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Let $A$ be a commutative ring of characteristic zero. Let $a_1,a_2,a_3,a_4 \in A$ be units such that $a_i^k\ne a_j^k$ for $i\ne j$ and $k=1,2$.

How to show that $a_1 a_2 a_3 + a_1 a_2 a_4 + a_1 a_3 a_4 + a_2 a_3 a_4 \ne 0$.

With a positive answer for this question, is that possible to generalize for $n$ units?

Remark: If necessary, $A$ could be considered a discrete valuation ring.

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Isn't $a_1 = 1$, $a_2 = 2$, $a_3 = 3$, $a_4 = -\frac{6}{11}$ (which works in any $\mathbb{Z}_p$ for $p \ne 11$) a counterexample ? –  Joel Cohen Dec 14 '11 at 20:56
    
@Joel Cohen: You are right, I didn't realize this solution... However, what happens if we assume that $A$ is a discrete valuation ring? –  Binai Dec 14 '11 at 23:01
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@user27264 : For any prime number $p$, set $\mathbb{Z}_{(p)} = \{\frac{a}{b} \in \mathbb{Q}, \ b \text{ coprime to } p\}$. This is a discrete valuation ring whose only prime ideal is generated by $p$ (and its completion is $\mathbb{Z}_p$). If $p \ne 11$, then $\mathbb{Z}_{(p)}$ contains $1, 2, 3$ and $-\frac{6}{11}$. For a counterexample $\mathbb{Z}_{(11)}$, just take $a_1 = 1$, $a_2 = 2$, $a_3 = 4$ and $a_4 = -\frac{4}{7}$. Now any discrete valuation ring of characteristic $0$ contains at least one of those ring, so you get a counterexample in all of them. –  Joel Cohen Dec 15 '11 at 2:33
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1 Answer

up vote 4 down vote accepted

$abc+abd+acd+bcd=abcd(a^{-1}+b^{-1}+c^{-1}+d^{-1})$ so lets make the sum of four units equal to zero. take some field, pick three random elements $x,y,z$ and let $w=-(x+y+z)$. these will satisfy your "distinctness" property in general. for example, $x,y,z,w=1,2,3,-6\in\mathbb{R}$ give $1/6-1/12-1/18-1/36=0$

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You are right, I didn't realize this solution... However, what happens if we assume that $A$ is a discrete valuation ring? –  Binai Dec 14 '11 at 23:01
1  
Let $A=\mathbb{F}_2[[x]]$ be the ring of formal power series over the field of two elements. Note that $A$ is a discrete valuation ring (ie a PID with only one nonzero maximal ideal). Since $U(A)=\{f(x)\in A\,\vert\,f(0)=1\}$, we have that the sum of any two units is a nonunit and that the sum of a nonunit and a unit is a unit. So, letting $f=1$, $g=1+x$, $h=1+x+x^3$ and $k=f+g+h=1+x^3$, we have four distinct units that satisfy $\alpha^i\not=\beta^j$ for $\alpha,\beta\in \{f,g,h,k\}$, $\alpha\not=\beta$ and $1\leq i,j\leq 2$. However, $f+g+h+k=0$. –  user5137 Dec 14 '11 at 23:13
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