Application of Hilbert's basis theorem in representation theory
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In Smalo: Degenerations of Representations of Associative Algebras, Milan J. Math., 2008 there is an application of Hilbert's basis theorem that I don't understand:
Two orders are defined on the set of $d$-dimensional modules over an algebra $Lambda$ finite dimensional over a field $k$. One by $Mleq_{operatorname{Hom}} N$ iff $dim operatorname{Hom}(X,M)leq dim operatorname{Hom}(X,N)$ for all $X$ and one by $Mleq_n N$ iff $dim operatorname{Hom}(Lambda^n/Lambda^nA,M)leq dim operatorname{Hom}(Lambda^n/Lambda^nA,N)$ for all $ntimes n$-matrices $A$. It is now claimed that from Hilbert's basis theorem for $n$ large enough (depending on $d$, but not on $M$ or $N$) one gets that $leq_n$ is equivalent to $leq_{operatorname{Hom}}$. Can somebody provide a more detailed argument?
ADDED by David E Speyer The problem here is that the set ${ (M,N) : M leq_n N }$ is neither Zariski closed nor Zariski open. (Take $Lambda = k[epsilon]/epsilon^2$ and $d=2$. So $mathrm{rep}_2 Lambda$ (in the notation of the paper) is the space of $2 times 2$ matrices with square zero. Then two matrices $rho$ and $sigma$ in $mathrm{rep}_2 Lambda$ obey $rho leq_1 sigma$ if and and only if either $sigma =0$ or $rho neq 0$.) If these spaces were Zariski closed, this would be an easy consequence of Hilbert's basis theorem but, as it is, I am stumped.
algebraic-geometry representation-theory
edited Oct 14 '14 at 18:18
David E Speyer
46.2k4127211
asked Mar 26 '12 at 14:17
Julian KuelshammerJulian Kuelshammer
7,59732767
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add a comment |
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In Smalo: Degenerations of Representations of Associative Algebras, Milan J. Math., 2008 there is an application of Hilbert's basis theorem that I don't understand:
Two orders are defined on the set of $d$-dimensional modules over an algebra $Lambda$ finite dimensional over a field $k$. One by $Mleq_{operatorname{Hom}} N$ iff $dim operatorname{Hom}(X,M)leq dim operatorname{Hom}(X,N)$ for all $X$ and one by $Mleq_n N$ iff $dim operatorname{Hom}(Lambda^n/Lambda^nA,M)leq dim operatorname{Hom}(Lambda^n/Lambda^nA,N)$ for all $ntimes n$-matrices $A$. It is now claimed that from Hilbert's basis theorem for $n$ large enough (depending on $d$, but not on $M$ or $N$) one gets that $leq_n$ is equivalent to $leq_{operatorname{Hom}}$. Can somebody provide a more detailed argument?
ADDED by David E Speyer The problem here is that the set ${ (M,N) : M leq_n N }$ is neither Zariski closed nor Zariski open. (Take $Lambda = k[epsilon]/epsilon^2$ and $d=2$. So $mathrm{rep}_2 Lambda$ (in the notation of the paper) is the space of $2 times 2$ matrices with square zero. Then two matrices $rho$ and $sigma$ in $mathrm{rep}_2 Lambda$ obey $rho leq_1 sigma$ if and and only if either $sigma =0$ or $rho neq 0$.) If these spaces were Zariski closed, this would be an easy consequence of Hilbert's basis theorem but, as it is, I am stumped.
algebraic-geometry representation-theory
edited Oct 14 '14 at 18:18
David E Speyer
46.2k4127211
asked Mar 26 '12 at 14:17
Julian KuelshammerJulian Kuelshammer
7,59732767
$endgroup$
$begingroup$
Edits remove two ambiguities in the original question (what finite dimensional means, and what $n$ is allowed to depend on), based on my reading of the article. Then added a paragraph to point out what doesn't work here.
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– David E Speyer
Oct 14 '14 at 18:20
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@DavidSpeyer. It is a general rule on SE sites that posts should not contain visible edit history. They should read smoothly.
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– TRiG
Jun 20 '16 at 11:01
add a comment |
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In Smalo: Degenerations of Representations of Associative Algebras, Milan J. Math., 2008 there is an application of Hilbert's basis theorem that I don't understand:
Two orders are defined on the set of $d$-dimensional modules over an algebra $Lambda$ finite dimensional over a field $k$. One by $Mleq_{operatorname{Hom}} N$ iff $dim operatorname{Hom}(X,M)leq dim operatorname{Hom}(X,N)$ for all $X$ and one by $Mleq_n N$ iff $dim operatorname{Hom}(Lambda^n/Lambda^nA,M)leq dim operatorname{Hom}(Lambda^n/Lambda^nA,N)$ for all $ntimes n$-matrices $A$. It is now claimed that from Hilbert's basis theorem for $n$ large enough (depending on $d$, but not on $M$ or $N$) one gets that $leq_n$ is equivalent to $leq_{operatorname{Hom}}$. Can somebody provide a more detailed argument?
