Mixing
Induced ring homomorphism by map on spectra
$begingroup$
I know that when we have a ring homomorphism $$ phi: Arightarrow B$$ this induces continuous map between the spectra $$ phi': operatorname{Spec} Brightarrow operatorname{Spec}A$$ which maps $p$ to $phi^{-1}(p)$. What I am wondering about is the other direction. Suppose we have a map between the spectra, is it possible to determine (explicitly) the induced map on the ring? Is it absolutely necessary to consider a morphism affine schemes to do this?
My particular question concerns showing that $A$ is finitely generated as a $mathbb C$-algebra, i.e. there exists a surjective morphism of rings $mathbb C[x_1,..., x_n]rightarrow A$, knowing that $Aotimes_{mathbb C} B$ is finitely generated (i.e. there exists a surjective morphism of rings $mathbb C[x_1,..., x_n]rightarrow Aotimes_{mathbb C} B $) using the fact that $operatorname{Spec}(Aotimes_mathbb C B)=operatorname{Spec} Atimes operatorname{Spec} B$. Or is this equality only true in the reduced and finitely generated case to begin with?
algebraic-geometry commutative-algebra
edited Jan 22 '16 at 20:34
user26857
39.3k124083
asked Jan 20 '16 at 23:22
user306194user306194
1427
$endgroup$
add a comment |
$begingroup$
I know that when we have a ring homomorphism $$ phi: Arightarrow B$$ this induces continuous map between the spectra $$ phi': operatorname{Spec} Brightarrow operatorname{Spec}A$$ which maps $p$ to $phi^{-1}(p)$. What I am wondering about is the other direction. Suppose we have a map between the spectra, is it possible to determine (explicitly) the induced map on the ring? Is it absolutely necessary to consider a morphism affine schemes to do this?
My particular question concerns showing that $A$ is finitely generated as a $mathbb C$-algebra, i.e. there exists a surjective morphism of rings $mathbb C[x_1,..., x_n]rightarrow A$, knowing that $Aotimes_{mathbb C} B$ is finitely generated (i.e. there exists a surjective morphism of rings $mathbb C[x_1,..., x_n]rightarrow Aotimes_{mathbb C} B $) using the fact that $operatorname{Spec}(Aotimes_mathbb C B)=operatorname{Spec} Atimes operatorname{Spec} B$. Or is this equality only true in the reduced and finitely generated case to begin with?
algebraic-geometry commutative-algebra
edited Jan 22 '16 at 20:34
user26857
39.3k124083
asked Jan 20 '16 at 23:22
user306194user306194
1427
$endgroup$
3
$begingroup$
Spec is neither faithful nor full, so given a continuous map on spectra, a map on rings inducing it will neither exist nor be unique in general. The "equality" you claim is false; the underlying topological space of the product of two varieties does not have the product topology. (All told, Spec is a very poorly behaved functor.)
$endgroup$
– Qiaochu Yuan
Jan 21 '16 at 0:40
2
$begingroup$
Anytime $Y$ is an affine scheme we have a bijective correspondance between $operatorname{Hom}(X,Y)$ and $operatorname{Hom}(mathcal{O}(Y), mathcal{O}(X))$, where we consider morphsms of locally ringed spaces rather than just continuous maps.
$endgroup$
– basket
Jan 21 '16 at 0:52
1
$begingroup$
An example: you can define a homomorphism $k[x] to k[y]/(y^2)$ either by sending $x$ to $y$ or $0$. The corresponding continuous maps on spectra can't tell the difference.
$endgroup$
– Hoot
Jan 21 '16 at 2:05
add a comment |
$begingroup$
I know that when we have a ring homomorphism $$ phi: Arightarrow B$$ this induces continuous map between the spectra $$ phi': operatorname{Spec} Brightarrow operatorname{Spec}A$$ which maps $p$ to $phi^{-1}(p)$. What I am wondering about is the other direction. Suppose we have a map between the spectra, is it possible to determine (explicitly) the induced map on the ring? Is it absolutely necessary to consider a morphism affine schemes to do this?
