Mixing
The distinguished open sets are affine subschemes
$begingroup$
If $Y=$ Spec$A$ is an affine scheme and $D(f)subseteq Y$ (with $fin A$) is a distinguished open, I want to show that $(D(f),mathscr O_{Y|D(f)})$ is an affine scheme. Below there is my attempt of proof, but I've found a little problem, please help me to fix it.
Preliminary notations
If $(X,mathscr O_X)$ and $(Y,mathscr O_Y)$ are two schemes then a morphism between them is simply a morphism of locally ringed spaces $(varphi, varphi^{flat})$.
$varphi: Xrightarrow Y$ is a continuous map and $varphi^{flat}: mathscr O_Yrightarrow{varphi_{ast}mathscr O}_X $ is a morphism of sheaves (of rings) on $X$ such that the corresponding map (by adjunction) $varphi^{#}:{varphi^{-1}mathscr O}_Yrightarrow mathscr O_X$ has the property that the induced map on stalks $varphi_x^{#}:mathscr O_{Y,varphi(x)}rightarrow mathscr O_{X,x}$ is a local homomorphism of local rings for every $xin X$. Practically to give a morphism of schemes I can exhibit a couple $(varphi, varphi^{flat})$ or, in alternative, a couple $(varphi, varphi^{#})$ (with the above notations) but in both cases the map $varphi_x^{#}$ should be a local homomorphism.
Data: $A$ is a ring, $Y=$ Spec$A$ and $X=$ Spec$A_f$ with $fin A$. Clearly $mathscr O_Y$ and $mathscr O_X$ are the relative structure sheaves.
Claim: $(D(f),mathscr O_{Y|D(f)})cong (X,mathscr O_X)$
My proof: By an important fact from commutative algebra we know that $X$ is homeomorphic to $D(f)$ by $varphi$; this homeomorphism $varphi$ is induced by the canonical map $Arightarrow A_f$. Now I will show a map $psi_x:mathscr O_{Y,varphi(x)}rightarrow mathscr O_{X,x}$ that is an isomorphism of local rings for any $xin X$. Then we have finished because we can say that exits a map $varphi^{#}$ such that $varphi^{#}_x=psi_x$ for any $xin X$.
Since $mathscr O_{Y,varphi(x)}=A_{varphi(x)}$ (remember $x$ is a prime ideal) and $mathscr O_{X,x}=(A_f)_x$, the isomorphism $psi_x$ is the obviuous one.
The problem: Is it enough to give the maps $varphi^{#}_x=psi_x$ at level of the stalks? Who does ensure that once $psi_x$ is given for every $x$, does exist $varphi^{#}$ such that $varphi^{#}_x=psi_x$?
Edit:
For a simpler proof, maybe, I should use the following fact
I can take the identity of $A_f$ (it represent the global section of $X$ and $Y$) and then show that the corresponding map between $X$ and $Y$ is an isomorphism of schemes.
algebraic-geometry affine-schemes
asked Jun 1 '13 at 14:38
DubiousDubious
3,31952675
$endgroup$
add a comment |
$begingroup$
If $Y=$ Spec$A$ is an affine scheme and $D(f)subseteq Y$ (with $fin A$) is a distinguished open, I want to show that $(D(f),mathscr O_{Y|D(f)})$ is an affine scheme. Below there is my attempt of proof, but I've found a little problem, please help me to fix it.
Preliminary notations
If $(X,mathscr O_X)$ and $(Y,mathscr O_Y)$ are two schemes then a morphism between them is simply a morphism of locally ringed spaces $(varphi, varphi^{flat})$.
$varphi: Xrightarrow Y$ is a continuous map and $varphi^{flat}: mathscr O_Yrightarrow{varphi_{ast}mathscr O}_X $ is a morphism of sheaves (of rings) on $X$ such that the corresponding map (by adjunction) $varphi^{#}:{varphi^{-1}mathscr O}_Yrightarrow mathscr O_X$ has the property that the induced map on stalks $varphi_x^{#}:mathscr O_{Y,varphi(x)}rightarrow mathscr O_{X,x}$ is a local homomorphism of local rings for every $xin X$. Practically to give a morphism of schemes I can exhibit a couple $(varphi, varphi^{flat})$ or, in alternative, a couple $(varphi, varphi^{#})$ (with the above notations) but in both cases the map $varphi_x^{#}$ should be a local homomorphism.
Data: $A$ is a ring, $Y=$ Spec$A$ and $X=$ Spec$A_f$ with $fin A$. Clearly $mathscr O_Y$ and $mathscr O_X$ are the relative structure sheaves.
Claim: $(D(f),mathscr O_{Y|D(f)})cong (X,mathscr O_X)$
My proof: By an important fact from commutative algebra we know that $X$ is homeomorphic to $D(f)$ by $varphi$; this homeomorphism $varphi$ is induced by the canonical map $Arightarrow A_f$. Now I will show a map $psi_x:mathscr O_{Y,varphi(x)}rightarrow mathscr O_{X,x}$ that is an isomorphism of local rings for any $xin X$. Then we have finished because we can say that exits a map $varphi^{#}$ such that $varphi^{#}_x=psi_x$ for any $xin X$.
Since $mathscr O_{Y,varphi(x)}=A_{varphi(x)}$ (remember $x$ is a prime ideal) and $mathscr O_{X,x}=(A_f)_x$, the isomorphism $psi_x$ is the obviuous one.
The problem: Is it enough to give the maps $varphi^{#}_x=psi_x$ at level of the stalks? Who does ensure that once $psi_x$ is given for every $x$, does exist $varphi^{#}$ such that $varphi^{#}_x=psi_x$?
