<s>[INST] Find the element in $\mathbb{Z}/143\mathbb{Z}$ whose image is $(\overline{10},\overline{11})$ under the Chinese remainder theorem Find the element in $\mathbb{Z}/143\mathbb{Z}$ whose image is $(\overline{10},\overline{11})$ in $\mathbb{Z}/11\mathbb{Z} \times \mathbb{Z}/13\mathbb{Z}$ under the Chinese remainder theorem

So I figured we must find an element $\overline{x}$ in $\mathbb{Z}/143\mathbb{Z}$ that satisfies:
$x \equiv 10  \pmod {11} $
$x \equiv 11  \pmod {13} $
Applying the Chinese remainder theorem gives us $x=76$.
Can I then simply conclude that the element we're looking for in $\mathbb{Z}/143\mathbb{Z}$ is $\overline{76}$?
 [/INST] By Bezout's identity we know that we can find $ x, y$ such that $ 11x + 13y = 1 $. Using the Euclidean algorithm, we have:
$$ 13 = 11\cdot 1 + 2 $$
$$ 11 = 2 \cdot 5 + 1 $$
$$ 2 = 2 \cdot 1 + 0 $$
and now we substitute back:
$$ 1 = 11 - 2 \cdot 5 = 11 - (13 - 11) \cdot 5 = 11 \cdot 6 - 13 \cdot 5 $$
This helps us because now we can explicitly produce the solution. Indeed, I claim that the solution is $ 11 \cdot 6 \cdot 11 - 13 \cdot 5 \cdot 10 = 76 $. We have that
$$ 11 \cdot 6 \cdot 11 - 13 \cdot 5 \cdot 10 \equiv 11 \cdot 6 \cdot 10 - 13 \cdot 5 \cdot 10 \equiv 10 \pmod{11} $$
and
$$ 11 \cdot 6 \cdot 11 - 13 \cdot 5 \cdot 10 \equiv 11 \cdot 6 \cdot 10 - 13 \cdot 5 \cdot 11 \equiv 11 \pmod{13} $$
Generally, if $ I $ and $ J $ are comaximal ideals of a ring $ R $, we can find $ i \in I $ and $ j \in J $ such that $ i + j = 1_R $. Then, the expression $ bi + aj $ lies in the cosets $ a + I $ and $ b + J $. In the case that $ R $ is an Euclidean domain, the Euclidean algorithm gives an easy way of computing $ i $ and $ j $ explicitly.
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