup$
– Arthur
Jan 29 at 18:58
$begingroup$
Thanks, it was a typo, I edited.
$endgroup$
– roi_saumon
Jan 29 at 19:12
1
$begingroup$
Did you really mean to change from “$/(x^2+y^4+zy)$” to “$/(x^4+y^4+zy)$” ?
$endgroup$
– Lubin
Jan 29 at 19:40
add a comment |
$begingroup$
$F[x_1,...,x_n]/(x^2+y^4+zy)$ 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 $x, y, z$.
$endgroup$
– Arthur
Jan 29 at 18:58
$begingroup$
Thanks, it was a typo, I edited.
$endgroup$
– roi_saumon
Jan 29 at 19:12
1
$begingroup$
Did you really mean to change from “$/(x^2+y^4+zy)$” to “$/(x^4+y^4+zy)$” ?
$endgroup$
– Lubin
Jan 29 at 19:40
$begingroup$
$F[x_1,...,x_n]/(x^2+y^4+zy)$ 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 $x, y, z$.
$endgroup$
– Arthur
Jan 29 at 18:58
$begingroup$
$F[x_1,...,x_n]/(x^2+y^4+zy)$ 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 $x, y, z$.
$endgroup$
– Arthur
Jan 29 at 18:58
$begingroup$
Thanks, it was a typo, I edited.
$endgroup$
– roi_saumon
Jan 29 at 19:12
$begingroup$
Thanks, it was a typo, I edited.
$endgroup$
– roi_saumon
Jan 29 at 19:12
1
1
$begingroup$
Did you really mean to change from “$/(x^2+y^4+zy)$” to “$/(x^4+y^4+zy)$” ?
$endgroup$
– Lubin
Jan 29 at 19:40
$begingroup$
Did you really mean to change from “$/(x^2+y^4+zy)$” to “$/(x^4+y^4+zy)$” ?
$endgroup$
– Lubin
Jan 29 at 19:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Jan 29 at 20:16
BernardBernard
124k741118
$endgroup$
$begingroup$
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 $F(y,z)[x]/(x^4+y^4+zy)$?
$endgroup$
– roi_saumon
Jan 29 at 20:39
$begingroup$
No, I mean $B=F(y,z)[x]/(x^4+y^4+zy)$ is integral over $A=F(y,z)$.
$endgroup$
– Bernard
Jan 29 at 20:42
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
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.
answered Jan 29 at 20:16
BernardBernard
124k741118
$endgroup$
$begingroup$
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 $F(y,z)[x]/(x^4+y^4+zy)$?
$endgroup$
– roi_saumon
Jan 29 at 20:39
$begingroup$
No, I mean $B=F(y,z)[x]/(x^4+y^4+zy)$ is integral over $A=F(y,z)$.
$endgroup$
– Bernard
Jan 29 at 20:42
add a comment |
$begingroup$
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.
answered Jan 29 at 20:16
BernardBernard
124k741118
$endgroup$
$begingroup$
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 $F(y,z)[x]/(x^4+y^4+zy)$?
$endgroup$
– roi_saumon
Jan 29 at 20:39
$begingroup$
No, I mean $B=F(y,z)[x]/(x^4+y^4+zy)$ is integral over $A=F(y,z)$.
$endgroup$
– Bernard
Jan 29 at 20:42
add a comment |
$begingroup$
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.
answered Jan 29 at 20:16
BernardBernard
124k741118
$endgroup$
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.
answered Jan 29 at 20:16
BernardBernard
124k741118
answered Jan 29 at 20:16
BernardBernard
124k741118
answered Jan 29 at 20:16
BernardBernard
124k741118
answered Jan 29 at 20:16
BernardBernard
124k741118
124k741118
$begingroup$
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 $F(y,z)[x]/(x^4+y^4+zy)$?
$endgroup$
– roi_saumon
Jan 29 at 20:39
$begingroup$
No, I mean $B=F(y,z)[x]/(x^4+y^4+zy)$ is integral over $A=F(y,z)$.
$endgroup$
– Bernard
Jan 29 at 20:42
add a comment |
$begingroup$
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 $F(y,z)[x]/(x^4+y^4+zy)$?
$endgroup$
– roi_saumon
Jan 29 at 20:39
$begingroup$
No, I mean $B=F(y,z)[x]/(x^4+y^4+zy)$ is integral over $A=F(y,z)$.
$endgroup$
– Bernard
Jan 29 at 20:42
$begingroup$
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 $F(y,z)[x]/(x^4+y^4+zy)$?
$endgroup$
– roi_saumon
Jan 29 at 20:39
$begingroup$
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 $F(y,z)[x]/(x^4+y^4+zy)$?
$endgroup$
– roi_saumon
Jan 29 at 20:39
$begingroup$
No, I mean $B=F(y,z)[x]/(x^4+y^4+zy)$ is integral over $A=F(y,z)$.
$endgroup$
– Bernard
Jan 29 at 20:42
$begingroup$
No, I mean $B=F(y,z)[x]/(x^4+y^4+zy)$ is integral over $A=F(y,z)$.
$endgroup$
– Bernard
Jan 29 at 20:42
add a comment |
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$begingroup$
$F[x_1,...,x_n]/(x^2+y^4+zy)$ 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 $x, y, z$.
$endgroup$
– Arthur
Jan 29 at 18:58
$begingroup$
Thanks, it was a typo, I edited.
$endgroup$
– roi_saumon
Jan 29 at 19:12
1
$begingroup$
Did you really mean to change from “$/(x^2+y^4+zy)$” to “$/(x^4+y^4+zy)$” ?
$endgroup$
– Lubin
Jan 29 at 19:40