Edit:
For a simpler proof, maybe, I should use the following fact
I can take the identity of $A_f$ (it represent the global section of $X$ and $Y$) and then show that the corresponding map between $X$ and $Y$ is an isomorphism of schemes.
algebraic-geometry affine-schemes
asked Jun 1 '13 at 14:38
DubiousDubious
3,31952675
$endgroup$
2
$begingroup$
The definition of morphism of schemes (which is really just morphism of locally ringed spaces) is not quite right. $phi$ should be a continuous map of the underlying top. spaces, and then the other map should be a map of sheaves.
$endgroup$
– Stephen
Jun 1 '13 at 20:18
$begingroup$
uhm, yes you are right!
$endgroup$
– Dubious
Jun 1 '13 at 21:04
$begingroup$
Edited! I had changed many times the notations and so the definition of morphism between schemes was terribly wrong.
$endgroup$
– Dubious
Jun 1 '13 at 21:12
add a comment |
$begingroup$
If $Y=$ Spec$A$ is an affine scheme and $D(f)subseteq Y$ (with $fin A$) is a distinguished open, I want to show that $(D(f),mathscr O_{Y|D(f)})$ is an affine scheme. Below there is my attempt of proof, but I've found a little problem, please help me to fix it.
Preliminary notations
If $(X,mathscr O_X)$ and $(Y,mathscr O_Y)$ are two schemes then a morphism between them is simply a morphism of locally ringed spaces $(varphi, varphi^{flat})$.
$varphi: Xrightarrow Y$ is a continuous map and $varphi^{flat}: mathscr O_Yrightarrow{varphi_{ast}mathscr O}_X $ is a morphism of sheaves (of rings) on $X$ such that the corresponding map (by adjunction) $varphi^{#}:{varphi^{-1}mathscr O}_Yrightarrow mathscr O_X$ has the property that the induced map on stalks $varphi_x^{#}:mathscr O_{Y,varphi(x)}rightarrow mathscr O_{X,x}$ is a local homomorphism of local rings for every $xin X$. Practically to give a morphism of schemes I can exhibit a couple $(varphi, varphi^{flat})$ or, in alternative, a couple $(varphi, varphi^{#})$ (with the above notations) but in both cases the map $varphi_x^{#}$ should be a local homomorphism.
Data: $A$ is a ring, $Y=$ Spec$A$ and $X=$ Spec$A_f$ with $fin A$. Clearly $mathscr O_Y$ and $mathscr O_X$ are the relative structure sheaves.
Claim: $(D(f),mathscr O_{Y|D(f)})cong (X,mathscr O_X)$
My proof: By an important fact from commutative algebra we know that $X$ is homeomorphic to $D(f)$ by $varphi$; this homeomorphism $varphi$ is induced by the canonical map $Arightarrow A_f$. Now I will show a map $psi_x:mathscr O_{Y,varphi(x)}rightarrow mathscr O_{X,x}$ that is an isomorphism of local rings for any $xin X$. Then we have finished because we can say that exits a map $varphi^{#}$ such that $varphi^{#}_x=psi_x$ for any $xin X$.
Since $mathscr O_{Y,varphi(x)}=A_{varphi(x)}$ (remember $x$ is a prime ideal) and $mathscr O_{X,x}=(A_f)_x$, the isomorphism $psi_x$ is the obviuous one.
The problem: Is it enough to give the maps $varphi^{#}_x=psi_x$ at level of the stalks? Who does ensure that once $psi_x$ is given for every $x$, does exist $varphi^{#}$ such that $varphi^{#}_x=psi_x$?
Edit:
For a simpler proof, maybe, I should use the following fact
I can take the identity of $A_f$ (it represent the global section of $X$ and $Y$) and then show that the corresponding map between $X$ and $Y$ is an isomorphism of schemes.
algebraic-geometry affine-schemes
asked Jun 1 '13 at 14:38
DubiousDubious
3,31952675
$endgroup$
If $Y=$ Spec$A$ is an affine scheme and $D(f)subseteq Y$ (with $fin A$) is a distinguished open, I want to show that $(D(f),mathscr O_{Y|D(f)})$ is an affine scheme. Below there is my attempt of proof, but I've found a little problem, please help me to fix it.
Preliminary notations
If $(X,mathscr O_X)$ and $(Y,mathscr O_Y)$ are two schemes then a morphism between them is simply a morphism of locally ringed spaces $(varphi, varphi^{flat})$.
$varphi: Xrightarrow Y$ is a continuous map and $varphi^{flat}: mathscr O_Yrightarrow{varphi_{ast}mathscr O}_X $ is a morphism of sheaves (of rings) on $X$ such that the corresponding map (by adjunction) $varphi^{#}:{varphi^{-1}mathscr O}_Yrightarrow mathscr O_X$ has the property that the induced map on stalks $varphi_x^{#}:mathscr O_{Y,varphi(x)}rightarrow mathscr O_{X,x}$ is a local homomorphism of local rings for every $xin X$. Practically to give a morphism of schemes I can exhibit a couple $(varphi, varphi^{flat})$ or, in alternative, a couple $(varphi, varphi^{#})$ (with the above notations) but in both cases the map $varphi_x^{#}$ should be a local homomorphism.
Data: $A$ is a ring, $Y=$ Spec$A$ and $X=$ Spec$A_f$ with $fin A$. Clearly $mathscr O_Y$ and $mathscr O_X$ are the relative structure sheaves.
