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 SpeyerThe 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.

**Answer**

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 $\leq_{\operatorname{Hom}}$.

**Attribution***Source : Link , Question Author : Julian Kuelshammer , Answer Author : Jeremy Rickard*