Isomorphism between localizations of graded ring $S_{(P)} cong [S_{(f)}]_{PS_f cap S_{(f)}}$
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I know that if $S$ is a graded ring, and $f$ is a homogeneous element of positive degree, then there is a bijection between the homogeneous prime ideals of the localization $S_f$ and the prime ideals of $S_{(f)}$, the subring of $S_f$ comprising the homogeneous elements of degree $0$ such as from this MSE question. How can I prove that this bijection gives rise to an isomorphism of the following rings: for a homogenous prime ideal $P$ of $S$ with $f notin P$, there is an isomorphism
$$S_{(P)} cong [S_{(f)}]_{P S_f cap S_{(f)}}.$$
Is this true? If yes, what is this isomorphism?
In other words, how do I lift the bijection between prime ideals mentioned above, to elements in the prime ideals related by this bijection. Here $S_{(P)}$ is the subring of degree $0$ elements of the localization $S_{P}$ like always.
algebraic-geometry commutative-algebra localization graded-rings
edited Nov 20 at 15:03
user26857
39.2k123882
asked Nov 19 at 23:59
ramanujan_dirac
266210
add a comment |
up vote
1
down vote
favorite
I know that if $S$ is a graded ring, and $f$ is a homogeneous element of positive degree, then there is a bijection between the homogeneous prime ideals of the localization $S_f$ and the prime ideals of $S_{(f)}$, the subring of $S_f$ comprising the homogeneous elements of degree $0$ such as from this MSE question. How can I prove that this bijection gives rise to an isomorphism of the following rings: for a homogenous prime ideal $P$ of $S$ with $f notin P$, there is an isomorphism
$$S_{(P)} cong [S_{(f)}]_{P S_f cap S_{(f)}}.$$
Is this true? If yes, what is this isomorphism?
In other words, how do I lift the bijection between prime ideals mentioned above, to elements in the prime ideals related by this bijection. Here $S_{(P)}$ is the subring of degree $0$ elements of the localization $S_{P}$ like always.
algebraic-geometry commutative-algebra localization graded-rings
edited Nov 20 at 15:03
user26857
39.2k123882
asked Nov 19 at 23:59
ramanujan_dirac
266210
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I know that if $S$ is a graded ring, and $f$ is a homogeneous element of positive degree, then there is a bijection between the homogeneous prime ideals of the localization $S_f$ and the prime ideals of $S_{(f)}$, the subring of $S_f$ comprising the homogeneous elements of degree $0$ such as from this MSE question. How can I prove that this bijection gives rise to an isomorphism of the following rings: for a homogenous prime ideal $P$ of $S$ with $f notin P$, there is an isomorphism
$$S_{(P)} cong [S_{(f)}]_{P S_f cap S_{(f)}}.$$
Is this true? If yes, what is this isomorphism?
In other words, how do I lift the bijection between prime ideals mentioned above, to elements in the prime ideals related by this bijection. Here $S_{(P)}$ is the subring of degree $0$ elements of the localization $S_{P}$ like always.
algebraic-geometry commutative-algebra localization graded-rings
edited Nov 20 at 15:03
user26857
39.2k123882
asked Nov 19 at 23:59
ramanujan_dirac
266210
I know that if $S$ is a graded ring, and $f$ is a homogeneous element of positive degree, then there is a bijection between the homogeneous prime ideals of the localization $S_f$ and the prime ideals of $S_{(f)}$, the subring of $S_f$ comprising the homogeneous elements of degree $0$ such as from this MSE question. How can I prove that this bijection gives rise to an isomorphism of the following rings: for a homogenous prime ideal $P$ of $S$ with $f notin P$, there is an isomorphism
$$S_{(P)} cong [S_{(f)}]_{P S_f cap S_{(f)}}.$$
Is this true? If yes, what is this isomorphism?
