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Consider a finite field $F$, let $p_n(x)=(x-x_1)(x-x_2)\cdots(x-x_n)$ and a polynomial $p_{n-1}(x)$ of order $n-1$ such that $p_n(x)=(x-x_1)p_{n-1}(x)$.

I wonder whether is it necessary that $p_{n-1}(x)=(x-x_2)(x-x_3)\cdots(x-x_n)$? It is clear that this holds for an infinite field, but I am not familiar to the finite field. Could anyone help or provide reference?

emacs drives me nuts
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stephenkk
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  • Factorization of polynomials into irreducibles is unique up to constant factors over any field, finite or not. Linear polynomials are obviously irreducible. The claim follows. Why do you think there would be a difference between finite/infinite here? – Jyrki Lahtonen Feb 13 '20 at 09:00
  • @JyrkiLahtonen, since we can get explicitly the coefficients by Lagrangian interpolation if there are enough points. – stephenkk Feb 13 '20 at 09:16
  • @JyrkiLahtonen another point is that a higher order polynomial on $F_2$ can be indeed a zero order one, e.g., $x^2+x=0$ for any $x$ in $F_2$, so even if two polynomials are the same their order may be different. – stephenkk Feb 13 '20 at 09:34
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    Your latter response is telling. You seem to be confused by the difference between a polynomial and a polynomial function. Study this and other threads linked to it. Over $\Bbb{F}_2$ the polynomial $x^2+x$ has degree two even though its value is zero at both elements of $\Bbb{F}_2$. In algebra we think of polynomials as finite formal sums like $\sum_i a_ix^i$ with $a_i$ coming from a field $F$ (or occasionally a ring). But that $x$ is an indeterminate. – Jyrki Lahtonen Feb 13 '20 at 10:03
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    (cont'd) We do not think of such a polynomial as a function from $F$ to $F$ gotten by "plugging in all the elements of $F$ into $x$". Two polynomials $f(x)=\sum_i a_ix^i$ and $g(x)=\sum_i b_ix^i$ are then defined to be equal if and only if $a_i=b_i$ for all $i$. No attention is necessarily given to the functions from $F$ to itself defined by $a\mapsto f(a)$, $a\mapsto g(a)$. Over a finite field those functions may be equal for different polynomials $f,g$. – Jyrki Lahtonen Feb 13 '20 at 10:07
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    Mind you, it takes a while to get used to this subtlety. Largely because we all first met polynomials over $\Bbb{R}$, an infinite field. Over an infinite field there isn't much difference between the formal point of view and the function point of view (for example by Lagrange interpolation as you observed). – Jyrki Lahtonen Feb 13 '20 at 10:09
  • Thanks for your response and patience, I have understood your point clearly. – stephenkk Feb 13 '20 at 10:26
  • Glad to hear thar. – Jyrki Lahtonen Feb 13 '20 at 12:29

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