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

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

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 − 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 • 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)

@Tigraan: 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 • 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)