In other words, how do I lift the bijection between prime ideals mentioned above, to elements in the prime ideals related by this bijection. Here $S_{(P)}$ is the subring of degree $0$ elements of the localization $S_{P}$ like always.
algebraic-geometry commutative-algebra localization graded-rings
algebraic-geometry commutative-algebra localization graded-rings
edited Nov 20 at 15:03
user26857
39.2k123882
asked Nov 19 at 23:59
ramanujan_dirac
266210
edited Nov 20 at 15:03
user26857
39.2k123882
asked Nov 19 at 23:59
ramanujan_dirac
266210
edited Nov 20 at 15:03
user26857
39.2k123882
edited Nov 20 at 15:03
user26857
39.2k123882
edited Nov 20 at 15:03
user26857
39.2k123882
39.2k123882
asked Nov 19 at 23:59
ramanujan_dirac
266210
asked Nov 19 at 23:59
ramanujan_dirac
266210
asked Nov 19 at 23:59
ramanujan_dirac
266210
266210
add a comment |
add a comment |
1 Answer
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up vote
3
down vote
accepted
I suspect you may need that $S$ is generated in degree 1, but perhaps I'm missing a way to fix the following argument.
Since $P$ is a homogeneous prime and $S$ is generated in degree 1, there is an element $lambda$ of degree one such that $lambdanotin P$.
Let $phi : S_{(P)}to [S_{(f)}]_{PS_fcap S_{(f)}}$ be defined by $$phileft(frac{s}{r}right)=frac{slambda^k/f^n}{rlambda^k/f^n},$$
where $n$ is large enough that $ndeg f ge deg s$ and $k$ is such that $deg s + k = n deg f$. To see that this is well defined, observe that if
$frac{s}{r} = frac{s'}{r'}$, then there is $hnotin P$ such that $h(r's-rs')=0$, and for appropriate choices of $ell,j,k,n,m,o$, we have
$$frac{hlambda^ell}{f^o}left(frac{r'lambda^j}{f^m}frac{slambda^k}{f^n}-frac{s'lambda^j}{f^m}frac{rlambda^k}{f^n}right)=frac{lambda^{ell+j+k}}{f^{o+m+n}}(h(r's-s'r))=0.$$
Thus $phi(s/r)=phi(s'/r')$.
To see that $phi$ is the desired isomorphism, note $PS_f$ is the set of elements $a$ of $S_f$ such that $af^nin P$ for some $n$. Thus if $s/f^mnotin PS_f$ with $sin S_{mdeg f}$, and if $rin S_{ndeg f}$, and if $ell=max{n,m}$ we can define the inverse by
$$psileft(frac{r/f^n}{s/f^m}right)=frac{rf^{ell-n}}{sf^{ell-m}},$$
which works since we know $sf^{ell-m}notin P$, since $s/f^m notin PS_f$, and
both numerator and denominator have degree $ell deg f$.
It shouldn't be hard to verify that $psi$ is also well defined and $phi$ and $psi$ are inverses.
answered Nov 20 at 22:56
jgon
11.6k21839
Thanks for the answer @jgon. I believe the extra hypothesis is not necessary, but I am not able to prove it otherwise. So I will wait a couple more days to see if I get any more answers before accepting your excellent answer.
– ramanujan_dirac
Nov 22 at 5:32
add a comment |
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1 Answer
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1 Answer
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active
oldest
votes
active
oldest
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active
oldest
votes
up vote
3
down vote
accepted
I suspect you may need that $S$ is generated in degree 1, but perhaps I'm missing a way to fix the following argument.
Since $P$ is a homogeneous prime and $S$ is generated in degree 1, there is an element $lambda$ of degree one such that $lambdanotin P$.
Let $phi : S_{(P)}to [S_{(f)}]_{PS_fcap S_{(f)}}$ be defined by $$phileft(frac{s}{r}right)=frac{slambda^k/f^n}{rlambda^k/f^n},$$
where $n$ is large enough that $ndeg f ge deg s$ and $k$ is such that $deg s + k = n deg f$. To see that this is well defined, observe that if
$frac{s}{r} = frac{s'}{r'}$, then there is $hnotin P$ such that $h(r's-rs')=0$, and for appropriate choices of $ell,j,k,n,m,o$, we have
$$frac{hlambda^ell}{f^o}left(frac{r'lambda^j}{f^m}frac{slambda^k}{f^n}-frac{s'lambda^j}{f^m}frac{rlambda^k}{f^n}right)=frac{lambda^{ell+j+k}}{f^{o+m+n}}(h(r's-s'r))=0.$$
Thus $phi(s/r)=phi(s'/r')$.
