Let $R$ be a finite commutative ring with $1$. Assume that $u$ is a $5$th root of unity in $R$ such that $u - 1$ is a unit in $R$.

From the following system,

$x_1 = x_2 + x_3 + x_4 + x_5, \\ ux_1 = u^3x_2 + u^2x_3 + u^4x_4 + x_5 \\ u^2x_1 = ux_2 + u^4x_3 + u^3x_4 + x_5 \\ u^3x_1 = u^4x_2 + ux_3 + u^2x_4 + x_5 \\ u^4x_1 = u^2x_2 + u^3x_3 + ux_4 + x_5 $

Can anyone show that the solution of this system over the ring $R$ is unique which is $(0,0,0,0,0)$?

I have already checked that this system has only one solution which is $(0,0,0,0,0)$ if we work over $\mathbb{R}$ or $\mathbb{C}$.


The assumption that $R$ is finite is unnecessary. Note that since $$0=u^5-1=(u-1)(u^4+u^3+u^2+u+1)$$ and $u-1$ is a unit, $u^4+u^3+u^2+u+1=0$. Also, since $$(u-1)^5=u^5-5u^4+10u^3-10u^2+5u-1=5(-u^4+2u^3+2u^2+u),$$ $5$ is a unit in $R$.

Now note that adding your five equations together makes all the coefficients but one become $u^4+u^3+u^2+u+1=0$, so we get $5x_5=0$ and thus $x_5=0$. If we multiply the second equation by $u$, the third equation by $u^2$, the fourth equation by $u^3$, and the fifth equation by $u^4$ and then add them up, we similarly get $x_4=0$. In the same way (multiplying the equations by powers of $u^2$, then powers of $u^3$, then powers of $u^4$), we get $x_3=0$, $x_2=0$, and $x_1=0$.

  • $\begingroup$ Thank you @Eric Wofsey. This helps a lot. $\endgroup$ – NongAm Sep 12 '17 at 14:10

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