My particular question concerns showing that $A$ is finitely generated as a $mathbb C$-algebra, i.e. there exists a surjective morphism of rings $mathbb C[x_1,..., x_n]rightarrow A$, knowing that $Aotimes_{mathbb C} B$ is finitely generated (i.e. there exists a surjective morphism of rings $mathbb C[x_1,..., x_n]rightarrow Aotimes_{mathbb C} B $) using the fact that $operatorname{Spec}(Aotimes_mathbb C B)=operatorname{Spec} Atimes operatorname{Spec} B$. Or is this equality only true in the reduced and finitely generated case to begin with?
algebraic-geometry commutative-algebra
edited Jan 22 '16 at 20:34
user26857
39.3k124083
asked Jan 20 '16 at 23:22
user306194user306194
1427
$endgroup$
I know that when we have a ring homomorphism $$ phi: Arightarrow B$$ this induces continuous map between the spectra $$ phi': operatorname{Spec} Brightarrow operatorname{Spec}A$$ which maps $p$ to $phi^{-1}(p)$. What I am wondering about is the other direction. Suppose we have a map between the spectra, is it possible to determine (explicitly) the induced map on the ring? Is it absolutely necessary to consider a morphism affine schemes to do this?
My particular question concerns showing that $A$ is finitely generated as a $mathbb C$-algebra, i.e. there exists a surjective morphism of rings $mathbb C[x_1,..., x_n]rightarrow A$, knowing that $Aotimes_{mathbb C} B$ is finitely generated (i.e. there exists a surjective morphism of rings $mathbb C[x_1,..., x_n]rightarrow Aotimes_{mathbb C} B $) using the fact that $operatorname{Spec}(Aotimes_mathbb C B)=operatorname{Spec} Atimes operatorname{Spec} B$. Or is this equality only true in the reduced and finitely generated case to begin with?
algebraic-geometry commutative-algebra
algebraic-geometry commutative-algebra
edited Jan 22 '16 at 20:34
user26857
39.3k124083
asked Jan 20 '16 at 23:22
user306194user306194
1427
edited Jan 22 '16 at 20:34
user26857
39.3k124083
asked Jan 20 '16 at 23:22
user306194user306194
1427
edited Jan 22 '16 at 20:34
user26857
39.3k124083
edited Jan 22 '16 at 20:34
user26857
39.3k124083
edited Jan 22 '16 at 20:34
user26857
39.3k124083
39.3k124083
asked Jan 20 '16 at 23:22
user306194user306194
1427
asked Jan 20 '16 at 23:22
user306194user306194
1427
asked Jan 20 '16 at 23:22
user306194user306194
1427
1427
3
$begingroup$
Spec is neither faithful nor full, so given a continuous map on spectra, a map on rings inducing it will neither exist nor be unique in general. The "equality" you claim is false; the underlying topological space of the product of two varieties does not have the product topology. (All told, Spec is a very poorly behaved functor.)
$endgroup$
– Qiaochu Yuan
Jan 21 '16 at 0:40
2
$begingroup$
Anytime $Y$ is an affine scheme we have a bijective correspondance between $operatorname{Hom}(X,Y)$ and $operatorname{Hom}(mathcal{O}(Y), mathcal{O}(X))$, where we consider morphsms of locally ringed spaces rather than just continuous maps.
$endgroup$
– basket
Jan 21 '16 at 0:52
1
$begingroup$
An example: you can define a homomorphism $k[x] to k[y]/(y^2)$ either by sending $x$ to $y$ or $0$. The corresponding continuous maps on spectra can't tell the difference.
$endgroup$
– Hoot
Jan 21 '16 at 2:05
add a comment |
3
$begingroup$
Spec is neither faithful nor full, so given a continuous map on spectra, a map on rings inducing it will neither exist nor be unique in general. The "equality" you claim is false; the underlying topological space of the product of two varieties does not have the product topology. (All told, Spec is a very poorly behaved functor.)
$endgroup$
– Qiaochu Yuan
Jan 21 '16 at 0:40
2
$begingroup$
Anytime $Y$ is an affine scheme we have a bijective correspondance between $operatorname{Hom}(X,Y)$ and $operatorname{Hom}(mathcal{O}(Y), mathcal{O}(X))$, where we consider morphsms of locally ringed spaces rather than just continuous maps.
$endgroup$
– basket
Jan 21 '16 at 0:52
1
$begingroup$
An example: you can define a homomorphism $k[x] to k[y]/(y^2)$ either by sending $x$ to $y$ or $0$. The corresponding continuous maps on spectra can't tell the difference.