Claim: $(D(f),mathscr O_{Y|D(f)})cong (X,mathscr O_X)$
My proof: By an important fact from commutative algebra we know that $X$ is homeomorphic to $D(f)$ by $varphi$; this homeomorphism $varphi$ is induced by the canonical map $Arightarrow A_f$. Now I will show a map $psi_x:mathscr O_{Y,varphi(x)}rightarrow mathscr O_{X,x}$ that is an isomorphism of local rings for any $xin X$. Then we have finished because we can say that exits a map $varphi^{#}$ such that $varphi^{#}_x=psi_x$ for any $xin X$.
Since $mathscr O_{Y,varphi(x)}=A_{varphi(x)}$ (remember $x$ is a prime ideal) and $mathscr O_{X,x}=(A_f)_x$, the isomorphism $psi_x$ is the obviuous one.
The problem: Is it enough to give the maps $varphi^{#}_x=psi_x$ at level of the stalks? Who does ensure that once $psi_x$ is given for every $x$, does exist $varphi^{#}$ such that $varphi^{#}_x=psi_x$?
Edit:
For a simpler proof, maybe, I should use the following fact
I can take the identity of $A_f$ (it represent the global section of $X$ and $Y$) and then show that the corresponding map between $X$ and $Y$ is an isomorphism of schemes.
algebraic-geometry affine-schemes
algebraic-geometry affine-schemes
asked Jun 1 '13 at 14:38
DubiousDubious
3,31952675
asked Jun 1 '13 at 14:38
DubiousDubious
3,31952675
edited Jun 1 '13 at 21:21
Dubious
asked Jun 1 '13 at 14:38
DubiousDubious
3,31952675
asked Jun 1 '13 at 14:38
DubiousDubious
3,31952675
asked Jun 1 '13 at 14:38
DubiousDubious
3,31952675
3,31952675
2
$begingroup$
The definition of morphism of schemes (which is really just morphism of locally ringed spaces) is not quite right. $phi$ should be a continuous map of the underlying top. spaces, and then the other map should be a map of sheaves.
$endgroup$
– Stephen
Jun 1 '13 at 20:18
$begingroup$
uhm, yes you are right!
$endgroup$
– Dubious
Jun 1 '13 at 21:04
$begingroup$
Edited! I had changed many times the notations and so the definition of morphism between schemes was terribly wrong.
$endgroup$
– Dubious
Jun 1 '13 at 21:12
add a comment |
2
$begingroup$
The definition of morphism of schemes (which is really just morphism of locally ringed spaces) is not quite right. $phi$ should be a continuous map of the underlying top. spaces, and then the other map should be a map of sheaves.
$endgroup$
– Stephen
Jun 1 '13 at 20:18
$begingroup$
uhm, yes you are right!
$endgroup$
– Dubious
Jun 1 '13 at 21:04
$begingroup$
Edited! I had changed many times the notations and so the definition of morphism between schemes was terribly wrong.
$endgroup$
– Dubious
Jun 1 '13 at 21:12
2
2
$begingroup$
The definition of morphism of schemes (which is really just morphism of locally ringed spaces) is not quite right. $phi$ should be a continuous map of the underlying top. spaces, and then the other map should be a map of sheaves.
$endgroup$
– Stephen
Jun 1 '13 at 20:18
$begingroup$
The definition of morphism of schemes (which is really just morphism of locally ringed spaces) is not quite right. $phi$ should be a continuous map of the underlying top. spaces, and then the other map should be a map of sheaves.
$endgroup$
– Stephen
Jun 1 '13 at 20:18
$begingroup$
uhm, yes you are right!
$endgroup$
– Dubious
Jun 1 '13 at 21:04
$begingroup$
uhm, yes you are right!
$endgroup$
– Dubious
Jun 1 '13 at 21:04
$begingroup$
Edited! I had changed many times the notations and so the definition of morphism between schemes was terribly wrong.
$endgroup$
– Dubious
Jun 1 '13 at 21:12
$begingroup$
Edited! I had changed many times the notations and so the definition of morphism between schemes was terribly wrong.
$endgroup$
– Dubious
Jun 1 '13 at 21:12
add a comment |
1 Answer
1
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oldest
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$begingroup$
You don't just want to prove that $D(f)$ is isomorphic to $mathrm{Spec}(A_f)$ as locally ringed spaces via some random isomorphism. The ring map $psi:Arightarrow A_f$ (localization map) induces a morphism of schemes $g=mathrm{Spec}(psi):X=mathrm{Spec}(A_f)rightarrow Y=mathrm{Spec}(A)$. As you seem to already know, the underlying map of topological spaces is a homeomorphism onto $D(f)$, which is an open subset of $Y$.
In general, a morphism of locally ringed spaces $g:Xrightarrow Y$ is called an open immersion if $g$ induces (on the underlying topological spaces) a homeomorphism onto an open subset of the target and $g_x^sharp:mathscr{O}_{Y,g(x)}rightarrowmathscr{O}_{X,x}$ is an isomorphism for each $xin X$. It is equivalent for there to exist an open subset $V$ of $Y$ and an isomorphism of locally ringed spaces $h:Xcong V$ such that $icirc h=g$, where $i:Vhookrightarrow Y$ is the canonical inclusion morphism and $V$ is regarded as a locally ringed space in the standard way, by restricting the structure sheaf of $Y$. If $g$ is an open immersion in the first sense, then $V=g(X)$ is open in $Y$, and there is a unique isomorphism $h:Xcong V$ with $icirc h=g$.
