Ideal generated by homogeneous polynomialsRing theory problem, about algebraic setnon-trivial common zero of...
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Ideal generated by homogeneous polynomials
Ring theory problem, about algebraic setnon-trivial common zero of polynomialsSequence of irreducible polynomials in $K[X_{1},…,X_{n}]$ generates a prime ideal?Is an ideal generated by multilinear polynomials of different degrees always radical?Confusing part of Grobner Basis section of Dummit and FooteA question about the proof of Hilbert's Basis TheoremThe ideal $(x_1,x_2,x_3,…)$ (with infinitely many variables) in the ring $K[x_1,x_2,x_3,…]$ is not finitely generated.Finitely Generated SetsExpress the identity element as a linear combination of $N^{th}$ powers of generators in a ringgenerating radical of an ideal with small degree polynomials
$begingroup$
Let $R=k[x_1,x_2,...,x_n]$, and $Ssubset R$ be the ring of invariant polynomials under the action of a finite subset $Gsubset operatorname{GL}(n,k)$. Let $M$ be the ideal of $S$ generated by all homogeneous elements of $S$ of positive degree.
The question is: show we can find homogeneous $f_1,f_2,...,f_kin S$ such that these $f_i$ generate the ideal $MRsubset R$.
My book leaves this part of the proof out as something trivial... However I'm lost why this should be obvious (or even why this could be done for the matter). My guess is that this shouldn't be a constructive proof. Any ideas to prove this claim?
abstract-algebra ring-theory
edited Mar 12 at 23:44
Andrews
1,2691421
asked Mar 12 at 22:58
davidhdavidh
3178
$endgroup$
add a comment |
$begingroup$
Let $R=k[x_1,x_2,...,x_n]$, and $Ssubset R$ be the ring of invariant polynomials under the action of a finite subset $Gsubset operatorname{GL}(n,k)$. Let $M$ be the ideal of $S$ generated by all homogeneous elements of $S$ of positive degree.
The question is: show we can find homogeneous $f_1,f_2,...,f_kin S$ such that these $f_i$ generate the ideal $MRsubset R$.
My book leaves this part of the proof out as something trivial... However I'm lost why this should be obvious (or even why this could be done for the matter). My guess is that this shouldn't be a constructive proof. Any ideas to prove this claim?
abstract-algebra ring-theory
edited Mar 12 at 23:44
Andrews
1,2691421
asked Mar 12 at 22:58
davidhdavidh
3178
$endgroup$
1
$begingroup$
It would suffice to show that $M$ is finitely generated.
$endgroup$
– Servaes
Mar 12 at 23:58
$begingroup$
@Servaes that makes sense. Thanks
$endgroup$
– davidh
Mar 13 at 0:17
$begingroup$
Perhaps, applying the Hilbert basis theorem to your ideal $MR$ first. Then averaging the generators with respect to $langle G rangle$, at least when the order of $langle G rangle$ is finite and invertible in $k$. The homogeneous parts of the resulting polynomials would be even a finite generating set of $M$ in $S$.
$endgroup$
– Orat
Mar 13 at 3:04
$begingroup$
Is "show we can find" asking for a constructive proof? I am pretty sure there is one, at least when $G$ is a group and $k$ has characteristic $0$ (see Noether's bound).
$endgroup$
– darij grinberg
Mar 13 at 16:01
add a comment |
$begingroup$
Let $R=k[x_1,x_2,...,x_n]$, and $Ssubset R$ be the ring of invariant polynomials under the action of a finite subset $Gsubset operatorname{GL}(n,k)$. Let $M$ be the ideal of $S$ generated by all homogeneous elements of $S$ of positive degree.
The question is: show we can find homogeneous $f_1,f_2,...,f_kin S$ such that these $f_i$ generate the ideal $MRsubset R$.
My book leaves this part of the proof out as something trivial... However I'm lost why this should be obvious (or even why this could be done for the matter). My guess is that this shouldn't be a constructive proof. Any ideas to prove this claim?
abstract-algebra ring-theory
edited Mar 12 at 23:44
Andrews
1,2691421
asked Mar 12 at 22:58
davidhdavidh
3178
$endgroup$
Let $R=k[x_1,x_2,...,x_n]$, and $Ssubset R$ be the ring of invariant polynomials under the action of a finite subset $Gsubset operatorname{GL}(n,k)$. Let $M$ be the ideal of $S$ generated by all homogeneous elements of $S$ of positive degree.
The question is: show we can find homogeneous $f_1,f_2,...,f_kin S$ such that these $f_i$ generate the ideal $MRsubset R$.
