Connecting Lagrange Interpolation and the Chinese Remainder Theorem in Principal Ideal Domain

The traditional Lagrange interpolating theorem looks like:

Let $a_1,\dots ,a_n$$a_1, \dots, a_n$ be a family of elements in $\mathbb{R}$$\mathbb{R}$, then there exists a polynomial $f(X)$$f(X)$ that satisfies $f(i)=a_i$$f(i) = a_i$.

As we will see, it is just a particular case of the Chinese Remainder Theorem.

Proof. By CRT we get that

$$\begin{array}{}\text{(1)}& \mathbb{R}[X]\twoheadrightarrow \frac{\mathbb{R}[X]}{(\prod _{i=1}^n(X-i))}\cong \prod _{i=1}^n\frac{\mathbb{R}[X]}{(X-i)}\cong {\mathbb{R}}^n.\end{array}$$

Hence there exists a polynomial $f(X)$$f(X)$ such that $f(X)\equiv a_i\mathrm{mod}(X-i)$$f(X) \equiv a_i \mod (X - i)$ for $i\in [n]$$i \in [n]$. $◻$$\square$

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So let us generalize it. Let $R$$R$ be a PID and consider $a_1\in k(x_1),\dots ,a_n\in k(x_n)$$a_1 \in k(x_1), \dots, a_n \in k(x_n)$, where ${\mathfrak{p}}_i\ne (0)$$\mathfrak{p}_i \neq (0)$.

Then there exists an $f\in R$$f \in R$ that satisfies $f(x_i)=a_i$$f(x_i) = a_i$.

Proof. By CRT we get that

$$\begin{array}{}\text{(2)}& \pi :R\twoheadrightarrow \frac{R}{\bigcap _{i=1}^n{\mathfrak{p}}_{x_i}}\cong \prod _{i=1}^n\frac{R}{{\mathfrak{p}}_{x_i}}\cong \prod _{i=1}^{n}k(x_i).\end{array}$$

Hence there exists an $f\in R$$f \in R$ such that $f(x_i)=a_i$$f(x_i) = a_i$.

To construct it, let $(p_{x_i})={\mathfrak{p}}_{x_i}$$(p_{x_i}) = \mathfrak{p}_{x_i}$. Consider $f_i^{\prime }=\prod _{j\ne i}{p}_{x_j}$$f'_i = \prod_{j \neq i} p_{x_j}$, so that $f_i^{\prime }(x_j)\ne 0⟺j=i$$f'_i(x_j) \neq 0 \iff j = i$. Then we can find $f_i(x_j)={\delta }_{i,j}$$f_i(x_j) = \delta_{i,j}$ by selecting a preimage of $\frac{1}{f_i^{\prime }(x_i)}$$\frac{1}{f'_i(x_i)}$ with respect to $R\twoheadrightarrow k(x_i)$$R \twoheadrightarrow k(x_i)$. Let $a_i^{\prime }$$a'_i$ be a preimage of $a_i$$a_i$ with respect to $R\twoheadrightarrow k(x_i)$$R \twoheadrightarrow k(x_i)$, then

$$\begin{array}{}\text{(3)}& f(x)=\sum _{i=1}^n{a}_i^{\prime }{f}_i(x),\end{array}$$

which satisfies $f(x_i)=a_i$$f(x_i) = a_i$ for all $i$$i$.