So you just need to prove that the map $g$ is an open immersion. You already have the topological part, so the sheaf part amounts to proving that for each prime ideal $mathfrak{q}$ of $A_f$, the map of stalks $g_mathfrak{q}^sharp:A_{psi^{-1}(mathfrak{q})}rightarrow (A_f)_{mathfrak{q}}$, is an isomorphism. Note that this map is given, by definition, by $a/smapstopsi(a)/psi(s)$, using the standard description of localizations as equivalence classes of fractions. Basic properties of localization tell you that $mathfrak{q}$ has the form $mathfrak{p}A_f$ for a unique prime ideal $mathfrak{p}$ of $A$ not containing $f$, i.e. $mathfrak{p}in D(f)$, and the inverse image $psi^{-1}(mathfrak{q})$ is $mathfrak{p}$. So you are reduced to the following claim: if $mathfrak{p}in D(f)$, then the canonical map induced by $psi$, $A_mathfrak{p}rightarrow(A_f)_{mathfrak{p}A_f}$, is an isomorphism.
You can prove this by explicitly manipulating fractions, but that's kind of messy in my opinion. I think it is better to observe that the map in question is defined solely in terms of the universal property of localization. Namely, the composite of localization maps $Arightarrow A_frightarrow (A_f)_{mathfrak{p}A_f}$ visibly sends $Asetminusmathfrak{p}$ into the units of the target ring, so there is a unique $A$-algebra map $A_mathfrak{p}rightarrow (A_f)_{mathfrak{p}A_f}$. This is $g_{mathfrak{p}A_f}^#$. Using similar considerations, you can, via the universal property of localization, produce a ring map in the other direction, and then verify, now using the uniqueness in the universal property, that the composite of the two maps in both directions is the identity on the appropriate ring, i.e., the maps are inverses of one another.
So, your approach is more or less correct in spirit, but somewhat out of order. It is not enough to show that the local rings of two locally ringed spaces match up, i.e., this does not yield a morphism in general (it is true that two morphisms which have the same effect on topological spaces and induce the same maps of local rings are equal, but you cannot necessarily start with maps on local rings and produce a morphism). You have to start with the morphism (which you have, induced by $psi$), and then prove that the canonical maps on stalks induced by $psi$ are isomorphisms.
answered Jun 1 '13 at 21:21
Keenan KidwellKeenan Kidwell
19.8k13472
$endgroup$
3
$begingroup$
Dear Keenan, +1 for this nice answer. @OP The isomorphism that $g_{mathfrak{p}A_f}^sharp$ that Keenan is asking you to prove is actually a specific case of the following more general fact in commutative algebra: Let $S,T$ be multiplicative sets in a commutative ring $A$ with $S subseteq T$ and let $varphi : A to S^{-1}A$ be the canonical morphism. Then $T^{-1}A cong varphi(T)^{-1}S^{-1}(A)$. In Kennan's answer we can take $S = {1,f,f^2,ldots}$ and $T = A - mathfrak{p}$.
$endgroup$
– user38268
Jun 1 '13 at 23:18
1
$begingroup$
This is a wonderfully clear explanation: +1.
$endgroup$
– Georges Elencwajg
Jun 2 '13 at 7:17
1
$begingroup$
Dear @Benja: your commment nicely puts Keenan's isomorphism in perspective: +1.
$endgroup$
– Georges Elencwajg
Jun 2 '13 at 7:25
$begingroup$
Dear @Georges, Thank you for the kind words.
$endgroup$
– Keenan Kidwell
Jun 5 '13 at 17:07
$begingroup$
@GeorgesElencwajg: Sir, can you kindly give any reference to the equivalence stated in second paragraph? I am in urgent need of knowing the proof.
$endgroup$
– yojusmath
Jul 13 '14 at 9:41
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You don't just want to prove that $D(f)$ is isomorphic to $mathrm{Spec}(A_f)$ as locally ringed spaces via some random isomorphism. The ring map $psi:Arightarrow A_f$ (localization map) induces a morphism of schemes $g=mathrm{Spec}(psi):X=mathrm{Spec}(A_f)rightarrow Y=mathrm{Spec}(A)$. As you seem to already know, the underlying map of topological spaces is a homeomorphism onto $D(f)$, which is an open subset of $Y$.
In general, a morphism of locally ringed spaces $g:Xrightarrow Y$ is called an open immersion if $g$ induces (on the underlying topological spaces) a homeomorphism onto an open subset of the target and $g_x^sharp:mathscr{O}_{Y,g(x)}rightarrowmathscr{O}_{X,x}$ is an isomorphism for each $xin X$. It is equivalent for there to exist an open subset $V$ of $Y$ and an isomorphism of locally ringed spaces $h:Xcong V$ such that $icirc h=g$, where $i:Vhookrightarrow Y$ is the canonical inclusion morphism and $V$ is regarded as a locally ringed space in the standard way, by restricting the structure sheaf of $Y$. If $g$ is an open immersion in the first sense, then $V=g(X)$ is open in $Y$, and there is a unique isomorphism $h:Xcong V$ with $icirc h=g$.
So you just need to prove that the map $g$ is an open immersion. You already have the topological part, so the sheaf part amounts to proving that for each prime ideal $mathfrak{q}$ of $A_f$, the map of stalks $g_mathfrak{q}^sharp:A_{psi^{-1}(mathfrak{q})}rightarrow (A_f)_{mathfrak{q}}$, is an isomorphism. Note that this map is given, by definition, by $a/smapstopsi(a)/psi(s)$, using the standard description of localizations as equivalence classes of fractions. Basic properties of localization tell you that $mathfrak{q}$ has the form $mathfrak{p}A_f$ for a unique prime ideal $mathfrak{p}$ of $A$ not containing $f$, i.e. $mathfrak{p}in D(f)$, and the inverse image $psi^{-1}(mathfrak{q})$ is $mathfrak{p}$. So you are reduced to the following claim: if $mathfrak{p}in D(f)$, then the canonical map induced by $psi$, $A_mathfrak{p}rightarrow(A_f)_{mathfrak{p}A_f}$, is an isomorphism.
