Let us consider different points $z_i=(x_i,y_i)$ in the plane where $i=1,\cdots n$.
Q Do there exist two variable polynomials $P_i(x,y)$ with minimal degree such that $P_i(z_j)=\delta_{ij}$?
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1$\begingroup$ This is clearly impossible unless all the $x_i$ are distinct and all the $y_i$ are distinct, in which case $$P_i(x, y) = \frac{\prod_{j \neq i}(x - x_i)(y - y_i)}{\prod_{j \neq i}(x_j - x_i)(y_j - y_i)}$$ Have I misunderstood something? $\endgroup$– Peter TaylorCommented Jan 25, 2022 at 15:54
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1$\begingroup$ Of course they exist: there exist some polynomials with prescribed values, among them there is a polynomial of minimal degree. Maybe you wanted to ask something different? $\endgroup$– Fedor PetrovCommented Jan 25, 2022 at 17:53
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1 Answer
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To get polynomials $P_i(x,y)$ such that $P_i(z_j)=\delta_{ij}$, you can write $$P_i(x,y)=\frac{\prod\limits_{j:j\ne i}[(x-x_j)^2+(y-y_j)^2]}{\prod\limits_{j:j\ne i}[(x_i-x_j)^2+(y_i-y_j)^2]}.$$
So, polynomials of minimal degree with such an interpolation property do exist.
answered Jan 25, 2022 at 17:49
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$\begingroup$ That is a great idea. How can we sure concerning with degree minimality? $\endgroup$– ABBCommented Jan 26, 2022 at 7:03
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$\begingroup$ @AliBagheri : Your question was "Do there exist two variable polynomials $P_i(x,y)$ with minimal degree such that $P_i(z_j)=\delta_{ij}$?" The answer above shows that the set of all polynomials $P_i(x,y)$ such that $P_i(z_j)=\delta_{ij}$ is nonempty. Selecting a polynomial of the minimal degree in this set of polynomials thus provides an affirmative answer to your question. Of course, one can be interested in an explicit construction of such a minimal-degree polynomial -- but that would be a question different from the one you asked. $\endgroup$ Commented Jan 26, 2022 at 14:28