Let $K$ be a field and $R=K[x_1,\ldots,x_n]=\bigoplus_{a\in\mathbb{N}^n}Kx^a$ the multigraded polynomial ring. Have finitely-generated multigraded $R$-modules been classified? Are they of the form $R^r\oplus\bigoplus_{i=1}^sR/Rx^{a_i}$ for some (unique?) $a_1,\ldots,a_s\in\mathbb{N}^n$?

I was thinking along the following lines. If an $\mathbb{N}^n$-graded $R$-module $M$ is generated by $v_1,\ldots,v_r$, then we may assume that every $v_i$ is homogenous of degree $a_i$ (otherwise each of these $v_i$ is a further finite combination of homogenous vectors). Let $R^{[a]}$ be the $R$-module $R$ with the grading shifted by $a\!\in\!\mathbb{N}^n$. Thus the map $R^r\!=\!\bigoplus_{i=1}^rR^{[a_i]}\rightarrow M$ that sends $e_i\mapsto v_i$ is a surjective graded morphism, so its kernel $A$ is a graded submodule and $M\cong R^r/A$, i.e. $$\textstyle{A=\bigoplus_{b\in\mathbb{N}^n}(R^r)_b\cap A=\bigoplus_{b\in\mathbb{N}^n}\prod_iKx^{a_i+b}\cap A}.$$ Hence $A$ is generated by $u_1,\ldots,u_s$ where every component of $u_i$ is $\alpha_{ji}x^{a_i+b_j}$ for some $\alpha_{ji}\!\in\!K$.

Thus our module is isomorphic to the cokernel of the matrix $$\left[\begin{matrix} \alpha_{11}x^{a_1+b_1} &\alpha_{12}x^{a_1+b_2}&\ldots&\alpha_{1s}x^{a_1+b_s}\\ \alpha_{21}x^{a_2+b_1} &\alpha_{22}x^{a_2+b_2}&\ldots&\alpha_{2s}x^{a_2+b_s}\\ \vdots&\vdots&\vdots&\vdots\\ \alpha_{r1}x^{a_r+b_1} &\alpha_{r2}x^{a_r+b_2}&\ldots&\alpha_{rs}x^{a_r+b_s}\\ \end{matrix}\right].$$ Now we can do row and column operations, without changing the isomorphism type of the module. But I'm not sure how the above can be transformed into $$\left[\begin{matrix} x^{c_1} &&&\\ &x^{c_2}&&\\ &&\ddots&\\ &&&x^{c_s}\\ \end{matrix}\right].$$

If my conjecture is not valid, I ask for a counterexample. For instance, if $n=2$ and $R=K[x,y]$, are $Coker\left[\begin{smallmatrix} x^2y &xy\\ & xy^2\\ \end{smallmatrix}\right]$ or $Coker\left[\begin{smallmatrix} x &y\\ \end{smallmatrix}\right]$ or $Coker\left[\begin{smallmatrix} x \\y \end{smallmatrix}\right]$ not of the above form?

Is there some other classification of $\mathbb{N}^n$-graded $K[x_1,\ldots,x_n]$-modules? Maybe they are of the form $\bigoplus_{i=1}^sR^{[a_i]}/\langle x^a; a\!\in\!A_i\rangle$ for some (unique?) $a_1,\ldots,a_s\!\in\!\mathbb{N}^n$ and $A_1,\ldots,A_s\!\subseteq\!\mathbb{N}^n$?

By Monomial Ideals (Herzog & Hibi, 2011), Dickson’s lemma 2.1.1, every $A_i$ may be finite. Also, elements of $A_i$ are assumed to be incomparable w.r.t. the componentwise partial order on $\mathbb{N}^n$.

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    $\begingroup$ Why the hell does this have a -3 vote count? (-2 now). Sometimes I really don't understand math.SE user's vote patterns. $\endgroup$ Commented Apr 6, 2015 at 17:13
  • $\begingroup$ @lenticcatachresis Thank you!!! I wondered myself... $\endgroup$
    – Leo
    Commented Apr 6, 2015 at 23:53
  • $\begingroup$ In the case $n\!=\!1$, all monomials are comparable, so the situation reduces to the classification of f.g. modules over a PID and the first statement holds. $\endgroup$
    – Leo
    Commented Apr 7, 2015 at 11:32

2 Answers 2


For $n=2$ let $M$ be the module with graded parts $$M_{i,j}=\begin{cases} K&\mbox{ if }(i,j)=(1,1),(1,0)\mbox{ or }(0,1)\\ 0&\mbox{ otherwise.} \end{cases}$$ and with $x_1$ and $x_2$ acting as isomorphisms $M_{0,1}\to M_{1,1}$ and $M_{1,0}\to M_{1,1}$ respectively.

Then your conjecture is not true for $M$.