You can prove this by explicitly manipulating fractions, but that's kind of messy in my opinion. I think it is better to observe that the map in question is defined solely in terms of the universal property of localization. Namely, the composite of localization maps $Arightarrow A_frightarrow (A_f)_{mathfrak{p}A_f}$ visibly sends $Asetminusmathfrak{p}$ into the units of the target ring, so there is a unique $A$-algebra map $A_mathfrak{p}rightarrow (A_f)_{mathfrak{p}A_f}$. This is $g_{mathfrak{p}A_f}^#$. Using similar considerations, you can, via the universal property of localization, produce a ring map in the other direction, and then verify, now using the uniqueness in the universal property, that the composite of the two maps in both directions is the identity on the appropriate ring, i.e., the maps are inverses of one another.
So, your approach is more or less correct in spirit, but somewhat out of order. It is not enough to show that the local rings of two locally ringed spaces match up, i.e., this does not yield a morphism in general (it is true that two morphisms which have the same effect on topological spaces and induce the same maps of local rings are equal, but you cannot necessarily start with maps on local rings and produce a morphism). You have to start with the morphism (which you have, induced by $psi$), and then prove that the canonical maps on stalks induced by $psi$ are isomorphisms.
answered Jun 1 '13 at 21:21
Keenan KidwellKeenan Kidwell
19.8k13472
$endgroup$
3
$begingroup$
Dear Keenan, +1 for this nice answer. @OP The isomorphism that $g_{mathfrak{p}A_f}^sharp$ that Keenan is asking you to prove is actually a specific case of the following more general fact in commutative algebra: Let $S,T$ be multiplicative sets in a commutative ring $A$ with $S subseteq T$ and let $varphi : A to S^{-1}A$ be the canonical morphism. Then $T^{-1}A cong varphi(T)^{-1}S^{-1}(A)$. In Kennan's answer we can take $S = {1,f,f^2,ldots}$ and $T = A - mathfrak{p}$.
$endgroup$
– user38268
Jun 1 '13 at 23:18
1
$begingroup$
This is a wonderfully clear explanation: +1.
$endgroup$
– Georges Elencwajg
Jun 2 '13 at 7:17
1
$begingroup$
Dear @Benja: your commment nicely puts Keenan's isomorphism in perspective: +1.
$endgroup$
– Georges Elencwajg
Jun 2 '13 at 7:25
$begingroup$
Dear @Georges, Thank you for the kind words.
$endgroup$
– Keenan Kidwell
Jun 5 '13 at 17:07
$begingroup$
@GeorgesElencwajg: Sir, can you kindly give any reference to the equivalence stated in second paragraph? I am in urgent need of knowing the proof.
$endgroup$
– yojusmath
Jul 13 '14 at 9:41
add a comment |
$begingroup$
You don't just want to prove that $D(f)$ is isomorphic to $mathrm{Spec}(A_f)$ as locally ringed spaces via some random isomorphism. The ring map $psi:Arightarrow A_f$ (localization map) induces a morphism of schemes $g=mathrm{Spec}(psi):X=mathrm{Spec}(A_f)rightarrow Y=mathrm{Spec}(A)$. As you seem to already know, the underlying map of topological spaces is a homeomorphism onto $D(f)$, which is an open subset of $Y$.
In general, a morphism of locally ringed spaces $g:Xrightarrow Y$ is called an open immersion if $g$ induces (on the underlying topological spaces) a homeomorphism onto an open subset of the target and $g_x^sharp:mathscr{O}_{Y,g(x)}rightarrowmathscr{O}_{X,x}$ is an isomorphism for each $xin X$. It is equivalent for there to exist an open subset $V$ of $Y$ and an isomorphism of locally ringed spaces $h:Xcong V$ such that $icirc h=g$, where $i:Vhookrightarrow Y$ is the canonical inclusion morphism and $V$ is regarded as a locally ringed space in the standard way, by restricting the structure sheaf of $Y$. If $g$ is an open immersion in the first sense, then $V=g(X)$ is open in $Y$, and there is a unique isomorphism $h:Xcong V$ with $icirc h=g$.
So you just need to prove that the map $g$ is an open immersion. You already have the topological part, so the sheaf part amounts to proving that for each prime ideal $mathfrak{q}$ of $A_f$, the map of stalks $g_mathfrak{q}^sharp:A_{psi^{-1}(mathfrak{q})}rightarrow (A_f)_{mathfrak{q}}$, is an isomorphism. Note that this map is given, by definition, by $a/smapstopsi(a)/psi(s)$, using the standard description of localizations as equivalence classes of fractions. Basic properties of localization tell you that $mathfrak{q}$ has the form $mathfrak{p}A_f$ for a unique prime ideal $mathfrak{p}$ of $A$ not containing $f$, i.e. $mathfrak{p}in D(f)$, and the inverse image $psi^{-1}(mathfrak{q})$ is $mathfrak{p}$. So you are reduced to the following claim: if $mathfrak{p}in D(f)$, then the canonical map induced by $psi$, $A_mathfrak{p}rightarrow(A_f)_{mathfrak{p}A_f}$, is an isomorphism.
