Understanding polynomial interpolation as a special case of Chinese Remainder Theorem As some background information:
Chinese Remainder Theorem: Let $R$ be a commutative ring and $\{I_k\}_{k\geq1}$ be pairwise coprime ideals in $R$. Chinese Remainder Theorem claims that $R/I_1\cap...\cap I_k\cong (R/I_1) \times...\times(R/I_k).$
Polynomial interpolation: claims that for pairwise distinct $\lambda_k$, $k=1,2,...,n$, $\lambda_i\in\mathbb{F}$ for some field. Then for $a_i$, $i=1,...,n,$ there exists a polynomial $p\in \mathbb{F}[x]$ such that $p(\lambda_i)=a_i$ for each i.
The proof I was given goes as follows:
Since $\lambda_k$ are pairwise distinct then $\langle X-\lambda_k\rangle$ are pairwise coprime. So we can apply CRT here to obtain: $\mathbb{F}[x]/\langle X-\lambda_1\rangle \cap...\cap \langle X-\lambda_n\rangle \cong (\mathbb{F}[x]/\langle X-\lambda_1\rangle) \times...\times(\mathbb{F}[x]/\langle X-\lambda_n\rangle).$ Then there is a polynomial $p(X)\in \mathbb{F}[x]/\langle X-\lambda_1\rangle\cap...\cap \langle X-\lambda_n\rangle$ such that p(X)$\in$ $a_i$+$\langle X-\lambda_i\rangle $ for each $i$.
My issue here lies within the last sentence (Italic one).
Firstly why does it even imply there exists $p(X)$ that has the declared property ($p(X)\in$ $a_i$+$\langle X-\lambda_i\rangle $ for each $i$)? Secondly why would such $p(X)$ has the required property in the theorem? Is it really as simple as $p(\lambda_k)\in a_i+\langle \lambda_i-\lambda_i\rangle=a_i?$
I hope I didnt explain this too messily and any help is much appreciated!
 A: Clearer by CRT, which implies the evaluation map $\, f(x)\mapsto (f(\lambda_1),\ldots,f(\lambda_n))$ from $\,F[x]$ to $F^n$ is surjective (= onto), with kernel $\,I={\large  \cap_i}(x-\lambda_i)\,[=\prod_i (x-\lambda_i)$ by here or here]. Being onto, there is some $\,f(x)\in F[x]\,$ which maps to $\,(a_1,\ldots,a_n),\,$ i.e $\,f(\lambda_i) = a_i,\,$ which is unique $\!\bmod I.\,$ Below shows how $\,\scr L = $ Lagrange interpolation is a special case of CRT.
$$\begin{align}f(\lambda_1)= a_1\!\!\!\overset{\rm\color{#0a0}{RT}\!\!\!}\iff\!  f(x)&\equiv a_1\!\!\!\!\pmod{\!x \!-\! \lambda_1}\\
\vdots\qquad\qquad\qquad &\ \ \vdots\qquad\qquad \vdots\\
f(\lambda_n) = a_n\smash{\overset{\rm\color{#0a0}{RT}\!}
\iff}\, f(x)&\equiv a_n\!\!\!\!\pmod{\!x \!-\!\lambda_n}\end{align}
\!\!\!\!\overset{\rm\small CRT\!}\iff\! f(x)\equiv {\scr L}^{a}_{\lambda}(x)\!\!\pmod{\!{\smash{\small \prod_i}} (x\!-\!\lambda_i)}$$
where we used  $\ {\rm\color{#0a0}{RT}}\!:\,\ f(\lambda) =  (f(x)\,\bmod\, x-\lambda)\,\ $  [Polynomial Remainder Theorem].
Remark $ $ Further, the Lagrange interpolation formula is a special case of a CRT formula generalizing the integer CRT formula to pairwise coprime moduli $p_i$ in $F[x]$ (or any PID), viz.
$\qquad \begin{align} &f \equiv a_1\pmod{p_1}\\
          &\ \ \ \ \vdots\qquad\qquad\ \ \vdots \\
          &f \equiv a_k\pmod{p_k}\end{align}$
$\iff \left\{\begin{align} f \,\equiv\, &\sum_i\, a_i\:\! P_i\:\! (\color{#c00}{P_i^{-1}\bmod p_i})\!\pmod{\!P}\\  &\ {\rm for}\ \ P =  p_1\cdots p_k,\ \ P_i = P/p_i\end{align}\right.$
Here we have that: $\ p_i = x\!-\!\lambda_i,\, $ $\, P_i = (x\!-\!\lambda_1)\cdots \rlap{\ \ ////}(x\!-\!\lambda_i)\cdots (x\!-\!\lambda_k)$
so $\,\color{#c00}{P_i^{-1}\bmod p_i} = \dfrac{1}{P_i}\bmod  x\!-\!\lambda_i  = \dfrac{1}{(\!\lambda_i\!-\!\lambda_1)\cdots \rlap{\ \ ////}(\lambda_i\!-\!\lambda_i)\cdots (\lambda_i\!-\!\lambda_k)} $
$${\rm so}\ \ f\, \equiv\, \sum_i a_i\,\dfrac{(x-\!\lambda_1)\cdots\:\!\rlap{\ \ ////}(x-\!\lambda_i)\cdots\:\! (x-\!\lambda_k)\ }{(\!\lambda_i\!-\!\lambda_1)\cdots \rlap{\ \ ////}(\lambda_i\!-\!\lambda_i)\cdots (\lambda_i\!-\!\lambda_k)}\ \ \ \ $$
so the above CRT formula specializes to the Lagrange interpolation formula (where again we used  ${\rm\color{#0a0}{RT}}$ = Poly Remainder Theorem to deduce $\,P_i\bmod x\!-\!\lambda_i\,$ equals $P_i$ evaluated at $\,x = \lambda_i)$.
A: For the second question: it's almost that simple, though we can't write $p(\lambda_i) \in a_i+\langle \lambda_i-\lambda_i\rangle$, as this doesn't make sense.
Rather, write it in this form: 
$$p(X)=q(X)(X-\lambda_i) +a_i$$
to see $p(\lambda_i) =a_i$.
For the first question, take the unique correspondent of 
$$(a_1+\langle X-\lambda_1\rangle, \ \dots,\ a_n+\langle X-\lambda_n\rangle) \ \in\ \prod_i\Bbb F[X]/\langle X-\lambda_i\rangle$$
along the given isomorphism. 
