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 $\mathbb{N}^n$.
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.