Wikipedia:Reference desk/Archives/Mathematics/2020 October 22

= October 22 =

Number of roots of polinomial in fields other than C
Hi,

if I am not mistaken, the fundermental theorem of Algebra states that there are at most $$deg(p)$$ roots of a polinomial $$p \in \mathbb {C}[X] $$. Does this still hold true for finite fields?

Thanks TheFibonacciEffect (talk) 16:04, 22 October 2020 (UTC)


 * First, that degree-$n$ polynomials have at most $n$ roots over any field is reasonably straightforward to show; whenever $a$ is a root, you can factor out $(x &minus; a)$. In fact, this much remains true for polynomials over any integral domain.
 * The FToA says that every complex polynomial has at least one complex root. Combining this with the above observation, it follows that every such polynomial has exactly $n$ roots (counted with multiplicity).  However, finite fields are not algebraically closed, so there exist polynomials over them with no roots.  In fact, finite fields will always have degree-2 polynonmials without roots; this can shown by a counting argument. –Deacon Vorbis (carbon &bull; videos) 16:25, 22 October 2020 (UTC)
 * For a simple example, $$X^2\equiv~3\!\!\!\pmod 4$$ has no integer solution, i.e. $$X^2-3$$ in $$\mathbb{Z}_4$$ has no root. Tigraan Click here to contact me 08:50, 23 October 2020 (UTC)
 * Thank you
 * - TheFibonacciEffect (talk) 11:14, 23 October 2020 (UTC)
 * But $$\mathbb{Z}_4$$ is not a field, no? Double sharp (talk) 13:49, 26 October 2020 (UTC)
 * I meant to say something here but forgot. No, it's not (it's not even an integral domain, which is equivalent to not being a field for finite objects anyway).  And the bit about having no more roots than degree can fail when you're not in an integral domain. For example, the polynomial $$p(x) = 2x^2 + 2x$$ has 4 roots in $$\mathbb{Z}/4.$$ –Deacon Vorbis (carbon &bull; videos) 14:16, 26 October 2020 (UTC)


 * In $$\mathbb{Z}_3$$, the polynomial $$X^2+1$$ has no roots. --Lambiam 21:37, 26 October 2020 (UTC)


 * Experimentally, it appears that the polynomial $$X^{n{-}1}{+}1,$$ $$n>1,$$ has roots in $$\mathbb{Z}/n$$ iff its degree $$n{-}1$$ is odd. In the latter case we always have the root $$X=n{-}1,$$ and usually this is the only root. But sometimes there are more: for example, $$X^{27}{+}1$$ has 3 roots in $$\mathbb{Z}/28$$ ($$X=3,$$ $$X=19,$$ $$X=27$$), $$X^{741}{+}1$$ has 39 roots in $$\mathbb{Z}/742, $$ and $$X^{945}{+}1$$ has 105 roots in $$\mathbb{Z}/946.$$ I see no obvious pattern, but no doubt someone has studied this. --Lambiam 08:05, 27 October 2020 (UTC)


 * If $$p$$ is an odd prime, the absence of roots of $$X^{p{-}1}{+}1$$ in $$\mathbb{Z}/p$$ follows from Fermat's little theorem. For composite odd moduli, their absence remains unexplained. --Lambiam 17:50, 28 October 2020 (UTC)
 * The Chinese remainder theorem covers the case of products of distinct odd primes.--Jasper Deng (talk) 03:53, 29 October 2020 (UTC)