$endgroup$
– Hoot
Jan 21 '16 at 2:05
3
3
$begingroup$
Spec is neither faithful nor full, so given a continuous map on spectra, a map on rings inducing it will neither exist nor be unique in general. The "equality" you claim is false; the underlying topological space of the product of two varieties does not have the product topology. (All told, Spec is a very poorly behaved functor.)
$endgroup$
– Qiaochu Yuan
Jan 21 '16 at 0:40
$begingroup$
Spec is neither faithful nor full, so given a continuous map on spectra, a map on rings inducing it will neither exist nor be unique in general. The "equality" you claim is false; the underlying topological space of the product of two varieties does not have the product topology. (All told, Spec is a very poorly behaved functor.)
$endgroup$
– Qiaochu Yuan
Jan 21 '16 at 0:40
2
2
$begingroup$
Anytime $Y$ is an affine scheme we have a bijective correspondance between $operatorname{Hom}(X,Y)$ and $operatorname{Hom}(mathcal{O}(Y), mathcal{O}(X))$, where we consider morphsms of locally ringed spaces rather than just continuous maps.
$endgroup$
– basket
Jan 21 '16 at 0:52
$begingroup$
Anytime $Y$ is an affine scheme we have a bijective correspondance between $operatorname{Hom}(X,Y)$ and $operatorname{Hom}(mathcal{O}(Y), mathcal{O}(X))$, where we consider morphsms of locally ringed spaces rather than just continuous maps.
$endgroup$
– basket
Jan 21 '16 at 0:52
1
1
$begingroup$
An example: you can define a homomorphism $k[x] to k[y]/(y^2)$ either by sending $x$ to $y$ or $0$. The corresponding continuous maps on spectra can't tell the difference.
$endgroup$
– Hoot
Jan 21 '16 at 2:05
$begingroup$
An example: you can define a homomorphism $k[x] to k[y]/(y^2)$ either by sending $x$ to $y$ or $0$. The corresponding continuous maps on spectra can't tell the difference.
$endgroup$
– Hoot
Jan 21 '16 at 2:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I prefer summerize and improve the answers of Qiaochu Yuan and basket!
Let $A$ and $B$ be rings, $X=operatorname{Spec}A$ and $Y=operatorname{Spec}B$; one can prove that the function
$$
Phi:(f,f^{sharp})inoperatorname{Hom}_{bf LocRingSp}((X,mathcal{O}_X),(Y,mathcal{O}_Y))to f^{sharp}(Y)inoperatorname{Hom}_{bf Ring}(B,A)
$$
is bijective; moreover, the morphism $(f,f^{sharp})$ is uniquely determined by $f^{sharp}(Y)$.
What I mean? Which is the improvement?
As you (user306194) affirm: given a morphism of rings (that is $f^{sharp}(Y)$ in my notation) you can construct a unique morphism of locally ringed spaces $(f,f^{sharp})$; but, given a continuous map $f$ between the support of locally ringed spaces, you can't define uniquely $f^{sharp}$!
For example: you read the previous comment of Hoot.
Obviously, all this works in the setting of schemes.
In general, let $A$ be a ring and let $Y$ be a scheme, $X=operatorname{Spec}A$; one can prove that the function
$$
Phi:(f,f^{sharp})inoperatorname{Hom}_{bf Sch}(Y,X)to f^{sharp}(X)inoperatorname{Hom}_{bf Ring}(A,mathcal{O}_Y(Y))
$$
is bijective.
Moreover in general: let $S$ be a scheme, let $X$ be an $S$-scheme with structure morphism $p$ and let $mathcal{A}$ be a quasi-coherent $mathcal{O}_S$-algebra; one can prove that the function
$$
Phi:(f,f^{sharp})inoperatorname{Hom}_{S-bf Sch}(X,operatorname{Spec}mathcal{A})to f^{sharp}(operatorname{Spec}mathcal{A})inoperatorname{Hom}_{mathcal{O}_S-bf Alg}(mathcal{A},p_{*}mathcal{O}_X)
$$
is bijective; where $operatorname{Spec}mathcal{A}$ is the relative spectrum of $mathcal{A}$.
For a reference, I council Bosch - Commutative Algebra and Algebraic geometry, Springer, sections 6.6 and 7.1.
