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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 $p\in ℂ[X]$. Does this still hold true for finite fields?

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

First, that degree-Template:Mvar polynomials have at most Template:Mvar roots over any field is reasonably straightforward to show; whenever Template:Mvar is a root, you can factor out Template:Math. In fact, this much remains true for polynomials over any integral domain.

Template:PbThe 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 Template:Mvar 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, $X^2\equiv 3(\mathrm{mod}4)$ 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)

Template:Ping 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)