Intersection of Spectra
Let $R, A$ be commutative rings and $I$ an ideal in $A$. Moreover, $A subseteq R$.
I know that in this case $$R otimes_A A/I= R/IR.$$
Since I want to use this statement for a proof (with the aim of obtaining $Spec(R/IR)$), I was wondering whether the following holds
$$Spec(A/I) cap Spec(R)=Spec(A/I otimes_A R).$$
So far, I've only found out that the intersection of two spectra is again a spectrum, but not how this spectrum would look like.
Thank you for your help. I don't necessarily want a proof of this statement, some reference for me on where to look this up would be great.
algebraic-geometry spectra
asked Nov 20 at 10:42
SallyOwens
405211
add a comment |
Let $R, A$ be commutative rings and $I$ an ideal in $A$. Moreover, $A subseteq R$.
I know that in this case $$R otimes_A A/I= R/IR.$$
Since I want to use this statement for a proof (with the aim of obtaining $Spec(R/IR)$), I was wondering whether the following holds
$$Spec(A/I) cap Spec(R)=Spec(A/I otimes_A R).$$
So far, I've only found out that the intersection of two spectra is again a spectrum, but not how this spectrum would look like.
Thank you for your help. I don't necessarily want a proof of this statement, some reference for me on where to look this up would be great.
algebraic-geometry spectra
asked Nov 20 at 10:42
SallyOwens
405211
2
The rule is ${rm spec}(A otimes_C B) = {rm spec}(A) times_{{rm spec}(C)} {rm spec}(B)$. Fibre products can be similar to intersections at times, but they're different operations in general.
– Trevor Gunn
Nov 20 at 11:00
Thank you - that's what I was looking for!
– SallyOwens
Nov 20 at 11:44
add a comment |
Let $R, A$ be commutative rings and $I$ an ideal in $A$. Moreover, $A subseteq R$.
I know that in this case $$R otimes_A A/I= R/IR.$$
Since I want to use this statement for a proof (with the aim of obtaining $Spec(R/IR)$), I was wondering whether the following holds
$$Spec(A/I) cap Spec(R)=Spec(A/I otimes_A R).$$
So far, I've only found out that the intersection of two spectra is again a spectrum, but not how this spectrum would look like.
Thank you for your help. I don't necessarily want a proof of this statement, some reference for me on where to look this up would be great.
algebraic-geometry spectra
asked Nov 20 at 10:42
SallyOwens
405211
Let $R, A$ be commutative rings and $I$ an ideal in $A$. Moreover, $A subseteq R$.
I know that in this case $$R otimes_A A/I= R/IR.$$
Since I want to use this statement for a proof (with the aim of obtaining $Spec(R/IR)$), I was wondering whether the following holds
$$Spec(A/I) cap Spec(R)=Spec(A/I otimes_A R).$$
So far, I've only found out that the intersection of two spectra is again a spectrum, but not how this spectrum would look like.
Thank you for your help. I don't necessarily want a proof of this statement, some reference for me on where to look this up would be great.
algebraic-geometry spectra
algebraic-geometry spectra
asked Nov 20 at 10:42
SallyOwens
405211
asked Nov 20 at 10:42
SallyOwens
405211
asked Nov 20 at 10:42
SallyOwens
405211
asked Nov 20 at 10:42
SallyOwens
405211
asked Nov 20 at 10:42
SallyOwens
405211
405211
2
The rule is ${rm spec}(A otimes_C B) = {rm spec}(A) times_{{rm spec}(C)} {rm spec}(B)$. Fibre products can be similar to intersections at times, but they're different operations in general.
– Trevor Gunn
Nov 20 at 11:00
Thank you - that's what I was looking for!
– SallyOwens
Nov 20 at 11:44
add a comment |
2
The rule is ${rm spec}(A otimes_C B) = {rm spec}(A) times_{{rm spec}(C)} {rm spec}(B)$. Fibre products can be similar to intersections at times, but they're different operations in general.
– Trevor Gunn
Nov 20 at 11:00
Thank you - that's what I was looking for!
– SallyOwens
Nov 20 at 11:44
2
2
The rule is ${rm spec}(A otimes_C B) = {rm spec}(A) times_{{rm spec}(C)} {rm spec}(B)$. Fibre products can be similar to intersections at times, but they're different operations in general.
– Trevor Gunn
Nov 20 at 11:00
The rule is ${rm spec}(A otimes_C B) = {rm spec}(A) times_{{rm spec}(C)} {rm spec}(B)$. Fibre products can be similar to intersections at times, but they're different operations in general.
– Trevor Gunn
Nov 20 at 11:00
Thank you - that's what I was looking for!
– SallyOwens
Nov 20 at 11:44
Thank you - that's what I was looking for!
– SallyOwens
Nov 20 at 11:44
add a comment |
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The rule is ${rm spec}(A otimes_C B) = {rm spec}(A) times_{{rm spec}(C)} {rm spec}(B)$. Fibre products can be similar to intersections at times, but they're different operations in general.
– Trevor Gunn
Nov 20 at 11:00
Thank you - that's what I was looking for!
– SallyOwens
Nov 20 at 11:44