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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 $M\leq_{\operatorname{Hom}} N$ iff $\dim \operatorname{Hom}(X,M)\leq \dim \operatorname{Hom}(X,N)$ for all $X$ and one by $M\leq_n N$ iff $\dim \operatorname{Hom}(\Lambda^n/\Lambda^nA,M)\leq \dim \operatorname{Hom}(\Lambda^n/\Lambda^nA,N)$ for all $n\times n$-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.

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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. – David Speyer Oct 14 '14 at 18:20
@DavidSpeyer. It is a general rule on SE sites that posts should not contain visible edit history. They should read smoothly. – TRiG Jun 20 at 11:01

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