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I am searching for some reference book for the following theorem in polynomials for citation. Could anybody provide me some details?

$\textbf{Theorem:}$ Let $p$ be a prime, and let $\{(x_1,y_1), \ldots ,(x_{t+1},y_{t+1})\}\subseteq\mathbb{Z}_p\times\mathbb{Z}_p$ to be a set of points whose $x_i$ values are all distinct. Then there is a unique degree-$t$ polynomial $f$ with coefficients from $\mathbb{Z}_p$ that satisfies $y_i \equiv_p f(x_i)$ for all $i$ (I would add to the theorem where $s=f(0)$).

P.S. Is $\mathbb{Z}_p$ notation the same with $GF(p)$?

Arturo Magidin
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Hunger Learn
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    $GF(p)$ is the galois-field with $p$ elements, so the same as $\mathbb Z_p$. In a field we have the property that a poylnomial with degree $n$ can have at most $n$ roots, hence the uniqueness follows easily. For the existence, you can use the Newton-approach or the Lagrange-interpolating-polynomial. Both lead to the unique polynomial doing the job. – Peter Jan 17 '22 at 08:59
  • A reference, among others, is the number theory book by Ireland and Rosen. – Dietrich Burde Jan 17 '22 at 09:10
  • this one https://link.springer.com/book/10.1007/978-1-4757-2103-4 ? – Hunger Learn Jan 17 '22 at 09:11
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    Since Lagrange interpolation is a special case of CRT = Chinese Remainder Theorem it suffices to cite a general proof of CRT, e.g. see see Prop. 1.10 in Atiyah & MacDonald's Intro to Commutative Algebra. – Bill Dubuque Jan 17 '22 at 10:38
  • @BillDubuque you mean that prop $1.10$ from the book is equivalent to the theorem above? – Hunger Learn Jan 17 '22 at 13:12
  • Not equivalent, rather such interpolation is a special case of CRT, e.g. see here and here and here – Bill Dubuque Jan 17 '22 at 13:34
  • @BillDubuque sorry but I do not get the similarities....sorry...it is my problem...but I don't get it...I mean the similarities from the book you propose with the theorem above... – Hunger Learn Jan 17 '22 at 13:43
  • I edited this answer to show how the CRT formula specializes to the Lagrange interpolation formula. – Bill Dubuque Jan 17 '22 at 15:24
  • @BillDubuque Thanks anyway!You are very helpful and your answers are very informative. – Hunger Learn Jan 17 '22 at 16:05
  • As you write it it makes no sense, because the elements of $\mathbb{Z}_p$ (equivalently, $\mathbb{F}_p$) are not ordered pairs. The set should be contained in $\mathbb{Z}_p\times\mathbb{Z}_p$. – Arturo Magidin Jan 17 '22 at 16:53
  • One should be careful with "$\mathbb{Z}_p$", and the notation may mean "the integers modulo $p$" (which in this case is the same as the field with $p$ elements, $GF(p)$ or $\mathbb{F}_p$), but could also mean "the $p$-adic integers". In this case, it plainly means "the integers modulo $p$", but one must be careful and explicit. – Arturo Magidin Jan 17 '22 at 16:55

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