# Math Help - Two more questions involving direct products

1. ## Two more questions involving direct products

1. Is Z x Z cyclic? [Here Z means (Z, +).]

2. Prove the Chinese remainder theorem: If $m_1,.....,m_n$ are positive integers such that $m_i$ and $m_j$ are relatively prime for i not equal to j, and $k_1,....,k_n$ are any integers, then there is an integer x such that x is congruent to $k_i(mod m_i)$ for 1 less than or equal to i less than or equal to n. (Hint: Consider the generator (1,1,....,1) for $Zm_1$ X......X $Zm_n.$)

2. Originally Posted by Janu42
1. Is Z x Z cyclic? [Here Z means (Z, +).]
no, however, it is generated by two elements, (1,0) and (0,1), i.e, <(1,0),(0,1)> = Z x Z

3. Originally Posted by Janu42
2. Prove the Chinese remainder theorem: If $m_1,.....,m_n$ are positive integers such that $m_i$ and $m_j$ are relatively prime for i not equal to j, and $k_1,....,k_n$ are any integers, then there is an integer x such that x is congruent to $k_i(mod m_i)$ for 1 less than or equal to i less than or equal to n. (Hint: Consider the generator (1,1,....,1) for $Zm_1$ X......X $Zm_n.$)
If $m_1,...,m_n$ are pairwise coprime then it means $G=\mathbb{Z}_{m_1}\times ... \times \mathbb{Z}_{m_n}$ is a cyclic group, furthermore, $([1]_{m_1},...,[1]_{m_n})$ is a generator for this group. Let $k_1,...,k_n\in \mathbb{Z}$ and consider $([k_1]_{m_1}, ... , [k_n]_{m_n}) \in G$. Therefore, there exists $x\in \mathbb{Z}$ so that $x([1]_{m_1},...,[1]_{m_n}) = ([k_1]_{m_1}, ... , [k_n]_{m_n})$, but that means $([x]_{m_1}, ... , [x]_{m_n}) = ([k_1]_{m_1}, ... , [k_n]_{m_n})$. Therefore, $[x] = [k_j]_{m_j}$ for all $1\leq j\leq n$. Thus, $x\equiv k_j(\bmod m_j)$.