In conclusion, I remember you that $operatorname{Spec}(Aotimes_{mathbb{C}}B)$ is canonically isomorphic to $operatorname{Spec}Atimes_{operatorname{Spec}mathbb{C}}operatorname{Spec}B$; usually, the Cartesian product of sets is very bad in the category of schemes.
From the same book, I council you the section 7.2 with any examples and exercises 3, 4 and 5.
answered Jan 22 '16 at 11:14
Armando j18eosArmando j18eos
2,63511328
$endgroup$
add a comment |
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$begingroup$
I prefer summerize and improve the answers of Qiaochu Yuan and basket!
Let $A$ and $B$ be rings, $X=operatorname{Spec}A$ and $Y=operatorname{Spec}B$; one can prove that the function
$$
Phi:(f,f^{sharp})inoperatorname{Hom}_{bf LocRingSp}((X,mathcal{O}_X),(Y,mathcal{O}_Y))to f^{sharp}(Y)inoperatorname{Hom}_{bf Ring}(B,A)
$$
is bijective; moreover, the morphism $(f,f^{sharp})$ is uniquely determined by $f^{sharp}(Y)$.
What I mean? Which is the improvement?
As you (user306194) affirm: given a morphism of rings (that is $f^{sharp}(Y)$ in my notation) you can construct a unique morphism of locally ringed spaces $(f,f^{sharp})$; but, given a continuous map $f$ between the support of locally ringed spaces, you can't define uniquely $f^{sharp}$!
For example: you read the previous comment of Hoot.
Obviously, all this works in the setting of schemes.
In general, let $A$ be a ring and let $Y$ be a scheme, $X=operatorname{Spec}A$; one can prove that the function
$$
Phi:(f,f^{sharp})inoperatorname{Hom}_{bf Sch}(Y,X)to f^{sharp}(X)inoperatorname{Hom}_{bf Ring}(A,mathcal{O}_Y(Y))
$$
is bijective.
Moreover in general: let $S$ be a scheme, let $X$ be an $S$-scheme with structure morphism $p$ and let $mathcal{A}$ be a quasi-coherent $mathcal{O}_S$-algebra; one can prove that the function
$$
Phi:(f,f^{sharp})inoperatorname{Hom}_{S-bf Sch}(X,operatorname{Spec}mathcal{A})to f^{sharp}(operatorname{Spec}mathcal{A})inoperatorname{Hom}_{mathcal{O}_S-bf Alg}(mathcal{A},p_{*}mathcal{O}_X)
$$
is bijective; where $operatorname{Spec}mathcal{A}$ is the relative spectrum of $mathcal{A}$.
For a reference, I council Bosch - Commutative Algebra and Algebraic geometry, Springer, sections 6.6 and 7.1.
In conclusion, I remember you that $operatorname{Spec}(Aotimes_{mathbb{C}}B)$ is canonically isomorphic to $operatorname{Spec}Atimes_{operatorname{Spec}mathbb{C}}operatorname{Spec}B$; usually, the Cartesian product of sets is very bad in the category of schemes.
From the same book, I council you the section 7.2 with any examples and exercises 3, 4 and 5.
answered Jan 22 '16 at 11:14
Armando j18eosArmando j18eos
2,63511328
$endgroup$
add a comment |
$begingroup$
I prefer summerize and improve the answers of Qiaochu Yuan and basket!
Let $A$ and $B$ be rings, $X=operatorname{Spec}A$ and $Y=operatorname{Spec}B$; one can prove that the function
$$
Phi:(f,f^{sharp})inoperatorname{Hom}_{bf LocRingSp}((X,mathcal{O}_X),(Y,mathcal{O}_Y))to f^{sharp}(Y)inoperatorname{Hom}_{bf Ring}(B,A)
$$
is bijective; moreover, the morphism $(f,f^{sharp})$ is uniquely determined by $f^{sharp}(Y)$.
What I mean? Which is the improvement?
As you (user306194) affirm: given a morphism of rings (that is $f^{sharp}(Y)$ in my notation) you can construct a unique morphism of locally ringed spaces $(f,f^{sharp})$; but, given a continuous map $f$ between the support of locally ringed spaces, you can't define uniquely $f^{sharp}$!