To see that $phi$ is the desired isomorphism, note $PS_f$ is the set of elements $a$ of $S_f$ such that $af^nin P$ for some $n$. Thus if $s/f^mnotin PS_f$ with $sin S_{mdeg f}$, and if $rin S_{ndeg f}$, and if $ell=max{n,m}$ we can define the inverse by
$$psileft(frac{r/f^n}{s/f^m}right)=frac{rf^{ell-n}}{sf^{ell-m}},$$
which works since we know $sf^{ell-m}notin P$, since $s/f^m notin PS_f$, and
both numerator and denominator have degree $ell deg f$.
It shouldn't be hard to verify that $psi$ is also well defined and $phi$ and $psi$ are inverses.
answered Nov 20 at 22:56
jgon
11.6k21839
Thanks for the answer @jgon. I believe the extra hypothesis is not necessary, but I am not able to prove it otherwise. So I will wait a couple more days to see if I get any more answers before accepting your excellent answer.
– ramanujan_dirac
Nov 22 at 5:32
add a comment |
up vote
3
down vote
accepted
I suspect you may need that $S$ is generated in degree 1, but perhaps I'm missing a way to fix the following argument.
Since $P$ is a homogeneous prime and $S$ is generated in degree 1, there is an element $lambda$ of degree one such that $lambdanotin P$.
Let $phi : S_{(P)}to [S_{(f)}]_{PS_fcap S_{(f)}}$ be defined by $$phileft(frac{s}{r}right)=frac{slambda^k/f^n}{rlambda^k/f^n},$$
where $n$ is large enough that $ndeg f ge deg s$ and $k$ is such that $deg s + k = n deg f$. To see that this is well defined, observe that if
$frac{s}{r} = frac{s'}{r'}$, then there is $hnotin P$ such that $h(r's-rs')=0$, and for appropriate choices of $ell,j,k,n,m,o$, we have
$$frac{hlambda^ell}{f^o}left(frac{r'lambda^j}{f^m}frac{slambda^k}{f^n}-frac{s'lambda^j}{f^m}frac{rlambda^k}{f^n}right)=frac{lambda^{ell+j+k}}{f^{o+m+n}}(h(r's-s'r))=0.$$
Thus $phi(s/r)=phi(s'/r')$.
To see that $phi$ is the desired isomorphism, note $PS_f$ is the set of elements $a$ of $S_f$ such that $af^nin P$ for some $n$. Thus if $s/f^mnotin PS_f$ with $sin S_{mdeg f}$, and if $rin S_{ndeg f}$, and if $ell=max{n,m}$ we can define the inverse by
$$psileft(frac{r/f^n}{s/f^m}right)=frac{rf^{ell-n}}{sf^{ell-m}},$$
which works since we know $sf^{ell-m}notin P$, since $s/f^m notin PS_f$, and
both numerator and denominator have degree $ell deg f$.
It shouldn't be hard to verify that $psi$ is also well defined and $phi$ and $psi$ are inverses.
answered Nov 20 at 22:56
jgon
11.6k21839
Thanks for the answer @jgon. I believe the extra hypothesis is not necessary, but I am not able to prove it otherwise. So I will wait a couple more days to see if I get any more answers before accepting your excellent answer.
– ramanujan_dirac
Nov 22 at 5:32
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
I suspect you may need that $S$ is generated in degree 1, but perhaps I'm missing a way to fix the following argument.
Since $P$ is a homogeneous prime and $S$ is generated in degree 1, there is an element $lambda$ of degree one such that $lambdanotin P$.
Let $phi : S_{(P)}to [S_{(f)}]_{PS_fcap S_{(f)}}$ be defined by $$phileft(frac{s}{r}right)=frac{slambda^k/f^n}{rlambda^k/f^n},$$
where $n$ is large enough that $ndeg f ge deg s$ and $k$ is such that $deg s + k = n deg f$. To see that this is well defined, observe that if
$frac{s}{r} = frac{s'}{r'}$, then there is $hnotin P$ such that $h(r's-rs')=0$, and for appropriate choices of $ell,j,k,n,m,o$, we have
$$frac{hlambda^ell}{f^o}left(frac{r'lambda^j}{f^m}frac{slambda^k}{f^n}-frac{s'lambda^j}{f^m}frac{rlambda^k}{f^n}right)=frac{lambda^{ell+j+k}}{f^{o+m+n}}(h(r's-s'r))=0.$$
Thus $phi(s/r)=phi(s'/r')$.