You can prove this by explicitly manipulating fractions, but that's kind of messy in my opinion. I think it is better to observe that the map in question is defined solely in terms of the universal property of localization. Namely, the composite of localization maps $Arightarrow A_frightarrow (A_f)_{mathfrak{p}A_f}$ visibly sends $Asetminusmathfrak{p}$ into the units of the target ring, so there is a unique $A$-algebra map $A_mathfrak{p}rightarrow (A_f)_{mathfrak{p}A_f}$. This is $g_{mathfrak{p}A_f}^#$. Using similar considerations, you can, via the universal property of localization, produce a ring map in the other direction, and then verify, now using the uniqueness in the universal property, that the composite of the two maps in both directions is the identity on the appropriate ring, i.e., the maps are inverses of one another.
So, your approach is more or less correct in spirit, but somewhat out of order. It is not enough to show that the local rings of two locally ringed spaces match up, i.e., this does not yield a morphism in general (it is true that two morphisms which have the same effect on topological spaces and induce the same maps of local rings are equal, but you cannot necessarily start with maps on local rings and produce a morphism). You have to start with the morphism (which you have, induced by $psi$), and then prove that the canonical maps on stalks induced by $psi$ are isomorphisms.
answered Jun 1 '13 at 21:21
Keenan KidwellKeenan Kidwell
19.8k13472
$endgroup$
3
$begingroup$
Dear Keenan, +1 for this nice answer. @OP The isomorphism that $g_{mathfrak{p}A_f}^sharp$ that Keenan is asking you to prove is actually a specific case of the following more general fact in commutative algebra: Let $S,T$ be multiplicative sets in a commutative ring $A$ with $S subseteq T$ and let $varphi : A to S^{-1}A$ be the canonical morphism. Then $T^{-1}A cong varphi(T)^{-1}S^{-1}(A)$. In Kennan's answer we can take $S = {1,f,f^2,ldots}$ and $T = A - mathfrak{p}$.
$endgroup$
– user38268
Jun 1 '13 at 23:18
1
$begingroup$
This is a wonderfully clear explanation: +1.
$endgroup$
– Georges Elencwajg
Jun 2 '13 at 7:17
1
$begingroup$
Dear @Benja: your commment nicely puts Keenan's isomorphism in perspective: +1.
$endgroup$
– Georges Elencwajg
Jun 2 '13 at 7:25
$begingroup$
Dear @Georges, Thank you for the kind words.
$endgroup$
– Keenan Kidwell
Jun 5 '13 at 17:07
$begingroup$
@GeorgesElencwajg: Sir, can you kindly give any reference to the equivalence stated in second paragraph? I am in urgent need of knowing the proof.
$endgroup$
– yojusmath
Jul 13 '14 at 9:41
add a comment |
$begingroup$
You don't just want to prove that $D(f)$ is isomorphic to $mathrm{Spec}(A_f)$ as locally ringed spaces via some random isomorphism. The ring map $psi:Arightarrow A_f$ (localization map) induces a morphism of schemes $g=mathrm{Spec}(psi):X=mathrm{Spec}(A_f)rightarrow Y=mathrm{Spec}(A)$. As you seem to already know, the underlying map of topological spaces is a homeomorphism onto $D(f)$, which is an open subset of $Y$.
In general, a morphism of locally ringed spaces $g:Xrightarrow Y$ is called an open immersion if $g$ induces (on the underlying topological spaces) a homeomorphism onto an open subset of the target and $g_x^sharp:mathscr{O}_{Y,g(x)}rightarrowmathscr{O}_{X,x}$ is an isomorphism for each $xin X$. It is equivalent for there to exist an open subset $V$ of $Y$ and an isomorphism of locally ringed spaces $h:Xcong V$ such that $icirc h=g$, where $i:Vhookrightarrow Y$ is the canonical inclusion morphism and $V$ is regarded as a locally ringed space in the standard way, by restricting the structure sheaf of $Y$. If $g$ is an open immersion in the first sense, then $V=g(X)$ is open in $Y$, and there is a unique isomorphism $h:Xcong V$ with $icirc h=g$.
So you just need to prove that the map $g$ is an open immersion. You already have the topological part, so the sheaf part amounts to proving that for each prime ideal $mathfrak{q}$ of $A_f$, the map of stalks $g_mathfrak{q}^sharp:A_{psi^{-1}(mathfrak{q})}rightarrow (A_f)_{mathfrak{q}}$, is an isomorphism. Note that this map is given, by definition, by $a/smapstopsi(a)/psi(s)$, using the standard description of localizations as equivalence classes of fractions. Basic properties of localization tell you that $mathfrak{q}$ has the form $mathfrak{p}A_f$ for a unique prime ideal $mathfrak{p}$ of $A$ not containing $f$, i.e. $mathfrak{p}in D(f)$, and the inverse image $psi^{-1}(mathfrak{q})$ is $mathfrak{p}$. So you are reduced to the following claim: if $mathfrak{p}in D(f)$, then the canonical map induced by $psi$, $A_mathfrak{p}rightarrow(A_f)_{mathfrak{p}A_f}$, is an isomorphism.
You can prove this by explicitly manipulating fractions, but that's kind of messy in my opinion. I think it is better to observe that the map in question is defined solely in terms of the universal property of localization. Namely, the composite of localization maps $Arightarrow A_frightarrow (A_f)_{mathfrak{p}A_f}$ visibly sends $Asetminusmathfrak{p}$ into the units of the target ring, so there is a unique $A$-algebra map $A_mathfrak{p}rightarrow (A_f)_{mathfrak{p}A_f}$. This is $g_{mathfrak{p}A_f}^#$. Using similar considerations, you can, via the universal property of localization, produce a ring map in the other direction, and then verify, now using the uniqueness in the universal property, that the composite of the two maps in both directions is the identity on the appropriate ring, i.e., the maps are inverses of one another.