My book leaves this part of the proof out as something trivial... However I'm lost why this should be obvious (or even why this could be done for the matter). My guess is that this shouldn't be a constructive proof. Any ideas to prove this claim?
abstract-algebra ring-theory
abstract-algebra ring-theory
edited Mar 12 at 23:44
Andrews
1,2691421
asked Mar 12 at 22:58
davidhdavidh
3178
edited Mar 12 at 23:44
Andrews
1,2691421
asked Mar 12 at 22:58
davidhdavidh
3178
edited Mar 12 at 23:44
Andrews
1,2691421
edited Mar 12 at 23:44
Andrews
1,2691421
edited Mar 12 at 23:44
Andrews
1,2691421
1,2691421
asked Mar 12 at 22:58
davidhdavidh
3178
asked Mar 12 at 22:58
davidhdavidh
3178
asked Mar 12 at 22:58
davidhdavidh
3178
3178
1
$begingroup$
It would suffice to show that $M$ is finitely generated.
$endgroup$
– Servaes
Mar 12 at 23:58
$begingroup$
@Servaes that makes sense. Thanks
$endgroup$
– davidh
Mar 13 at 0:17
$begingroup$
Perhaps, applying the Hilbert basis theorem to your ideal $MR$ first. Then averaging the generators with respect to $langle G rangle$, at least when the order of $langle G rangle$ is finite and invertible in $k$. The homogeneous parts of the resulting polynomials would be even a finite generating set of $M$ in $S$.
$endgroup$
– Orat
Mar 13 at 3:04
$begingroup$
Is "show we can find" asking for a constructive proof? I am pretty sure there is one, at least when $G$ is a group and $k$ has characteristic $0$ (see Noether's bound).
$endgroup$
– darij grinberg
Mar 13 at 16:01
add a comment |
1
$begingroup$
It would suffice to show that $M$ is finitely generated.
$endgroup$
– Servaes
Mar 12 at 23:58
$begingroup$
@Servaes that makes sense. Thanks
$endgroup$
– davidh
Mar 13 at 0:17
$begingroup$
Perhaps, applying the Hilbert basis theorem to your ideal $MR$ first. Then averaging the generators with respect to $langle G rangle$, at least when the order of $langle G rangle$ is finite and invertible in $k$. The homogeneous parts of the resulting polynomials would be even a finite generating set of $M$ in $S$.
$endgroup$
– Orat
Mar 13 at 3:04
$begingroup$
Is "show we can find" asking for a constructive proof? I am pretty sure there is one, at least when $G$ is a group and $k$ has characteristic $0$ (see Noether's bound).
$endgroup$
– darij grinberg
Mar 13 at 16:01
1
1
$begingroup$
It would suffice to show that $M$ is finitely generated.
$endgroup$
– Servaes
Mar 12 at 23:58
$begingroup$
It would suffice to show that $M$ is finitely generated.
$endgroup$
– Servaes
Mar 12 at 23:58
$begingroup$
@Servaes that makes sense. Thanks
$endgroup$
– davidh
Mar 13 at 0:17
$begingroup$
@Servaes that makes sense. Thanks
$endgroup$
– davidh
Mar 13 at 0:17
$begingroup$
Perhaps, applying the Hilbert basis theorem to your ideal $MR$ first. Then averaging the generators with respect to $langle G rangle$, at least when the order of $langle G rangle$ is finite and invertible in $k$. The homogeneous parts of the resulting polynomials would be even a finite generating set of $M$ in $S$.
$endgroup$
– Orat
Mar 13 at 3:04
$begingroup$
Perhaps, applying the Hilbert basis theorem to your ideal $MR$ first. Then averaging the generators with respect to $langle G rangle$, at least when the order of $langle G rangle$ is finite and invertible in $k$. The homogeneous parts of the resulting polynomials would be even a finite generating set of $M$ in $S$.
$endgroup$
– Orat
Mar 13 at 3:04
$begingroup$
Is "show we can find" asking for a constructive proof? I am pretty sure there is one, at least when $G$ is a group and $k$ has characteristic $0$ (see Noether's bound).
$endgroup$
– darij grinberg
Mar 13 at 16:01
$begingroup$
Is "show we can find" asking for a constructive proof? I am pretty sure there is one, at least when $G$ is a group and $k$ has characteristic $0$ (see Noether's bound).
$endgroup$
– darij grinberg
Mar 13 at 16:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $S$ be a subring of a noetherian ring $R$ and let $B$ be a set of generators of the ideal $I$ of $S$. Then the ideal $IR$ is generated by finitely many elements from $B$.