For example: you read the previous comment of Hoot.
Obviously, all this works in the setting of schemes.
In general, let $A$ be a ring and let $Y$ be a scheme, $X=operatorname{Spec}A$; one can prove that the function
$$
Phi:(f,f^{sharp})inoperatorname{Hom}_{bf Sch}(Y,X)to f^{sharp}(X)inoperatorname{Hom}_{bf Ring}(A,mathcal{O}_Y(Y))
$$
is bijective.
Moreover in general: let $S$ be a scheme, let $X$ be an $S$-scheme with structure morphism $p$ and let $mathcal{A}$ be a quasi-coherent $mathcal{O}_S$-algebra; one can prove that the function
$$
Phi:(f,f^{sharp})inoperatorname{Hom}_{S-bf Sch}(X,operatorname{Spec}mathcal{A})to f^{sharp}(operatorname{Spec}mathcal{A})inoperatorname{Hom}_{mathcal{O}_S-bf Alg}(mathcal{A},p_{*}mathcal{O}_X)
$$
is bijective; where $operatorname{Spec}mathcal{A}$ is the relative spectrum of $mathcal{A}$.
For a reference, I council Bosch - Commutative Algebra and Algebraic geometry, Springer, sections 6.6 and 7.1.
In conclusion, I remember you that $operatorname{Spec}(Aotimes_{mathbb{C}}B)$ is canonically isomorphic to $operatorname{Spec}Atimes_{operatorname{Spec}mathbb{C}}operatorname{Spec}B$; usually, the Cartesian product of sets is very bad in the category of schemes.
From the same book, I council you the section 7.2 with any examples and exercises 3, 4 and 5.
answered Jan 22 '16 at 11:14
Armando j18eosArmando j18eos
2,63511328
$endgroup$
add a comment |
$begingroup$
I prefer summerize and improve the answers of Qiaochu Yuan and basket!
Let $A$ and $B$ be rings, $X=operatorname{Spec}A$ and $Y=operatorname{Spec}B$; one can prove that the function
$$
Phi:(f,f^{sharp})inoperatorname{Hom}_{bf LocRingSp}((X,mathcal{O}_X),(Y,mathcal{O}_Y))to f^{sharp}(Y)inoperatorname{Hom}_{bf Ring}(B,A)
$$
is bijective; moreover, the morphism $(f,f^{sharp})$ is uniquely determined by $f^{sharp}(Y)$.
What I mean? Which is the improvement?
As you (user306194) affirm: given a morphism of rings (that is $f^{sharp}(Y)$ in my notation) you can construct a unique morphism of locally ringed spaces $(f,f^{sharp})$; but, given a continuous map $f$ between the support of locally ringed spaces, you can't define uniquely $f^{sharp}$!
For example: you read the previous comment of Hoot.
Obviously, all this works in the setting of schemes.
In general, let $A$ be a ring and let $Y$ be a scheme, $X=operatorname{Spec}A$; one can prove that the function
$$
Phi:(f,f^{sharp})inoperatorname{Hom}_{bf Sch}(Y,X)to f^{sharp}(X)inoperatorname{Hom}_{bf Ring}(A,mathcal{O}_Y(Y))
$$
is bijective.
Moreover in general: let $S$ be a scheme, let $X$ be an $S$-scheme with structure morphism $p$ and let $mathcal{A}$ be a quasi-coherent $mathcal{O}_S$-algebra; one can prove that the function
$$
Phi:(f,f^{sharp})inoperatorname{Hom}_{S-bf Sch}(X,operatorname{Spec}mathcal{A})to f^{sharp}(operatorname{Spec}mathcal{A})inoperatorname{Hom}_{mathcal{O}_S-bf Alg}(mathcal{A},p_{*}mathcal{O}_X)
$$
is bijective; where $operatorname{Spec}mathcal{A}$ is the relative spectrum of $mathcal{A}$.
For a reference, I council Bosch - Commutative Algebra and Algebraic geometry, Springer, sections 6.6 and 7.1.
In conclusion, I remember you that $operatorname{Spec}(Aotimes_{mathbb{C}}B)$ is canonically isomorphic to $operatorname{Spec}Atimes_{operatorname{Spec}mathbb{C}}operatorname{Spec}B$; usually, the Cartesian product of sets is very bad in the category of schemes.
