Talk:Lagrange's Theorem (Number Theory)

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Generalize $\Z_p$ to integral domains?

To cite [1]:

Integral domains have no zero divisors, and consequently allow one to solve equations of
the form $x(y − z) = 0$ in the way one expects: either $x = 0$ or else $y = z$. (This is precisely
the cancellation law for multiplication discussed earlier.
Theorem 1.7. Let $p(X)$ be a polynomial over an integral domain $R$ with degree $n > 0$.
Then the polynomial equation $p(x) = 0$ has at most $n = \deg(p(X))$ roots in $R$.

The proof is the same as Lagrange's Theorem (Number Theory) for $\Z_p$. --Hbghlyj (talk) 23:37, 17 March 2024 (UTC)

Feel free to write another page for this. Please do not amend the statement of the theorem of this page, of course, for obvious reasons. --prime mover (talk) 20:25, 18 March 2024 (UTC)
Though the proof is almost the same, it will need Polynomial Factor Theorem for integral domains. But the current Polynomial Factor Theorem is only for fields. So we need to write another page of Polynomial Factor Theorem for integral domains first. --Hbghlyj (talk) 21:54, 18 March 2024 (UTC)
That is how this site works. One finds the dissertation of what one wants to write about in the literature. One then uses that source work by building up the material from first principles for that work.
Then, when one has worked through the book to the level of the result in it, that piece of work is ready to be presented.
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