# Having trouble interpreting this Algebraic statement

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.

• What is $\mathbb F$? – mathcounterexamples.net Sep 16 '19 at 13:04
• $\mathbb F$ seems to be an arbitrary field, it’s not explicit. – 7l-l04 Sep 16 '19 at 13:09

$$L_x$$ 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.

• Very clear. Thank you. – 7l-l04 Sep 16 '19 at 13:29
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))