How to understand whether two distinguished open sets are isomorphic
$begingroup$
Let $R=k[x_1,...,x_n]$ be the polynomial ring over an algebraically closed field $k$ and let $f,gin R$. Assume that $f$ and $g$ are irreducible. How can I understand whether $k[x_1,...,x_n,frac{1}{f}]$ and $k[x_1,...,x_n,frac{1}{g}]$ are isomorphic?
Alternatively, consider the open sets $D(f)={pinmathbb{A}^nmid f(p)neq 0}$ and $D(g)={pinmathbb{A}^nmid g(p)neq 0}$. Is there an easy way to deduce whether $D(f)cong D(g)$?
That amounts of checking $V(tf-1)cong V(tg-1)$ in $k[x_1,...,x_n,t]$ and I believe that this should not be an easy problem.
algebraic-geometry
asked Mar 9 '18 at 12:01
LeventLevent
2,729925
$endgroup$
add a comment |
$begingroup$
Let $R=k[x_1,...,x_n]$ be the polynomial ring over an algebraically closed field $k$ and let $f,gin R$. Assume that $f$ and $g$ are irreducible. How can I understand whether $k[x_1,...,x_n,frac{1}{f}]$ and $k[x_1,...,x_n,frac{1}{g}]$ are isomorphic?
Alternatively, consider the open sets $D(f)={pinmathbb{A}^nmid f(p)neq 0}$ and $D(g)={pinmathbb{A}^nmid g(p)neq 0}$. Is there an easy way to deduce whether $D(f)cong D(g)$?
That amounts of checking $V(tf-1)cong V(tg-1)$ in $k[x_1,...,x_n,t]$ and I believe that this should not be an easy problem.
algebraic-geometry
asked Mar 9 '18 at 12:01
LeventLevent
2,729925
$endgroup$
$begingroup$
I think the only way this can happen is if there is an isomorphism $phi:k[x_1,ldots, x_n]$ to itself such that $phi(f)=g$.
$endgroup$
– Mohan
Jan 12 at 1:43
$begingroup$
Some one-variable examples: $k[x,1/x] simeq k[x,1/x^2]$. But what about $k[x,1/f]$ and $k[x,1/g]$ when $f(x)=x(x-1)(x-2)$ and $g(x)=x(x-1)(x-3)$?
$endgroup$
– Matt F.
Jan 15 at 14:48
add a comment |
$begingroup$
Let $R=k[x_1,...,x_n]$ be the polynomial ring over an algebraically closed field $k$ and let $f,gin R$. Assume that $f$ and $g$ are irreducible. How can I understand whether $k[x_1,...,x_n,frac{1}{f}]$ and $k[x_1,...,x_n,frac{1}{g}]$ are isomorphic?
Alternatively, consider the open sets $D(f)={pinmathbb{A}^nmid f(p)neq 0}$ and $D(g)={pinmathbb{A}^nmid g(p)neq 0}$. Is there an easy way to deduce whether $D(f)cong D(g)$?
That amounts of checking $V(tf-1)cong V(tg-1)$ in $k[x_1,...,x_n,t]$ and I believe that this should not be an easy problem.
algebraic-geometry
asked Mar 9 '18 at 12:01
LeventLevent
2,729925
$endgroup$
Let $R=k[x_1,...,x_n]$ be the polynomial ring over an algebraically closed field $k$ and let $f,gin R$. Assume that $f$ and $g$ are irreducible. How can I understand whether $k[x_1,...,x_n,frac{1}{f}]$ and $k[x_1,...,x_n,frac{1}{g}]$ are isomorphic?
Alternatively, consider the open sets $D(f)={pinmathbb{A}^nmid f(p)neq 0}$ and $D(g)={pinmathbb{A}^nmid g(p)neq 0}$. Is there an easy way to deduce whether $D(f)cong D(g)$?
That amounts of checking $V(tf-1)cong V(tg-1)$ in $k[x_1,...,x_n,t]$ and I believe that this should not be an easy problem.
algebraic-geometry
algebraic-geometry
asked Mar 9 '18 at 12:01
LeventLevent
2,729925
asked Mar 9 '18 at 12:01
LeventLevent
2,729925
asked Mar 9 '18 at 12:01
LeventLevent
2,729925
asked Mar 9 '18 at 12:01
LeventLevent
2,729925
asked Mar 9 '18 at 12:01
LeventLevent
2,729925
2,729925
$begingroup$
I think the only way this can happen is if there is an isomorphism $phi:k[x_1,ldots, x_n]$ to itself such that $phi(f)=g$.
$endgroup$
– Mohan
Jan 12 at 1:43
$begingroup$
Some one-variable examples: $k[x,1/x] simeq k[x,1/x^2]$. But what about $k[x,1/f]$ and $k[x,1/g]$ when $f(x)=x(x-1)(x-2)$ and $g(x)=x(x-1)(x-3)$?
$endgroup$
– Matt F.
Jan 15 at 14:48
add a comment |
$begingroup$
I think the only way this can happen is if there is an isomorphism $phi:k[x_1,ldots, x_n]$ to itself such that $phi(f)=g$.
$endgroup$
– Mohan
Jan 12 at 1:43
$begingroup$
Some one-variable examples: $k[x,1/x] simeq k[x,1/x^2]$. But what about $k[x,1/f]$ and $k[x,1/g]$ when $f(x)=x(x-1)(x-2)$ and $g(x)=x(x-1)(x-3)$?
$endgroup$
– Matt F.
Jan 15 at 14:48
$begingroup$
I think the only way this can happen is if there is an isomorphism $phi:k[x_1,ldots, x_n]$ to itself such that $phi(f)=g$.
$endgroup$
– Mohan
Jan 12 at 1:43
$begingroup$
I think the only way this can happen is if there is an isomorphism $phi:k[x_1,ldots, x_n]$ to itself such that $phi(f)=g$.
$endgroup$
– Mohan
Jan 12 at 1:43
$begingroup$
Some one-variable examples: $k[x,1/x] simeq k[x,1/x^2]$. But what about $k[x,1/f]$ and $k[x,1/g]$ when $f(x)=x(x-1)(x-2)$ and $g(x)=x(x-1)(x-3)$?
$endgroup$
– Matt F.
Jan 15 at 14:48
$begingroup$
Some one-variable examples: $k[x,1/x] simeq k[x,1/x^2]$. But what about $k[x,1/f]$ and $k[x,1/g]$ when $f(x)=x(x-1)(x-2)$ and $g(x)=x(x-1)(x-3)$?
$endgroup$
– Matt F.
Jan 15 at 14:48
add a comment |
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$begingroup$
I think the only way this can happen is if there is an isomorphism $phi:k[x_1,ldots, x_n]$ to itself such that $phi(f)=g$.
$endgroup$
– Mohan
Jan 12 at 1:43
$begingroup$
Some one-variable examples: $k[x,1/x] simeq k[x,1/x^2]$. But what about $k[x,1/f]$ and $k[x,1/g]$ when $f(x)=x(x-1)(x-2)$ and $g(x)=x(x-1)(x-3)$?
$endgroup$
– Matt F.
Jan 15 at 14:48