So, your approach is more or less correct in spirit, but somewhat out of order. It is not enough to show that the local rings of two locally ringed spaces match up, i.e., this does not yield a morphism in general (it is true that two morphisms which have the same effect on topological spaces and induce the same maps of local rings are equal, but you cannot necessarily start with maps on local rings and produce a morphism). You have to start with the morphism (which you have, induced by $psi$), and then prove that the canonical maps on stalks induced by $psi$ are isomorphisms.
answered Jun 1 '13 at 21:21
Keenan KidwellKeenan Kidwell
19.8k13472
$endgroup$
You don't just want to prove that $D(f)$ is isomorphic to $mathrm{Spec}(A_f)$ as locally ringed spaces via some random isomorphism. The ring map $psi:Arightarrow A_f$ (localization map) induces a morphism of schemes $g=mathrm{Spec}(psi):X=mathrm{Spec}(A_f)rightarrow Y=mathrm{Spec}(A)$. As you seem to already know, the underlying map of topological spaces is a homeomorphism onto $D(f)$, which is an open subset of $Y$.
In general, a morphism of locally ringed spaces $g:Xrightarrow Y$ is called an open immersion if $g$ induces (on the underlying topological spaces) a homeomorphism onto an open subset of the target and $g_x^sharp:mathscr{O}_{Y,g(x)}rightarrowmathscr{O}_{X,x}$ is an isomorphism for each $xin X$. It is equivalent for there to exist an open subset $V$ of $Y$ and an isomorphism of locally ringed spaces $h:Xcong V$ such that $icirc h=g$, where $i:Vhookrightarrow Y$ is the canonical inclusion morphism and $V$ is regarded as a locally ringed space in the standard way, by restricting the structure sheaf of $Y$. If $g$ is an open immersion in the first sense, then $V=g(X)$ is open in $Y$, and there is a unique isomorphism $h:Xcong V$ with $icirc h=g$.
So you just need to prove that the map $g$ is an open immersion. You already have the topological part, so the sheaf part amounts to proving that for each prime ideal $mathfrak{q}$ of $A_f$, the map of stalks $g_mathfrak{q}^sharp:A_{psi^{-1}(mathfrak{q})}rightarrow (A_f)_{mathfrak{q}}$, is an isomorphism. Note that this map is given, by definition, by $a/smapstopsi(a)/psi(s)$, using the standard description of localizations as equivalence classes of fractions. Basic properties of localization tell you that $mathfrak{q}$ has the form $mathfrak{p}A_f$ for a unique prime ideal $mathfrak{p}$ of $A$ not containing $f$, i.e. $mathfrak{p}in D(f)$, and the inverse image $psi^{-1}(mathfrak{q})$ is $mathfrak{p}$. So you are reduced to the following claim: if $mathfrak{p}in D(f)$, then the canonical map induced by $psi$, $A_mathfrak{p}rightarrow(A_f)_{mathfrak{p}A_f}$, is an isomorphism.
You can prove this by explicitly manipulating fractions, but that's kind of messy in my opinion. I think it is better to observe that the map in question is defined solely in terms of the universal property of localization. Namely, the composite of localization maps $Arightarrow A_frightarrow (A_f)_{mathfrak{p}A_f}$ visibly sends $Asetminusmathfrak{p}$ into the units of the target ring, so there is a unique $A$-algebra map $A_mathfrak{p}rightarrow (A_f)_{mathfrak{p}A_f}$. This is $g_{mathfrak{p}A_f}^#$. Using similar considerations, you can, via the universal property of localization, produce a ring map in the other direction, and then verify, now using the uniqueness in the universal property, that the composite of the two maps in both directions is the identity on the appropriate ring, i.e., the maps are inverses of one another.
So, your approach is more or less correct in spirit, but somewhat out of order. It is not enough to show that the local rings of two locally ringed spaces match up, i.e., this does not yield a morphism in general (it is true that two morphisms which have the same effect on topological spaces and induce the same maps of local rings are equal, but you cannot necessarily start with maps on local rings and produce a morphism). You have to start with the morphism (which you have, induced by $psi$), and then prove that the canonical maps on stalks induced by $psi$ are isomorphisms.
answered Jun 1 '13 at 21:21
Keenan KidwellKeenan Kidwell
19.8k13472
edited Jan 11 at 22:14
answered Jun 1 '13 at 21:21
Keenan KidwellKeenan Kidwell
19.8k13472
answered Jun 1 '13 at 21:21
Keenan KidwellKeenan Kidwell
19.8k13472
answered Jun 1 '13 at 21:21
Keenan KidwellKeenan Kidwell
19.8k13472
19.8k13472
3
$begingroup$
Dear Keenan, +1 for this nice answer. @OP The isomorphism that $g_{mathfrak{p}A_f}^sharp$ that Keenan is asking you to prove is actually a specific case of the following more general fact in commutative algebra: Let $S,T$ be multiplicative sets in a commutative ring $A$ with $S subseteq T$ and let $varphi : A to S^{-1}A$ be the canonical morphism. Then $T^{-1}A cong varphi(T)^{-1}S^{-1}(A)$. In Kennan's answer we can take $S = {1,f,f^2,ldots}$ and $T = A - mathfrak{p}$.
