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 $(\sharp)$?

  • $\begingroup$ First, you have sign problems. or example, the first should be $(a_2,-a_1, 0,\ldots, 0)$. I would suggest you try the case with $m=2$ first and see the difference when $a_1,a_2$ have no common factor and when they do. $\endgroup$ – Mohan Sep 22 '17 at 14:45
  • $\begingroup$ When $m=2$, it will be sufficient if $(a_1,a_2) = 1$. For $m > 2$, I have no idea. $\endgroup$ – Rinmyaku Sep 22 '17 at 15:47
  • $\begingroup$ If you have tried $m=3$ in addition, you would have probably realized that it is never true, at least when you assume that all the $a_i(0,0)=0$. (Of course, if one of them is non-zero, it is a unit and you can easily reduce the case to $m-1$ equations) $\endgroup$ – Mohan Sep 22 '17 at 17:06
  • $\begingroup$ I am wondering how 0ne can get the 2 generators for m = 3. I might suppose that a_1 = (X_1)^m + ... is a monic polynomial with respect to X_1 and a_2, a_3 are non-monic polynomial of degrees strictly less than m with respect to X_1. I cannot get easy generators under this assumption. $\endgroup$ – Rinmyaku Sep 22 '17 at 18:29
  • $\begingroup$ There are no easy ways to get the $m-1$ generators of the kernel in general. $\endgroup$ – Mohan Sep 22 '17 at 19:23

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.

  • $\begingroup$ Thank you, but I would like to ask Mohan why (m-1)× $\endgroup$ – Pierre MATSUMI Sep 24 '17 at 6:41

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