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In Dan Boneh's PLONK Video - https://www.youtube.com/watch?v=vxyoPM2m7Yg he refers to the Prod Check Gadget

$\omega \in F_p$ is a primitive $k$-th root of unity ($\omega^{k-1} = 1$)

$H = \{1, \omega, \omega^2, ..., \omega^{k-1}\} \subseteq F_p$

Prod Check Gadget is used to prove that $\prod_{a \in H} f(a) = c$

I was trying to construct a polynomial $f(x)$ which there is true.

I used $p=17$ i.e $f \in F_{17}$

$\omega = 4$ is the primitive $4$th root of unity in $F_{17}$.

So $H = \{1, 4, 16, 13\}$

Let $c = 5$

Now, I try to find a polynomial $f$ such that $f(a) = 5$ for all $a \in H$ using Lagrange Polynomial Interpolation.

sage: F17 = GF(17)
sage: R17.<x> = PolynomialRing(F17)
sage: w = F17(4)
sage: c = F17(5)
sage: points = [(w^0, c), (w, c), (w^2, c), (w^3, c)]
sage: R17.lagrange_polynomial(points)
5

So Lagrange Interpolation gives returns $f$ as the trivial polynomial $f(x) = 5$

Any value of $c$ returns the polynomial as $f(x) = c$

So I am unable figure what is the point of this gadget?

Or am I doing something wrong in my toy example?

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ProdCheck gadget is used for checking the product of values of the polynomial over a multiplicative subgroup of the field is equal to some constant or not.

In your toy example, you want to find a polynomial $f$ such that $f(a)=5$ for all $a\in H$. But Prodcheck gadget is for checking $\prod\limits_{a\in H}f(a)=c$ or not.

For the Lagrange polynomial computation over $F_{17}$ , sage outputs $5$ because I think it will be the constant polynomial $f(x)=5$

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  • $\begingroup$ You are right. Brain fade moment for me! However, Is there a way, I can create a polynomial where $\prod\limits_{a\in H}f(a)=c$ in $F_{17}$ with $w$ as the primitive 4th root of unity $\endgroup$
    – user93353
    Apr 6, 2023 at 11:41
  • $\begingroup$ You can construct a polynomial $f$ like this: choose $c_0,c_1,c_2,c_3$ such that $c=c_0c_1c_2c_3$ and construct $f$ using Lagrange interpolation such that $f(w^i)=c_i.$ $\endgroup$
    – Manas Jana
    Apr 6, 2023 at 12:20

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