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Let's consider the polynomial $m(x)$ over a field $\mathbb{Z}_{3}$. We know that $[m(x)]_{m(x)}=m(a)=0$.

Now $m(x)=x^{3}+1$; in my lectures slide, it's said at this point that:

$m(a)=0$ implies that $a^{3}=2$.

What this means ?

Moreover, in the slides about the modulo arithmetic with polynomials it's said that, when we work with polynomial whose coefficients are modulo n then the negative coefficients equals to the positive ones. But why ?

AleWolf
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1 Answers1

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We are working in modulo 3, so if $m(a) = 0$, we have $a^3 \equiv -1 \equiv 2 \pmod 3$. So this means $a^3 = 2$.

auscrypt
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  • Wait, we are working in modulo m(x).... not modulo 3.... – AleWolf May 23 '19 at 17:12
  • @AleQuercia Yes, but m(x) is defined over the field $\mathbb{Z}_3$, so we can still use such equivalences. – auscrypt May 23 '19 at 17:14
  • I would like to study this in a book, which are correct the keywords for these arguments ? So i can search on a volume – AleWolf May 23 '19 at 17:17
  • @AleQuercia I'm not quite sure what keywords would be used for these arguments to search. If you want justification, $\mathbb{Z}_3$ is a subfield of the field $\mathbb{Z_3}[x]/m(x)$. – auscrypt May 23 '19 at 17:20
  • Does this mean that we can work modulo m(x) and modulo 3 simultaneously ? :S – AleWolf May 23 '19 at 17:23
  • @AleQuercia Yes, modulo 3 is part of working in modulo m(x) since m(x) is already taken modulo 3. – auscrypt May 23 '19 at 17:24
  • Thanks so much, i've understood everything except one thing. Why can we just simplify modulo 3 the coefficients of the polynomials, what assures us that we don't mess up things doing so ? – AleWolf May 23 '19 at 17:38
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    The coefficients are elements of $\mathbb{Z}_3$. They are equivalence classes modulo $3$. You aren't "simlifying modulo $3$ the coefficients". The coefficients are taken modulo 3. – Arturo Magidin May 23 '19 at 17:44