Mixing
Projective nullstellensatz correspondence: how does this maximal ideal come into play?
$begingroup$
I know the affine Nullstellensatz:
If $I$ is a proper ideal of $k[X_1, dots, X_n]$, then $V_a(I) neq emptyset$ and for all ideals $I_a(V_a(I)) = rad(I) = sqrt{I}$.
and this gives a correspondence
$${algebraic sets in mathbb{A}^n(k)} leftrightarrow {radical ideals in k[X_1, dots, X_n]}$$
I'm reading Fulton's algebraic book, and they prove the projective Nullstellensatz:
For a homogeneous ideal in $k[X_1, dots, X_{n+1}]$
$V_p(I) = emptyset iff exists N geq 1: {F in k[X_1, dots, X_{n+1}] mid F monomial, deg(F) geq N} subseteq I$
and if $V_p(I) neq emptyset$, then $I_p(V_p(I)) = rad(I)$.
After the proof, without any explanation, there is mentioned that there is a correspondence:
$${algebraic sets in mathbb{P}^n(k)} leftrightarrow {radical, homogeneous ideals in k[X_1, dots, X_n]} setminus {M}$$
where $M = (X_1, dots, X_{n+1})$ (this ideal is maximal).
How does one deduce this last correspondence?
I can see that if $V_p(I) = emptyset$ and $I$ is a proper ideal, then $rad I = (X_1, dots, X_n)$. Maybe this is relevant.
abstract-algebra algebraic-geometry
asked Jan 8 at 13:17
Math_QEDMath_QED
7,54831450
$endgroup$
add a comment |
$begingroup$
I know the affine Nullstellensatz:
If $I$ is a proper ideal of $k[X_1, dots, X_n]$, then $V_a(I) neq emptyset$ and for all ideals $I_a(V_a(I)) = rad(I) = sqrt{I}$.
and this gives a correspondence
$${algebraic sets in mathbb{A}^n(k)} leftrightarrow {radical ideals in k[X_1, dots, X_n]}$$
I'm reading Fulton's algebraic book, and they prove the projective Nullstellensatz:
For a homogeneous ideal in $k[X_1, dots, X_{n+1}]$
$V_p(I) = emptyset iff exists N geq 1: {F in k[X_1, dots, X_{n+1}] mid F monomial, deg(F) geq N} subseteq I$
and if $V_p(I) neq emptyset$, then $I_p(V_p(I)) = rad(I)$.
After the proof, without any explanation, there is mentioned that there is a correspondence:
$${algebraic sets in mathbb{P}^n(k)} leftrightarrow {radical, homogeneous ideals in k[X_1, dots, X_n]} setminus {M}$$
where $M = (X_1, dots, X_{n+1})$ (this ideal is maximal).
How does one deduce this last correspondence?
I can see that if $V_p(I) = emptyset$ and $I$ is a proper ideal, then $rad I = (X_1, dots, X_n)$. Maybe this is relevant.
abstract-algebra algebraic-geometry
asked Jan 8 at 13:17
Math_QEDMath_QED
7,54831450
$endgroup$
$begingroup$
Lift algebraic sets to cones of $A^{n+1}(k)$. Those homogeneous ideals corresponds to cones of $A^{n+1}(k)$ as each point of $P^n(k)$ corresponds to a ray in $A^{n+1}(k)$.
$endgroup$
– user45765
Jan 8 at 14:22
add a comment |
$begingroup$
I know the affine Nullstellensatz:
If $I$ is a proper ideal of $k[X_1, dots, X_n]$, then $V_a(I) neq emptyset$ and for all ideals $I_a(V_a(I)) = rad(I) = sqrt{I}$.
and this gives a correspondence
$${algebraic sets in mathbb{A}^n(k)} leftrightarrow {radical ideals in k[X_1, dots, X_n]}$$
I'm reading Fulton's algebraic book, and they prove the projective Nullstellensatz:
For a homogeneous ideal in $k[X_1, dots, X_{n+1}]$
$V_p(I) = emptyset iff exists N geq 1: {F in k[X_1, dots, X_{n+1}] mid F monomial, deg(F) geq N} subseteq I$
and if $V_p(I) neq emptyset$, then $I_p(V_p(I)) = rad(I)$.