To see that $phi$ is the desired isomorphism, note $PS_f$ is the set of elements $a$ of $S_f$ such that $af^nin P$ for some $n$. Thus if $s/f^mnotin PS_f$ with $sin S_{mdeg f}$, and if $rin S_{ndeg f}$, and if $ell=max{n,m}$ we can define the inverse by
$$psileft(frac{r/f^n}{s/f^m}right)=frac{rf^{ell-n}}{sf^{ell-m}},$$
which works since we know $sf^{ell-m}notin P$, since $s/f^m notin PS_f$, and
both numerator and denominator have degree $ell deg f$.
It shouldn't be hard to verify that $psi$ is also well defined and $phi$ and $psi$ are inverses.
answered Nov 20 at 22:56
jgon
11.6k21839
I suspect you may need that $S$ is generated in degree 1, but perhaps I'm missing a way to fix the following argument.
Since $P$ is a homogeneous prime and $S$ is generated in degree 1, there is an element $lambda$ of degree one such that $lambdanotin P$.
Let $phi : S_{(P)}to [S_{(f)}]_{PS_fcap S_{(f)}}$ be defined by $$phileft(frac{s}{r}right)=frac{slambda^k/f^n}{rlambda^k/f^n},$$
where $n$ is large enough that $ndeg f ge deg s$ and $k$ is such that $deg s + k = n deg f$. To see that this is well defined, observe that if
$frac{s}{r} = frac{s'}{r'}$, then there is $hnotin P$ such that $h(r's-rs')=0$, and for appropriate choices of $ell,j,k,n,m,o$, we have
$$frac{hlambda^ell}{f^o}left(frac{r'lambda^j}{f^m}frac{slambda^k}{f^n}-frac{s'lambda^j}{f^m}frac{rlambda^k}{f^n}right)=frac{lambda^{ell+j+k}}{f^{o+m+n}}(h(r's-s'r))=0.$$
Thus $phi(s/r)=phi(s'/r')$.
To see that $phi$ is the desired isomorphism, note $PS_f$ is the set of elements $a$ of $S_f$ such that $af^nin P$ for some $n$. Thus if $s/f^mnotin PS_f$ with $sin S_{mdeg f}$, and if $rin S_{ndeg f}$, and if $ell=max{n,m}$ we can define the inverse by
$$psileft(frac{r/f^n}{s/f^m}right)=frac{rf^{ell-n}}{sf^{ell-m}},$$
which works since we know $sf^{ell-m}notin P$, since $s/f^m notin PS_f$, and
both numerator and denominator have degree $ell deg f$.
It shouldn't be hard to verify that $psi$ is also well defined and $phi$ and $psi$ are inverses.
answered Nov 20 at 22:56
jgon
11.6k21839
answered Nov 20 at 22:56
jgon
11.6k21839
answered Nov 20 at 22:56
jgon
11.6k21839
answered Nov 20 at 22:56
jgon
11.6k21839
11.6k21839
Thanks for the answer @jgon. I believe the extra hypothesis is not necessary, but I am not able to prove it otherwise. So I will wait a couple more days to see if I get any more answers before accepting your excellent answer.
– ramanujan_dirac
Nov 22 at 5:32
add a comment |
Thanks for the answer @jgon. I believe the extra hypothesis is not necessary, but I am not able to prove it otherwise. So I will wait a couple more days to see if I get any more answers before accepting your excellent answer.
– ramanujan_dirac
Nov 22 at 5:32
Thanks for the answer @jgon. I believe the extra hypothesis is not necessary, but I am not able to prove it otherwise. So I will wait a couple more days to see if I get any more answers before accepting your excellent answer.
– ramanujan_dirac
Nov 22 at 5:32
Thanks for the answer @jgon. I believe the extra hypothesis is not necessary, but I am not able to prove it otherwise. So I will wait a couple more days to see if I get any more answers before accepting your excellent answer.
– ramanujan_dirac
Nov 22 at 5:32
add a comment |
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