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$\DeclareMathOperator\GL{GL}$Let $R$ be a commutative ring, let $R[n] := R[M_d^{\oplus n}]$ be the polynomial ring on $nd^2$ variables corresponding to the coordinates of $n$-many $d\times d$ matrices. Let these matrix variables be $X_1,\ldots,X_n$. For an $d\times d$ matrix $A$, let $c_k(A)$ be the coefficient of $T^k$ in the characteristic polynomial $\det(A-TI)$.

The group scheme $\GL_{d,R}$ acts on $R[n]$ by simultaneous conjugation on the $d$ matrices. Clearly for any product $X_{i_1}X_{i_2}\cdots X_{i_r}$ (where $i_j\in[1\ldots n], r\ge 1$), the function $c_k(X_{i_1}\cdots X_{i_r})\in R[n]$ is invariant under $\GL_{d,R}$.

Is it true that for any commutative ring $R$, $R[n]^{\GL_{d,R}}$ is generated as an $R$-algebra by the functions $c_k(X_{i_1}\cdots X_{i_r})$?

Does anyone have a reference for this?

Remark - The statement over $\mathbb{C}$ is a classical result of Sibirski and Procesi. This was later extended to the case $R = \mathbb{Z}$ and $R$ any algebraically closed field by Donkin in Invariants of several matrices. In Concini-Procesi's The invariant theory of matrices, they also seem to obtain the result when $R$ is an infinite field.

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    $\begingroup$ Could you give a reference to the classical result? MathSciNet does not seem to recognise Sibirski. $\endgroup$
    – LSpice
    Sep 6 '21 at 23:57
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    $\begingroup$ FWIW: This is also part of Theorem 1.10 in Corrado De Concini, Claudio Procesi, The Invariant Theory of Matrices, AMS 2017. But I'm not quite sure what they require of the base ring $R$; possibly it has to be a field. (But algebraic closedness can certainly not be useful; the claim is of the form "a vector lies in the span of a bunch of other vectors", so must hold in any prime field if it holds in an extension.) $\endgroup$ Sep 7 '21 at 1:50
  • $\begingroup$ @LSpice The Sibirski result is cited in the introduction to Donkin's paper, though I have not checked it. Here's the citation: "Sibirski, K.S.: On unitary and orthogonal matrix invariants. Dokl. Akad. Nauk. SSSR 172, 1, 40-43 (1967)" $\endgroup$ Sep 7 '21 at 2:14
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    $\begingroup$ It might be worth adding the [quivers] tag, as this is asking about the invariants for the one-vertex, d-loop quiver. $\endgroup$
    – user44191
    Sep 7 '21 at 4:13
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    $\begingroup$ It must be true because Donkin shows one has a good filtration over $\mathbb Z$. That implies base change for the group scheme invariants. Infinite fields are only needed when one wants to confuse a group $G(k)$ with the group scheme $G_k$. $\endgroup$ Sep 7 '21 at 7:35
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It is true. The standard reference is the Book by Jantzen, Representations of Algebraic Groups, Second edition. In particular we need the Appendix `Chapter B', and the base change Proposition in part I, 4.18. By Donkin the $G_{\mathbb Z}$ module $M={\mathbb Z}[n]$ has good filtration. So $H^1(G_{\mathbb Z},M)$ vanishes. And $H^0(G_{\mathbb Z},M)$ just means $M^{G_{\mathbb Z}}$. The base change Proposition tells that the map $M^{G_{\mathbb Z}}\otimes R\to (M\otimes R)^{G_{R}}$ is an isomorphism. The result follows.

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  • $\begingroup$ Thank you! That is very helpful! When you say that Donkin proves that $M = \mathbb{Z}[n]$ has a good filtration, are you referring to his arguments in $\S3$ (on p399)? He doesn't seem to explicitly claim the existence of a good filtration over $\mathbb{Z}$ - he only seems to discuss it over alg. closed fields; is the result over $\mathbb{Z}$ implied by his arguments? (I am not especially familiar with representation theory so this is not clear to me) $\endgroup$ Sep 9 '21 at 5:45
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    $\begingroup$ Chapter B explains that it indeed is implied by the result over fields. You may also need that field extensions do not matter for good filtrations. So talking about algebraically closed fields is just meant to put the reader at ease. $\endgroup$ Sep 9 '21 at 7:36
  • $\begingroup$ From chapter B, we know that $M$ having a good filtration implies that the higher Ext groups $Ext^i_G(V(\lambda),M)$ vanish for $i > 0$ and $\lambda$ dominant, whereas we want to see that $H^1(G,M)$ vanishes. Is the trivial representation a direct summand of $V(\lambda)$? $\endgroup$ Sep 10 '21 at 18:10
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    $\begingroup$ @stupid_question_bot The trivial representation is $V(0)$ and $0$ is dominant. $\endgroup$ Sep 12 '21 at 6:06
  • $\begingroup$ If we replace $\mathbb{Z}[n]$ with the coordinate ring of the product of $n$ copies of $SL_2$, do you know if this coordinate ring (with the same conjugation action by $GL_2$) should have a good filtration? I think I can see it over fields, but it's less clear to me how to deduce it over $\mathbb{Z}$. The lemma in Appendix B.9 only applies to $k$-finite $GL_2$-modules; while it's obvious how one can filter $\mathbb{Z}[n]$ w/ $k$-finite $G$-stable pieces, I don't see an obvious analog for the coordinate ring of $SL_2$. $\endgroup$ Sep 14 '21 at 4:13
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For the benefit of myself and other novices in the area, I wanted to add some details to Wilberd's excellent answer, citing Donkin's Invariants of several matrices and Jantzen's book Representations of Algebraic Groups as appropriate.

