Solutions of the linear equation $a_1Y_1 + \cdots+ a_mY_m = 0$

Question

Let us consider a two-variable formal power series ring $A_2 \colon= K[[X_1,X_2]]$ over a field $K$. It is known that solutions for a linear equation

$$(\sharp) \quad a_1Y_1 + \cdots + a_mY_m = 0$$ such that $a_1,\ldots,a_m \in A_2$ is generated by $(m-1)$ numbers of certain solutions under the condition $Y_1,\ldots,Y_m \in A_2$.

By the way, there are following obvious solutions for $(\sharp)$: \begin{align} & \alpha_{1,2} = (a_2,-a_1,0,\ldots,0) \\ & \ldots \\ & \alpha_{i,j} = (0,\ldots,-a_j,0,\ldots,0,a_i,0,\ldots,0) \\ & \ldots \\ & \alpha_{m-1,m} = (0,\ldots,a_m,-a_{m-1}). \end{align}

Q. Under what conditions on $a_1,\ldots,a_m$, do these solutions $α_{i,j}$ generate all solutions of $(\sharp)$?

Answer 1

I will assume that all $a_i$ are in the maximal ideal, since otherwise, easily you can write down all solutions. If the $a_i$s have a common divisor, clearly you can cancel it in your equation and thus also assume that they have no common divisors.

Then, you have a map $A_2^m\to A_2$ given by the $a_i$s and you are interested in the kernel. Homological algebra immediately tells you that the kernel is $A_2^{m-1}$ and the map $A_2^{m-1}\to A_2^m$ is given by an $m-1\times m$ matrix $M$. It can be shown that after a change of basis if necessary, the $m$ $m-1\times m-1$ minors of $M$ are precisely the $a_i$s. This is the best you can do in general.