In Smalø: 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 $\Lambda$ that is finite dimensional over a field. 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 $d$) 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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    $\begingroup$ 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. $\endgroup$ Commented Oct 14, 2014 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
    Commented Jun 20, 2016 at 11:01
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    $\begingroup$ The question was recently edited to roll back most of an edit from over five years ago, on the grounds that it was "quite harmful". I don't want to get into an edit war, but I don't see what was harmful about that old edit. In fact, I think it was a helpful addition of context to indicate how Hilbert's basis theorem might be relevant to a question like this, and to explain why one simple approach to applying it doesn't work in this case. $\endgroup$ Commented Jun 21, 2020 at 21:52
  • $\begingroup$ A few questions: what does it mean for $M,N$ to be $d$-dimensional over $\Lambda$? Second, what is $X$ (a finitely generated $\Lambda$ module?) in the definition of $\leq_{Hom}$? $\endgroup$
    – Aphelli
    Commented Jun 24, 2020 at 7:20
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    $\begingroup$ I restored the edit. It was there for years, without a very strong and clear-cut reason it won't be removed now. $\endgroup$
    – quid
    Commented Jun 28, 2020 at 19:46

1 Answer 1


I found the following, in the introduction of the 2015 PhD thesis of Nils Nornes, a student of Smalø:

In [7], cowritten with my fellow student Tore A. Forbregd and our adviser Professor Sverre O. Smalø, we show that $\leq_{d^3}$ always is a partial order on $\operatorname{mod}_d\Lambda$. It seems like for large enough $n$, $\leq_n$ is equivalent to $\leq_{\operatorname{Hom}}$. In the paper we claimed this as a fact, but we did not give a proof. When the reviewer requested a proof, we realized that the proof we had in mind was incomplete. We decided to remove the statement, but unfortunately we wrote it twice and deleted it once, so it still appears in the published version. While we have not found a proof, we have not found any counterexample either. In fact, in all examples we have looked at, $\leq_n$ is either not a partial order or equivalent to $\leq_{\operatorname{Hom}}$. We still have no examples where $\leq_n$ is a partial order but is different from $\leq_{\operatorname{Hom}}$.


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