ADDED by David E Speyer The problem here is that the set ${ (M,N) : M leq_n N }$ is neither Zariski closed nor Zariski open. (Take $Lambda = k[epsilon]/epsilon^2$ and $d=2$. So $mathrm{rep}_2 Lambda$ (in the notation of the paper) is the space of $2 times 2$ matrices with square zero. Then two matrices $rho$ and $sigma$ in $mathrm{rep}_2 Lambda$ obey $rho leq_1 sigma$ if and and only if either $sigma =0$ or $rho neq 0$.) If these spaces were Zariski closed, this would be an easy consequence of Hilbert's basis theorem but, as it is, I am stumped.
algebraic-geometry representation-theory
edited Oct 14 '14 at 18:18
David E Speyer
46.2k4127211
asked Mar 26 '12 at 14:17
Julian KuelshammerJulian Kuelshammer
7,59732767
$endgroup$
In Smalo: Degenerations of Representations of Associative Algebras, Milan J. Math., 2008 there is an application of Hilbert's basis theorem that I don't understand:
Two orders are defined on the set of $d$-dimensional modules over an algebra $Lambda$ finite dimensional over a field $k$. One by $Mleq_{operatorname{Hom}} N$ iff $dim operatorname{Hom}(X,M)leq dim operatorname{Hom}(X,N)$ for all $X$ and one by $Mleq_n N$ iff $dim operatorname{Hom}(Lambda^n/Lambda^nA,M)leq dim operatorname{Hom}(Lambda^n/Lambda^nA,N)$ for all $ntimes n$-matrices $A$. It is now claimed that from Hilbert's basis theorem for $n$ large enough (depending on $d$, but not on $M$ or $N$) one gets that $leq_n$ is equivalent to $leq_{operatorname{Hom}}$. Can somebody provide a more detailed argument?
ADDED by David E Speyer The problem here is that the set ${ (M,N) : M leq_n N }$ is neither Zariski closed nor Zariski open. (Take $Lambda = k[epsilon]/epsilon^2$ and $d=2$. So $mathrm{rep}_2 Lambda$ (in the notation of the paper) is the space of $2 times 2$ matrices with square zero. Then two matrices $rho$ and $sigma$ in $mathrm{rep}_2 Lambda$ obey $rho leq_1 sigma$ if and and only if either $sigma =0$ or $rho neq 0$.) If these spaces were Zariski closed, this would be an easy consequence of Hilbert's basis theorem but, as it is, I am stumped.
algebraic-geometry representation-theory
algebraic-geometry representation-theory
edited Oct 14 '14 at 18:18
David E Speyer
46.2k4127211
asked Mar 26 '12 at 14:17
Julian KuelshammerJulian Kuelshammer
7,59732767
edited Oct 14 '14 at 18:18
David E Speyer
46.2k4127211
asked Mar 26 '12 at 14:17
Julian KuelshammerJulian Kuelshammer
7,59732767
edited Oct 14 '14 at 18:18
David E Speyer
46.2k4127211
edited Oct 14 '14 at 18:18
David E Speyer
46.2k4127211
edited Oct 14 '14 at 18:18
David E Speyer
46.2k4127211
46.2k4127211
asked Mar 26 '12 at 14:17
Julian KuelshammerJulian Kuelshammer
7,59732767
asked Mar 26 '12 at 14:17
Julian KuelshammerJulian Kuelshammer
7,59732767
asked Mar 26 '12 at 14:17
Julian KuelshammerJulian Kuelshammer
7,59732767
7,59732767
$begingroup$
Edits remove two ambiguities in the original question (what finite dimensional means, and what $n$ is allowed to depend on), based on my reading of the article. Then added a paragraph to point out what doesn't work here.
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– David E Speyer
Oct 14 '14 at 18:20
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@DavidSpeyer. It is a general rule on SE sites that posts should not contain visible edit history. They should read smoothly.
$endgroup$
– TRiG
Jun 20 '16 at 11:01
add a comment |
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Edits remove two ambiguities in the original question (what finite dimensional means, and what $n$ is allowed to depend on), based on my reading of the article. Then added a paragraph to point out what doesn't work here.
$endgroup$
– David E Speyer
Oct 14 '14 at 18:20
$begingroup$
@DavidSpeyer. It is a general rule on SE sites that posts should not contain visible edit history. They should read smoothly.
$endgroup$
– TRiG
Jun 20 '16 at 11:01
$begingroup$
Edits remove two ambiguities in the original question (what finite dimensional means, and what $n$ is allowed to depend on), based on my reading of the article. Then added a paragraph to point out what doesn't work here.
$endgroup$
– David E Speyer
Oct 14 '14 at 18:20
$begingroup$
Edits remove two ambiguities in the original question (what finite dimensional means, and what $n$ is allowed to depend on), based on my reading of the article. Then added a paragraph to point out what doesn't work here.
$endgroup$
– David E Speyer
Oct 14 '14 at 18:20
$begingroup$
@DavidSpeyer. It is a general rule on SE sites that posts should not contain visible edit history. They should read smoothly.
$endgroup$
– TRiG
Jun 20 '16 at 11:01
$begingroup$
@DavidSpeyer. It is a general rule on SE sites that posts should not contain visible edit history. They should read smoothly.
$endgroup$
– TRiG
Jun 20 '16 at 11:01
add a comment |
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protected by Community♦ Dec 12 '18 at 9:59
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
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Edits remove two ambiguities in the original question (what finite dimensional means, and what $n$ is allowed to depend on), based on my reading of the article. Then added a paragraph to point out what doesn't work here.
$endgroup$
– David E Speyer
Oct 14 '14 at 18:20
$begingroup$
@DavidSpeyer. It is a general rule on SE sites that posts should not contain visible edit history. They should read smoothly.
$endgroup$
– TRiG
Jun 20 '16 at 11:01