2
$\begingroup$

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 $(\sharp)$?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.