Homogeneous prime ideals and grade zero elements
$begingroup$
Let $R = bigoplus_{d in mathbb{N}_0}R^{(d)} $ be a graded ring and for homogeneous ideals $I$, let $V_{proj;R}(I) = {p supseteq I mid p text{ is a homogeneous prime ideal and } p nsupseteq R_+}$ where $R_+ = bigoplus_{d in mathbb{N}} R^{(d)}$ is the irrelevant ideal.
I want to show that $V_{proj;R}(I) = V_{proj;R}(I cap R_+)$.
I have already shown that $I = bigoplus_{d in mathbb{N}_0} I cap R^{(d)}$ holds, but I currently fail to see why the grade zero elements are uniquely determined by the elements of higher grade and am therefore looking for hints on this question.
algebraic-geometry commutative-algebra
asked Dec 7 '18 at 15:05
PLOPLO
674
$endgroup$
add a comment |
$begingroup$
Let $R = bigoplus_{d in mathbb{N}_0}R^{(d)} $ be a graded ring and for homogeneous ideals $I$, let $V_{proj;R}(I) = {p supseteq I mid p text{ is a homogeneous prime ideal and } p nsupseteq R_+}$ where $R_+ = bigoplus_{d in mathbb{N}} R^{(d)}$ is the irrelevant ideal.
I want to show that $V_{proj;R}(I) = V_{proj;R}(I cap R_+)$.
I have already shown that $I = bigoplus_{d in mathbb{N}_0} I cap R^{(d)}$ holds, but I currently fail to see why the grade zero elements are uniquely determined by the elements of higher grade and am therefore looking for hints on this question.
algebraic-geometry commutative-algebra
asked Dec 7 '18 at 15:05
PLOPLO
674
$endgroup$
1
$begingroup$
If P is a prime ideal and $Icap Jsubseteq P$ then what can conclude? Furthermore, what if $J$ is not contained in P?
$endgroup$
– user26857
Dec 7 '18 at 16:32
add a comment |
$begingroup$
Let $R = bigoplus_{d in mathbb{N}_0}R^{(d)} $ be a graded ring and for homogeneous ideals $I$, let $V_{proj;R}(I) = {p supseteq I mid p text{ is a homogeneous prime ideal and } p nsupseteq R_+}$ where $R_+ = bigoplus_{d in mathbb{N}} R^{(d)}$ is the irrelevant ideal.
I want to show that $V_{proj;R}(I) = V_{proj;R}(I cap R_+)$.
I have already shown that $I = bigoplus_{d in mathbb{N}_0} I cap R^{(d)}$ holds, but I currently fail to see why the grade zero elements are uniquely determined by the elements of higher grade and am therefore looking for hints on this question.
algebraic-geometry commutative-algebra
asked Dec 7 '18 at 15:05
PLOPLO
674
$endgroup$
Let $R = bigoplus_{d in mathbb{N}_0}R^{(d)} $ be a graded ring and for homogeneous ideals $I$, let $V_{proj;R}(I) = {p supseteq I mid p text{ is a homogeneous prime ideal and } p nsupseteq R_+}$ where $R_+ = bigoplus_{d in mathbb{N}} R^{(d)}$ is the irrelevant ideal.
I want to show that $V_{proj;R}(I) = V_{proj;R}(I cap R_+)$.
I have already shown that $I = bigoplus_{d in mathbb{N}_0} I cap R^{(d)}$ holds, but I currently fail to see why the grade zero elements are uniquely determined by the elements of higher grade and am therefore looking for hints on this question.
algebraic-geometry commutative-algebra
algebraic-geometry commutative-algebra
asked Dec 7 '18 at 15:05
PLOPLO
674
asked Dec 7 '18 at 15:05
PLOPLO
674
asked Dec 7 '18 at 15:05
PLOPLO
674
asked Dec 7 '18 at 15:05
PLOPLO
674
asked Dec 7 '18 at 15:05
PLOPLO
674
674
1
$begingroup$
If P is a prime ideal and $Icap Jsubseteq P$ then what can conclude? Furthermore, what if $J$ is not contained in P?
$endgroup$
– user26857
Dec 7 '18 at 16:32
add a comment |
1
$begingroup$
If P is a prime ideal and $Icap Jsubseteq P$ then what can conclude? Furthermore, what if $J$ is not contained in P?
$endgroup$
– user26857
Dec 7 '18 at 16:32
1
1
$begingroup$
If P is a prime ideal and $Icap Jsubseteq P$ then what can conclude? Furthermore, what if $J$ is not contained in P?
$endgroup$
– user26857
Dec 7 '18 at 16:32
$begingroup$
If P is a prime ideal and $Icap Jsubseteq P$ then what can conclude? Furthermore, what if $J$ is not contained in P?