After the proof, without any explanation, there is mentioned that there is a correspondence:
$${algebraic sets in mathbb{P}^n(k)} leftrightarrow {radical, homogeneous ideals in k[X_1, dots, X_n]} setminus {M}$$
where $M = (X_1, dots, X_{n+1})$ (this ideal is maximal).
How does one deduce this last correspondence?
I can see that if $V_p(I) = emptyset$ and $I$ is a proper ideal, then $rad I = (X_1, dots, X_n)$. Maybe this is relevant.
abstract-algebra algebraic-geometry
asked Jan 8 at 13:17
Math_QEDMath_QED
7,54831450
$endgroup$
I know the affine Nullstellensatz:
If $I$ is a proper ideal of $k[X_1, dots, X_n]$, then $V_a(I) neq emptyset$ and for all ideals $I_a(V_a(I)) = rad(I) = sqrt{I}$.
and this gives a correspondence
$${algebraic sets in mathbb{A}^n(k)} leftrightarrow {radical ideals in k[X_1, dots, X_n]}$$
I'm reading Fulton's algebraic book, and they prove the projective Nullstellensatz:
For a homogeneous ideal in $k[X_1, dots, X_{n+1}]$
$V_p(I) = emptyset iff exists N geq 1: {F in k[X_1, dots, X_{n+1}] mid F monomial, deg(F) geq N} subseteq I$
and if $V_p(I) neq emptyset$, then $I_p(V_p(I)) = rad(I)$.
After the proof, without any explanation, there is mentioned that there is a correspondence:
$${algebraic sets in mathbb{P}^n(k)} leftrightarrow {radical, homogeneous ideals in k[X_1, dots, X_n]} setminus {M}$$
where $M = (X_1, dots, X_{n+1})$ (this ideal is maximal).
How does one deduce this last correspondence?
I can see that if $V_p(I) = emptyset$ and $I$ is a proper ideal, then $rad I = (X_1, dots, X_n)$. Maybe this is relevant.
abstract-algebra algebraic-geometry
abstract-algebra algebraic-geometry
asked Jan 8 at 13:17
Math_QEDMath_QED
7,54831450
asked Jan 8 at 13:17
Math_QEDMath_QED
7,54831450
asked Jan 8 at 13:17
Math_QEDMath_QED
7,54831450
asked Jan 8 at 13:17
Math_QEDMath_QED
7,54831450
asked Jan 8 at 13:17
Math_QEDMath_QED
7,54831450
7,54831450
$begingroup$
Lift algebraic sets to cones of $A^{n+1}(k)$. Those homogeneous ideals corresponds to cones of $A^{n+1}(k)$ as each point of $P^n(k)$ corresponds to a ray in $A^{n+1}(k)$.
$endgroup$
– user45765
Jan 8 at 14:22
add a comment |
$begingroup$
Lift algebraic sets to cones of $A^{n+1}(k)$. Those homogeneous ideals corresponds to cones of $A^{n+1}(k)$ as each point of $P^n(k)$ corresponds to a ray in $A^{n+1}(k)$.
$endgroup$
– user45765
Jan 8 at 14:22
$begingroup$
Lift algebraic sets to cones of $A^{n+1}(k)$. Those homogeneous ideals corresponds to cones of $A^{n+1}(k)$ as each point of $P^n(k)$ corresponds to a ray in $A^{n+1}(k)$.
$endgroup$
– user45765
Jan 8 at 14:22
$begingroup$
Lift algebraic sets to cones of $A^{n+1}(k)$. Those homogeneous ideals corresponds to cones of $A^{n+1}(k)$ as each point of $P^n(k)$ corresponds to a ray in $A^{n+1}(k)$.
$endgroup$
– user45765
Jan 8 at 14:22
add a comment |
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$begingroup$
Lift algebraic sets to cones of $A^{n+1}(k)$. Those homogeneous ideals corresponds to cones of $A^{n+1}(k)$ as each point of $P^n(k)$ corresponds to a ray in $A^{n+1}(k)$.
$endgroup$
– user45765
Jan 8 at 14:22