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I’m struggling to interpret the meaning of the following statement.

Let the multiplicative subgroup $H \subset \mathbb{F} $ have order $n+1$, and let $x\in H$. The polynomial $L_x$ of degree at most $n$ that vanishes on $H$ \ $\{x\}$ and has $f(x)=1$, with some constant $c_x$ has representation of the form $$L_x(X) = \frac{c_x(X^{n+1}-1)}{X-x}$$

Although I feel uncertain where exactly my confusion lies, I don’t believe I understand what is meant by “$L_x$ vanishes on $H$ \ $\{x\}$,” and by extension, I don’t follow that $L_x$ has that representation.

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  • $\begingroup$ What is $\mathbb F$? $\endgroup$ – mathcounterexamples.net Sep 16 '19 at 13:04
  • $\begingroup$ $\mathbb F$ seems to be an arbitrary field, it’s not explicit. $\endgroup$ – 7l-l04 Sep 16 '19 at 13:09
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$L_x$ vanishes on $H \setminus \{x\}$ means that for all $y \in H \setminus \{x\}$, $L_x(y) = 0$.

As $\mathbb F$ is supposed to be a field, if we denote $\{x, y_1, \dots,y_n\}$ the elements of $H$, $(X-y_1) \dots(X-y_n)$ divides $L_x$ as $L_x$ vanishes on $H \setminus \{x\}$. Therefore the degree of $L_x$ is at least equal to $n$.

Now, $\frac{(X^{n+1}-1)}{X-x}$ also vanishes on $H \setminus\{x\}$ as the order of all elements of $H \setminus\{x\}$ divides the order of $H$, i.e. $n+1$.

Hence, it exists $c_x \in \mathbb F$ such that

$$L_x(X) = \frac{c_x(X^{n+1}-1)}{X-x}$$ as desired.

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  • $\begingroup$ Very clear. Thank you. $\endgroup$ – 7l-l04 Sep 16 '19 at 13:29
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Using Lagrange's theorem both monic polynomials ∏_{r∈H}(X-r) and Xⁿ⁺¹-1 have degree n+1 and H as their exact set of roots. Therefore Xⁿ⁺¹-1=∏_{r∈H}(X-r). Similarly, both L_{x}(X) and ∏_{r∈H/{x}}(X-r) have degree n and H/{x} as their exact set of roots yielding

L_{x}(X)=c_{x}⋅∏_{r∈H/{x}}(X-r)=c_{x}⋅((Xⁿ⁺¹-1)/(X-x))

Moreover,

1  = L_{x}(x)=c_{x}⋅(d/(dX))(Xⁿ⁺¹-1)|_{X=x}=c_{x}(n+1)⋅Xⁿ|_{X=x}
 = c_{x}(n+1)⋅xⁿ
 ∴ c_{x}=(1/((n+1)xⁿ))
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