From the same book, I council you the section 7.2 with any examples and exercises 3, 4 and 5.
answered Jan 22 '16 at 11:14
Armando j18eosArmando j18eos
2,63511328
$endgroup$
I prefer summerize and improve the answers of Qiaochu Yuan and basket!
Let $A$ and $B$ be rings, $X=operatorname{Spec}A$ and $Y=operatorname{Spec}B$; one can prove that the function
$$
Phi:(f,f^{sharp})inoperatorname{Hom}_{bf LocRingSp}((X,mathcal{O}_X),(Y,mathcal{O}_Y))to f^{sharp}(Y)inoperatorname{Hom}_{bf Ring}(B,A)
$$
is bijective; moreover, the morphism $(f,f^{sharp})$ is uniquely determined by $f^{sharp}(Y)$.
What I mean? Which is the improvement?
As you (user306194) affirm: given a morphism of rings (that is $f^{sharp}(Y)$ in my notation) you can construct a unique morphism of locally ringed spaces $(f,f^{sharp})$; but, given a continuous map $f$ between the support of locally ringed spaces, you can't define uniquely $f^{sharp}$!
For example: you read the previous comment of Hoot.
Obviously, all this works in the setting of schemes.
In general, let $A$ be a ring and let $Y$ be a scheme, $X=operatorname{Spec}A$; one can prove that the function
$$
Phi:(f,f^{sharp})inoperatorname{Hom}_{bf Sch}(Y,X)to f^{sharp}(X)inoperatorname{Hom}_{bf Ring}(A,mathcal{O}_Y(Y))
$$
is bijective.
Moreover in general: let $S$ be a scheme, let $X$ be an $S$-scheme with structure morphism $p$ and let $mathcal{A}$ be a quasi-coherent $mathcal{O}_S$-algebra; one can prove that the function
$$
Phi:(f,f^{sharp})inoperatorname{Hom}_{S-bf Sch}(X,operatorname{Spec}mathcal{A})to f^{sharp}(operatorname{Spec}mathcal{A})inoperatorname{Hom}_{mathcal{O}_S-bf Alg}(mathcal{A},p_{*}mathcal{O}_X)
$$
is bijective; where $operatorname{Spec}mathcal{A}$ is the relative spectrum of $mathcal{A}$.
For a reference, I council Bosch - Commutative Algebra and Algebraic geometry, Springer, sections 6.6 and 7.1.
In conclusion, I remember you that $operatorname{Spec}(Aotimes_{mathbb{C}}B)$ is canonically isomorphic to $operatorname{Spec}Atimes_{operatorname{Spec}mathbb{C}}operatorname{Spec}B$; usually, the Cartesian product of sets is very bad in the category of schemes.
From the same book, I council you the section 7.2 with any examples and exercises 3, 4 and 5.
answered Jan 22 '16 at 11:14
Armando j18eosArmando j18eos
2,63511328
edited Jan 23 '16 at 18:27
answered Jan 22 '16 at 11:14
Armando j18eosArmando j18eos
2,63511328
answered Jan 22 '16 at 11:14
Armando j18eosArmando j18eos
2,63511328
answered Jan 22 '16 at 11:14
Armando j18eosArmando j18eos
2,63511328
2,63511328
add a comment |
add a comment |
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3
$begingroup$
Spec is neither faithful nor full, so given a continuous map on spectra, a map on rings inducing it will neither exist nor be unique in general. The "equality" you claim is false; the underlying topological space of the product of two varieties does not have the product topology. (All told, Spec is a very poorly behaved functor.)
$endgroup$
– Qiaochu Yuan
Jan 21 '16 at 0:40
2
$begingroup$
Anytime $Y$ is an affine scheme we have a bijective correspondance between $operatorname{Hom}(X,Y)$ and $operatorname{Hom}(mathcal{O}(Y), mathcal{O}(X))$, where we consider morphsms of locally ringed spaces rather than just continuous maps.
$endgroup$
– basket
Jan 21 '16 at 0:52
1
$begingroup$
An example: you can define a homomorphism $k[x] to k[y]/(y^2)$ either by sending $x$ to $y$ or $0$. The corresponding continuous maps on spectra can't tell the difference.
$endgroup$
– Hoot
Jan 21 '16 at 2:05