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Let $\{f_i\}$ be a system of linear equations in $X_1,...,X_n$ with coefficients in $R = \mathbb Z/p^e \mathbb Z$ (i.e. modulo a primepower).

Assume

  1. there is a unique solution $a_i$.
  2. $f_1 = u g_1$ where $u$ is a non-unit, i.e. $p\mid u$.

Is it true that $\{f_i\mid i>1\}$ permits only the same unique solution? Or equivalently: is $f_1$ linearly dependent on the other equations?

Intuitively, it seems obvious. For if we substitute variables such that $g_1 = (Y_1 - a)$ then $Y_1$ is determined by $f_1$ up to some multiple of $p$. But there's no way the other equations could determine what multiple without determining what $Y_1$ is itself, since the only non-units are all multiples of $p$. But that is, of course, not very solid mathematics.

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    Indeed I do! Editted the question.2011-01-09

1 Answers 1

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Suppose for simplicity that the first equation is $p^tx_1 = 0$, so that $x_1 = kp^{e-t}$. Assume to the contrary that the system without this equation has more than one solution. Look at the projection $P$ of the set of all solutions on $x_1$. It must contain exactly one member of the form $kp^{e-t}$. Since it is linear, this must in fact be $0$. There is also another member $rp^{e-t-s}$ which is not a multiple of $p^{e-t}$ (here $(r,p)=1$ and $s \geq 1$). Again, since the solution space is linear we get that $P$ contains $rp^{e-t} \neq 0$, contradiction.

We can always transform the first equation to this form by simple substitution. Take the lowest power $p^t$ ($t \geq 1$) dividing any of the coefficients of $f_1$, say $y_1$. So the equation can be written as $p^t(y_1 + \cdots) = 0$. Now let $x_1 = y_1 + \cdots$. Since $y_1 = x_1 - \cdots$, we can replace $y_1$ with $x_1$. This transformation is one-to-one and so preserves all the required properties.

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    Tha$n$ks, that's more or less the "solidification" of my intuition that I was looking for, I guess.2011-01-10