$endgroup$
– user26857
Dec 7 '18 at 16:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As user26857 hinted at, if $I cap R_+ subset mathfrak{p}$, then also $I cdot R_+ subset mathfrak{p}$. Hence either $I subset mathfrak{p}$ or $R_+ subset mathfrak{p}$, because $mathfrak{p}$ is prime. The latter is excluded for all primes in $V_{text{proj }R}(I cap R_+)$, so we see that $mathfrak{p} in V(I) Leftrightarrow mathfrak{p} in V(I cap R_+)$.
answered Dec 7 '18 at 20:11
red_trumpetred_trumpet
853219
$endgroup$
$begingroup$
Thanks, I found this out too after reading his comment. I really missed this one detail.
$endgroup$
– PLO
Dec 7 '18 at 20:16
add a comment |
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1 Answer
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$begingroup$
As user26857 hinted at, if $I cap R_+ subset mathfrak{p}$, then also $I cdot R_+ subset mathfrak{p}$. Hence either $I subset mathfrak{p}$ or $R_+ subset mathfrak{p}$, because $mathfrak{p}$ is prime. The latter is excluded for all primes in $V_{text{proj }R}(I cap R_+)$, so we see that $mathfrak{p} in V(I) Leftrightarrow mathfrak{p} in V(I cap R_+)$.
answered Dec 7 '18 at 20:11
red_trumpetred_trumpet
853219
$endgroup$
$begingroup$
Thanks, I found this out too after reading his comment. I really missed this one detail.
$endgroup$
– PLO
Dec 7 '18 at 20:16
add a comment |
$begingroup$
As user26857 hinted at, if $I cap R_+ subset mathfrak{p}$, then also $I cdot R_+ subset mathfrak{p}$. Hence either $I subset mathfrak{p}$ or $R_+ subset mathfrak{p}$, because $mathfrak{p}$ is prime. The latter is excluded for all primes in $V_{text{proj }R}(I cap R_+)$, so we see that $mathfrak{p} in V(I) Leftrightarrow mathfrak{p} in V(I cap R_+)$.
answered Dec 7 '18 at 20:11
red_trumpetred_trumpet
853219
$endgroup$
$begingroup$
Thanks, I found this out too after reading his comment. I really missed this one detail.
$endgroup$
– PLO
Dec 7 '18 at 20:16
add a comment |
$begingroup$
As user26857 hinted at, if $I cap R_+ subset mathfrak{p}$, then also $I cdot R_+ subset mathfrak{p}$. Hence either $I subset mathfrak{p}$ or $R_+ subset mathfrak{p}$, because $mathfrak{p}$ is prime. The latter is excluded for all primes in $V_{text{proj }R}(I cap R_+)$, so we see that $mathfrak{p} in V(I) Leftrightarrow mathfrak{p} in V(I cap R_+)$.
answered Dec 7 '18 at 20:11
red_trumpetred_trumpet
853219
$endgroup$
As user26857 hinted at, if $I cap R_+ subset mathfrak{p}$, then also $I cdot R_+ subset mathfrak{p}$. Hence either $I subset mathfrak{p}$ or $R_+ subset mathfrak{p}$, because $mathfrak{p}$ is prime. The latter is excluded for all primes in $V_{text{proj }R}(I cap R_+)$, so we see that $mathfrak{p} in V(I) Leftrightarrow mathfrak{p} in V(I cap R_+)$.
answered Dec 7 '18 at 20:11
red_trumpetred_trumpet
853219
answered Dec 7 '18 at 20:11
red_trumpetred_trumpet
853219
answered Dec 7 '18 at 20:11
red_trumpetred_trumpet
853219
answered Dec 7 '18 at 20:11
red_trumpetred_trumpet
853219
853219
$begingroup$
Thanks, I found this out too after reading his comment. I really missed this one detail.
$endgroup$
– PLO
Dec 7 '18 at 20:16
add a comment |
$begingroup$
Thanks, I found this out too after reading his comment. I really missed this one detail.
$endgroup$
– PLO
Dec 7 '18 at 20:16
$begingroup$
Thanks, I found this out too after reading his comment. I really missed this one detail.
$endgroup$
– PLO
Dec 7 '18 at 20:16
$begingroup$
Thanks, I found this out too after reading his comment. I really missed this one detail.
$endgroup$
– PLO
Dec 7 '18 at 20:16
add a comment |
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$begingroup$
If P is a prime ideal and $Icap Jsubseteq P$ then what can conclude? Furthermore, what if $J$ is not contained in P?
$endgroup$
– user26857
Dec 7 '18 at 16:32