I have read a text where it says:
" The Chinese remainder theorem states that:
$$\begin{align*}
x &\equiv a_{1}\pmod{m_1}\\
x &\equiv a_{2}\pmod{m_2}\\
&\vdots\\
x &\equiv a_{n}\pmod{m_n}\\
\end{align*}$$
It has a solution as long as the modules $m_1, . . . , m_n$ are primes two to two.
In more modern terms, this same statement can be expressed like this:
If $m_1, . . . , m_n$ are integers pairwise coprime and $m = m_1 · · · m_n$, we have the isomorphism of rings
$$\begin{align*} \mathbb Z/m\mathbb Z\cong (\mathbb Z/m_1\mathbb Z) \times · · · \times(\mathbb Z/m_n\mathbb Z) \end{align*}$$
given by
$$\begin{align*}[a]\mapsto ([a], · · ·, [a])\end{align*}"$$
My question in this regard is, why does "expressed in more modern terms" is equivalent to there being an $x$ that solves the linear system of congruences-
Best Answer
We wish to find a solution for $$\begin{align*} x &\equiv a_{1}\pmod{m_1}\\ x &\equiv a_{2}\pmod{m_2}\\ &\vdots\\ x &\equiv a_{n}\pmod{m_n}\\ \end{align*}$$
Now $(\overline{a_1},\overline{a_2},\cdots, \overline{a_n})\in\prod_{i=1}^n\mathbb{Z}/m_i\mathbb{Z}$. So since $\overline{a}\mapsto (\overline{a},\overline{a},\cdots,\overline{a})$ is an isomorphism, then there exists $k\in\mathbb{Z}$ such that $\overline{k}\mapsto (\overline{a_1},\overline{a_2},\cdots, \overline{a_n})$. On the other hand, since $\overline{k}$ is mapped to $(\overline{k},\overline{k},\cdots, \overline{k})$, $(\overline{k},\overline{k},\cdots, \overline{k})=(\overline{a_1},\overline{a_2},\cdots, \overline{a_n})$. In particular, $$\begin{align*} k &\equiv a_{1}\pmod{m_1}\\ k &\equiv a_{2}\pmod{m_2}\\ &\vdots\\ k &\equiv a_{n}\pmod{m_n}\\ \end{align*}$$
Moreover, by injectivity, it must be the case that if $y\in\mathbb{Z}$ satisfies these congruences, then $y\equiv k\bmod m_1m_2\cdots m_n$,