Prove (using Chinese remainder theorem) that a polynomial of degree $n$ (over the field of real numbers) can be uniquely determined from its values at the $n + 1$ point.
I can prove it by Lagrange polynomial or some simple principles - but I have to use the Chinese theorem, which usage in this proof isn't clear for me.
Let $a_0,, a_1,\dots, a_n$ the $n+1$ points, and observe the remainder of the division of a polynomial $p$ by a linear polynomial$X-a_i$ is $p(a_i)$.
Consider the isomorphism of the Chinese remainder theorem: \begin{align} \mathbf R[X]\!\!\bigm/\!\bigl((X-a_0)(X-a_1)\dotsm(X-a_n)\bigr)&\longrightarrow\mathbf R[X]/(X-a_0)\times\mathbf R[X]/(X-a_1)\dotsm\mathbf R[X]/(X-a_n)\\ p\bmod(X-a_0)(X-a_1)\dotsm(X-a_n)&\longmapsto (p(a_0)\bmod X-a_0,\,p(a_1)\bmod X-a_1 ,\dots,p(a_n)\bmod X-a_n) \end{align} This isomorphism means that for each list of $n+1$ values $(\alpha_0,\alpha_1,\dots\alpha_n)$, there exists a polynomial $p$ such that $p(a_i)=\alpha_i$ for each $i=0,1,\dots, n\;$ (surjectivity), and this polynomial is unique up to a (polynomial) multiple of $(X-a_0)(X-a_1)\dotsm(X-a_n)\;$ (injectivity).
Now, as this modulus has degree $n+1$, the canonical representative of the congruence class of $p$, which is the remainder , has degree $\le n$.
Note: The solution does not not necessarily has degree $n$, depending on the relations between the lists $(a_0,, a_1,\dots, a_n)$ and $(\alpha_0,\alpha_1,\dots\alpha_n)$.