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Let $b_1, \ldots, b_n \in \mathbb{N}$. For $x, y \in \mathbb{Z}^n$, define $x \cdot y$ as $\newcommand{\lcm}{\operatorname{lcm}}$ $$x \cdot y = \left(\sum_{i=1}^n (x_i \text{ mod } b_i)(y_i \text{ mod } b_i) \text{ mod } b_i\right) \text{ mod } \lcm(b_1, \ldots, b_n)$$

Informally, each multiplication $x_iy_i$ is carried over $\mathbb{Z}/b_i\mathbb{Z}$ and the sum is carried over $\mathbb{Z}/ \lcm(b_1, \ldots, b_n)\mathbb{Z}$.

My question is the following: does $x \cdot (y + z) \equiv x \cdot y + x \cdot z \quad (\text{mod} \lcm(b_1, \ldots, b_n))$ hold?

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You have a sum on $i$ with $i$ not appearing in the summand. If that's not what you intended, please edit. – Gerry Myerson Sep 14 '11 at 22:26
Thanks, I've fixed it. – Li-thi Sep 14 '11 at 22:27
up vote 1 down vote accepted

For $n>1$, no. The individual $\mathbb{Z}/b_i\mathbb{Z}$'s will cycle around to $0$ under addition before the same occurs in the broader group of $\mathbb{Z}/\lcm(b_1,\dots,b_n)\mathbb{Z}$. Namely, let $e_1=(1,0,0,\dots)$. Then

$$e_1\cdot e_1\equiv 1\mod\ell $$ but $$(e_1+\cdots+e_1)\cdot e_1\equiv(b_1e_1)\cdot e_1\equiv0\cdot e_1\equiv0\mod\ell $$ while $e_1\cdot e_1+\cdots+e_1\cdot e_1$ with $b_1$ terms is congruent to $b_1(e_1\cdot e_1)\equiv b_1\not\equiv 0 \mod\ell$.

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