If $n>2$ (possibly also $n=2$) then this is a "wild" problem: classifying such modules is at least as hard as classifying pairs of square matrices up to simultaneous conjugacy.

Edit: This last claim is not too complicated to see for $n=5$:

Let $V=K^d$ and let $A,B$ be $d\times d$ matrices over $K$. Define a multigraded $K[x_1,\dots,x_5]$-module $N=N(A,B)$ with

$$ N_{00000}=V\oplus V,$$ $$N_{10000}=N_{01000}=N_{00100}=N_{00010}=N_{00001}=V,$$ and all other components zero, so the action of each $x_i$ is determined by a map $V\oplus V\to V$, given by the matrices $$\begin{align} X_1&=\begin{pmatrix}I_n&0\end{pmatrix}\\ X_2&=\begin{pmatrix}0&I_n\end{pmatrix}\\ X_3&=\begin{pmatrix}I_n&I_n\end{pmatrix}\\ X_4&=\begin{pmatrix}I_n&A\end{pmatrix}\\ X_5&=\begin{pmatrix}I_n&B\end{pmatrix}\\ \end{align}$$

An easy computation shows that an isomorphism $N(A,B)\cong N(A',B')$ is determined by a single automorphism of $V$, given by a matrix $T$ such that $T^{-1}AT=A'$ and $T^{-1}BT=B'$, so the classification of this particular class of multigraded modules up to isomorphism is just the classification of pairs of square matrices up to simultaneous conjugacy.

  • $\begingroup$ Your module is $Kx\oplus Ky\oplus Kxy$ $\cong$ $R^{[(1,0)]}\!/\langle x,y\rangle \oplus R^{[(0,1)]}\!/\langle x,y\rangle \oplus R^{[(1,1)]}\!/\langle x,y\rangle$ $\cong$ $R^{[(0,1)]}\!/\langle x,y\rangle \oplus R^{[(1,0)]}\!/\langle x,y^2\rangle$, so it is of the second form, which is not unique. Could you elaborate why this problem is at least as hard as classifying pairs of square matrices up to simultaneous conjugacy? $\endgroup$
    – Leo
    Commented Apr 7, 2015 at 18:31
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    $\begingroup$ @Leon No, all three modules are non-isomorphic. The annihilator of your first module contains both $x$ and $y$, the annihilator of your second module contains $x$ but not $y$, and the annihilator of my module contains neither $x$ nor $y$. $\endgroup$ Commented Apr 8, 2015 at 11:46
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    $\begingroup$ @Leon I've added a justification for my claim about wildness in the case $n=5$. It's a little more complicated to describe in the cases $n=3$ and $n=4$. $\endgroup$ Commented Apr 8, 2015 at 11:47

Carlsson and Zomorodian (2009) gave a classification theorem for finitely generated multigraded modules over $R=k[X_1, \ldots, X_n]$ in their paper on multiparameter persistence. The motivation for their study is that the category of discrete multiparameter persistence modules is equivalent to the category $\mathcal{C}$ of finitely generated multigraded $R$-modules.

The classification is a little complicated - in particular, there is no discrete complete parametrisation as in the $n=1$ case. I will try to describe their work briefly.

A free object $F \in \mathcal{C}$ has as its basis a multiset $\xi(F) := \{(x_1, \alpha_1), \ldots, (x_m, \alpha_m)\} \subset \mathbb{Z}^n \times \mathbb{N}$, such that $F$ is the direct sum $\bigoplus_{(x, \alpha) \in X} \Sigma^x R^\alpha$ (where $\Sigma^x$ denotes a degree shift by $x$). For arbitrary $M$ in $\mathcal{C}$ the classification proceeds by considering the generating set and relations.

  1. Let $\rho(M)$ denote the multigraded $k$-vector space $k \otimes_R M$, where $k$ has the $R$-module structure such that the $x_i$ act trivially. There is a free object $F$ in $\mathcal{C}$ and an epimorphism $\pi_M: F \to M$ such that $1 \otimes_R \pi : \rho(F) \to \rho(M)$ is an isomorphism of multigraded $k$-vector spaces, and $(F, \pi_M)$ is unique upto isomorphism.
  2. For every $M \in \mathcal{C}$ we have two invariants $\xi_0(M) := \xi(\rho(M))$ and $\xi_1(M) := \xi(\rho(\ker(\pi_M))$.
  3. The isomorphism class of $M$ is the isomorphism class of $F/L$ where $L$ is any member of the orbit of $G_F := \operatorname{Aut}(F)$ acting on the set of subobjects $S := \{L \subset F \mid \xi_0(L) = \xi_1(M)\}$.

In general $\operatorname{Aut}(F)$ is algebraic group and $S$ is a quasiprojective variety so the orbit space might not have a nice structure.


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