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Conclude that the difference of the polynomials from

  • $X^{\phi{(n)}}-[1]$
  • $\prod_{[a] \epsilon (Z /_n Z)^x}(X-[a])$

is a polynomial of degree strictly less than $\phi{(n)}$ and that it has $\phi{(n)}$ distinct roots.Can you also conclude also that $X^{\phi{(n)}}-[1]=\prod_{[a] \epsilon (Z /_n Z)^x}(X-[a])$?

I say yes but how can I prove it? Have already proven that each of the polynomials above have exactly $\phi{(n)}$ roots and the elements belong to $(Z/_n Z)^x$.

Username Unknown
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    Keep in mind that $x^2 - 1$ has four roots modulo $8$, despite being a nonzero polynomial of degree strictly less than 4, as a counterexample to the "obvious" rationale. –  Jun 07 '14 at 01:04
  • Aside: \times gives $\times$, which is good for specifying unit groups if you don't want to use $*$. –  Jun 07 '14 at 01:07
  • @Hurkyl Can I guess that a=1? – Username Unknown Jun 07 '14 at 01:10
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    this is false for $n=8$, the second polynomial is $X^4 + 6X^2 + 1$ and the first is $X^4 + 7$. – mercio Jun 12 '14 at 10:50

1 Answers1

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For $n=8$ you get some funny business don't you? $x^4 + 6x^2 + 1$ and $x^4 + 7$ both vanish at the elements of $(\mathbb{Z}/8\mathbb{Z})^\times$.

In fact, $\boxed{ x^2 + 7=0}$ for $x \in \{ 1,3,5,7\}$.

$$ \begin{array}{c|r|c|c} x & x^2 & 7 & x^2 + 7\\ \hline 1 & 1=1 & 7 & 0 \\ 3 & 9=1 & 7 & 0\\ 5 & 25=1 & 7 & 0\\ 7 & 49=1 & 7 & 0 \end{array} $$

cactus314
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