We work over a base ring $R$.

First, one checks that Donkin's definition of a "good filtration" on a $G$-module $V$ coincides with Jantzens (defined in II, 4.16). The latter uses the isomorphism $H^i(M) := R^i\text{Ind}^G_B(V)\cong H^i(G/B,\mathcal{L}(M))$ (Jantzen I, 5.12), where $B\subset G$ is a Borel, $M$ a $B$-module, and $\mathcal{L}(M)$ denotes the quasi-coherent sheaf on $G/B$ associated to $M$ (I, 5.8). If $M = R$ with trivial $G$-action, then $\mathcal{L}(R)$ is the structure sheaf $\mathcal{O}_{G/B}$.

Now write $R[n] = \bigoplus_{d\ge 0}R[n]_d$, where $R[n]_d$ consists of the polynomials of total degree $d$. One easily checks that each $R[n]_d$ is a $\text{GL}_{n,R}$-module. We claim that the $\mathbb{Z}$-finite $G = \text{GL}_{n,\mathbb{Z}}$-modules $\mathbb{Z}[n]_{\le D} := \bigoplus_{0\le d\le D} \mathbb{Z}[n]_d$ have good filtrations for any $D\ge 0$. We use the following Lemma

Lemma (Appendix B.9 of Jantzen) Let $G$ be a split reductive algebraic group, and let $T$ be a maximal torus. Suppose that $R$ is a principal ideal domain. Let $M$ be a $G$-module which is free of finite rank over $R$. As explained in B.4, let $V(\lambda) := H^0(-w_0\lambda)^*)$ be the Weyl module, where $w_0$ is the longest element of the Weyl group (II, 1.5). The following are equivalent:

(i) $M$ has a good filtration.

(ii) $Ext^i_G(V(\lambda),M) = 0$ for all $\lambda\in X(T)_+$ and all $i > 0$

(iii) $Ext^1_G(V(\lambda),M) = 0$ for all $\lambda\in X(T)_+$.

(iv) For each maximal ideal $\mathfrak{m}$ in $k$, the $G_{R/\mathfrak{m}}$-module $M\otimes R/\mathfrak{m}$ has a good filtration.

By (iv), it suffices to check that the $G_{\mathbb{F}_p}$-modules $\mathbb{F}_p[n]_{\le D}$ all have good filtrations. Using flat base change for $G$-module cohomology (see Jantzen I, 4.13), (ii) (or (iii)) implies that it suffices to check the existence of good filtrations over algebraically closed fields $k$. By a result of Donkin (Donkin The normality of closures of conjugacy classes of matrices Proposition 1.2a (i) and (ii)), good filtrations behave well with respect to direct summands, so it suffices to check that $\overline{\mathbb{F}_p}[n]$ have good filtrations. This is done in $\S3$ of Donkin's Invariants of several matrices. Thus we conclude that $\mathbb{Z}[n]_{\le D}$ has a good filtration for any $D\ge 0$.

Let $T$ be a maximal torus of $G = GL_{n,\mathbb{Z}}$, and let $\lambda = 0\in X(T)$, then $\lambda$ is dominant and we have $V(0) = H^0(G/B,\mathcal{L}(0)) = H^0(G/B,\mathcal{O}_{G/B})$. Since $G/B$ is smooth proper over $\mathbb{Z}$ (use Jantzen I, 5.6(9) and I,5.7 to reduce to the classical case over algebraically closed fields), Stein factorization implies that $V(0) = \mathbb{Z}$ (with trivial $G$-action). By (ii) or (iii), we conclude that $Ext^1_G(\mathbb{Z},M) = H^1(G,M) = 0$. By a "universal coefficient theorem" (see Jantzen I, 4.18), vanishing of $H^1(G,M)$ implies that for any ring $R$, $\mathbb{Z}[n]^{GL_{2,\mathbb{Z}}}\otimes R = R[n]^{GL_{2,R}}$, as desired.

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