Because: by assumption $IR$ is finitely generated. Let $c_1,ldots ,c_rin IR$ be a set of generators. Then
$c_k=sumlimits_{i=1}^{m_k}r_ib_{ki},; r_iin R, b_{ki}in B.$
Then $IR$ is generated by the finite set $B_0$ consisting of the elements $b_{ki}$, since obviously $Jsubseteq IR$ for the ideal $J$ of $R$ generated by $B_0$. On the other hand $c_kin J$ for all $k$, hence $IRsubseteq J$.
answered Mar 13 at 15:43
Hagen KnafHagen Knaf
6,9071318
$endgroup$
add a comment |
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$begingroup$
Let $S$ be a subring of a noetherian ring $R$ and let $B$ be a set of generators of the ideal $I$ of $S$. Then the ideal $IR$ is generated by finitely many elements from $B$.
Because: by assumption $IR$ is finitely generated. Let $c_1,ldots ,c_rin IR$ be a set of generators. Then
$c_k=sumlimits_{i=1}^{m_k}r_ib_{ki},; r_iin R, b_{ki}in B.$
Then $IR$ is generated by the finite set $B_0$ consisting of the elements $b_{ki}$, since obviously $Jsubseteq IR$ for the ideal $J$ of $R$ generated by $B_0$. On the other hand $c_kin J$ for all $k$, hence $IRsubseteq J$.
answered Mar 13 at 15:43
Hagen KnafHagen Knaf
6,9071318
$endgroup$
add a comment |
$begingroup$
Let $S$ be a subring of a noetherian ring $R$ and let $B$ be a set of generators of the ideal $I$ of $S$. Then the ideal $IR$ is generated by finitely many elements from $B$.
Because: by assumption $IR$ is finitely generated. Let $c_1,ldots ,c_rin IR$ be a set of generators. Then
$c_k=sumlimits_{i=1}^{m_k}r_ib_{ki},; r_iin R, b_{ki}in B.$
Then $IR$ is generated by the finite set $B_0$ consisting of the elements $b_{ki}$, since obviously $Jsubseteq IR$ for the ideal $J$ of $R$ generated by $B_0$. On the other hand $c_kin J$ for all $k$, hence $IRsubseteq J$.
answered Mar 13 at 15:43
Hagen KnafHagen Knaf
6,9071318
$endgroup$
add a comment |
$begingroup$
Let $S$ be a subring of a noetherian ring $R$ and let $B$ be a set of generators of the ideal $I$ of $S$. Then the ideal $IR$ is generated by finitely many elements from $B$.
Because: by assumption $IR$ is finitely generated. Let $c_1,ldots ,c_rin IR$ be a set of generators. Then
$c_k=sumlimits_{i=1}^{m_k}r_ib_{ki},; r_iin R, b_{ki}in B.$
Then $IR$ is generated by the finite set $B_0$ consisting of the elements $b_{ki}$, since obviously $Jsubseteq IR$ for the ideal $J$ of $R$ generated by $B_0$. On the other hand $c_kin J$ for all $k$, hence $IRsubseteq J$.
answered Mar 13 at 15:43
Hagen KnafHagen Knaf
6,9071318
$endgroup$
Let $S$ be a subring of a noetherian ring $R$ and let $B$ be a set of generators of the ideal $I$ of $S$. Then the ideal $IR$ is generated by finitely many elements from $B$.
Because: by assumption $IR$ is finitely generated. Let $c_1,ldots ,c_rin IR$ be a set of generators. Then
$c_k=sumlimits_{i=1}^{m_k}r_ib_{ki},; r_iin R, b_{ki}in B.$
Then $IR$ is generated by the finite set $B_0$ consisting of the elements $b_{ki}$, since obviously $Jsubseteq IR$ for the ideal $J$ of $R$ generated by $B_0$. On the other hand $c_kin J$ for all $k$, hence $IRsubseteq J$.
answered Mar 13 at 15:43
Hagen KnafHagen Knaf
6,9071318
answered Mar 13 at 15:43
Hagen KnafHagen Knaf
6,9071318
answered Mar 13 at 15:43
Hagen KnafHagen Knaf
6,9071318
answered Mar 13 at 15:43
Hagen KnafHagen Knaf
6,9071318
6,9071318
add a comment |
add a comment |
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1
$begingroup$
It would suffice to show that $M$ is finitely generated.
$endgroup$
– Servaes
Mar 12 at 23:58
$begingroup$
@Servaes that makes sense. Thanks
$endgroup$
– davidh
Mar 13 at 0:17
$begingroup$
Perhaps, applying the Hilbert basis theorem to your ideal $MR$ first. Then averaging the generators with respect to $langle G rangle$, at least when the order of $langle G rangle$ is finite and invertible in $k$. The homogeneous parts of the resulting polynomials would be even a finite generating set of $M$ in $S$.
$endgroup$
– Orat
Mar 13 at 3:04
$begingroup$
Is "show we can find" asking for a constructive proof? I am pretty sure there is one, at least when $G$ is a group and $k$ has characteristic $0$ (see Noether's bound).
$endgroup$
– darij grinberg
Mar 13 at 16:01