Intersection of Spectra











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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







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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















up vote
2
down vote

favorite
1












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







share|cite|improve this question


















  • 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













up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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















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