$endgroup$
– user38268
Jun 1 '13 at 23:18
1
$begingroup$
This is a wonderfully clear explanation: +1.
$endgroup$
– Georges Elencwajg
Jun 2 '13 at 7:17
1
$begingroup$
Dear @Benja: your commment nicely puts Keenan's isomorphism in perspective: +1.
$endgroup$
– Georges Elencwajg
Jun 2 '13 at 7:25
$begingroup$
Dear @Georges, Thank you for the kind words.
$endgroup$
– Keenan Kidwell
Jun 5 '13 at 17:07
$begingroup$
@GeorgesElencwajg: Sir, can you kindly give any reference to the equivalence stated in second paragraph? I am in urgent need of knowing the proof.
$endgroup$
– yojusmath
Jul 13 '14 at 9:41
add a comment |
3
$begingroup$
Dear Keenan, +1 for this nice answer. @OP The isomorphism that $g_{mathfrak{p}A_f}^sharp$ that Keenan is asking you to prove is actually a specific case of the following more general fact in commutative algebra: Let $S,T$ be multiplicative sets in a commutative ring $A$ with $S subseteq T$ and let $varphi : A to S^{-1}A$ be the canonical morphism. Then $T^{-1}A cong varphi(T)^{-1}S^{-1}(A)$. In Kennan's answer we can take $S = {1,f,f^2,ldots}$ and $T = A - mathfrak{p}$.
$endgroup$
– user38268
Jun 1 '13 at 23:18
1
$begingroup$
This is a wonderfully clear explanation: +1.
$endgroup$
– Georges Elencwajg
Jun 2 '13 at 7:17
1
$begingroup$
Dear @Benja: your commment nicely puts Keenan's isomorphism in perspective: +1.
$endgroup$
– Georges Elencwajg
Jun 2 '13 at 7:25
$begingroup$
Dear @Georges, Thank you for the kind words.
$endgroup$
– Keenan Kidwell
Jun 5 '13 at 17:07
$begingroup$
@GeorgesElencwajg: Sir, can you kindly give any reference to the equivalence stated in second paragraph? I am in urgent need of knowing the proof.
$endgroup$
– yojusmath
Jul 13 '14 at 9:41
3
3
$begingroup$
Dear Keenan, +1 for this nice answer. @OP The isomorphism that $g_{mathfrak{p}A_f}^sharp$ that Keenan is asking you to prove is actually a specific case of the following more general fact in commutative algebra: Let $S,T$ be multiplicative sets in a commutative ring $A$ with $S subseteq T$ and let $varphi : A to S^{-1}A$ be the canonical morphism. Then $T^{-1}A cong varphi(T)^{-1}S^{-1}(A)$. In Kennan's answer we can take $S = {1,f,f^2,ldots}$ and $T = A - mathfrak{p}$.
$endgroup$
– user38268
Jun 1 '13 at 23:18
$begingroup$
Dear Keenan, +1 for this nice answer. @OP The isomorphism that $g_{mathfrak{p}A_f}^sharp$ that Keenan is asking you to prove is actually a specific case of the following more general fact in commutative algebra: Let $S,T$ be multiplicative sets in a commutative ring $A$ with $S subseteq T$ and let $varphi : A to S^{-1}A$ be the canonical morphism. Then $T^{-1}A cong varphi(T)^{-1}S^{-1}(A)$. In Kennan's answer we can take $S = {1,f,f^2,ldots}$ and $T = A - mathfrak{p}$.
$endgroup$
– user38268
Jun 1 '13 at 23:18
1
1
$begingroup$
This is a wonderfully clear explanation: +1.
$endgroup$
– Georges Elencwajg
Jun 2 '13 at 7:17
$begingroup$
This is a wonderfully clear explanation: +1.
$endgroup$
– Georges Elencwajg
Jun 2 '13 at 7:17
1
1
$begingroup$
Dear @Benja: your commment nicely puts Keenan's isomorphism in perspective: +1.
$endgroup$
– Georges Elencwajg
Jun 2 '13 at 7:25
$begingroup$
Dear @Benja: your commment nicely puts Keenan's isomorphism in perspective: +1.
$endgroup$
– Georges Elencwajg
Jun 2 '13 at 7:25
$begingroup$
Dear @Georges, Thank you for the kind words.
$endgroup$
– Keenan Kidwell
Jun 5 '13 at 17:07
$begingroup$
Dear @Georges, Thank you for the kind words.
$endgroup$
– Keenan Kidwell
Jun 5 '13 at 17:07
$begingroup$
@GeorgesElencwajg: Sir, can you kindly give any reference to the equivalence stated in second paragraph? I am in urgent need of knowing the proof.
$endgroup$
– yojusmath
Jul 13 '14 at 9:41
$begingroup$
@GeorgesElencwajg: Sir, can you kindly give any reference to the equivalence stated in second paragraph? I am in urgent need of knowing the proof.
$endgroup$
– yojusmath
Jul 13 '14 at 9:41
add a comment |
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The definition of morphism of schemes (which is really just morphism of locally ringed spaces) is not quite right. $phi$ should be a continuous map of the underlying top. spaces, and then the other map should be a map of sheaves.
$endgroup$
– Stephen
Jun 1 '13 at 20:18
$begingroup$
uhm, yes you are right!
$endgroup$
– Dubious
Jun 1 '13 at 21:04
$begingroup$
Edited! I had changed many times the notations and so the definition of morphism between schemes was terribly wrong.
$endgroup$
– Dubious
Jun 1 '13 at 21:12