0

I've been looking at this proof that every prime $p$ has a primitive root, attributed to Legendre:

If $p=2$ then $g=1$ is a primitive root. Let us assume that $p>2$ is prime and let $n$ be the least universal exponent for $p$, i.e. $n$ is the smallest positive integer such that $x^n \equiv 1 \text{ (mod } p)$, for all non-zero $x\in\mathbb{Z}/p\mathbb{Z}$. Notice that, in particular by the Lemma, there is some element $g\in\mathbb{Z}/p\mathbb{Z}$ such that $g^n\equiv1$ but $g^m\not\equiv1\text{ (mod }p)$ for any $m<n$, i.e. the multiplicative order of $g$ is precisely $n$. Also, notice that by Fermat’s little theorem, $n\leq p−1$.

Now, the polynomial $f(x)=x^n−1$ has at most $n$ roots over the field $\mathbb{Z}/p\mathbb{Z}$ (see this entry), and $f(x)\equiv0\text{ (mod }p)$ for all non-zero $x\text{ (mod }p)$. Thus $n\geq p−1$. Hence, $n=p−1$ and $g$ is of exact order $p−1$, therefore $g$ is a primitive root.

It seems to me that if the conjecture in the question title is true then the double inequality argument could be simplified to a single equality using the Fundamental Theorem of Algebra. Is it?

mjc
  • 2,249
  • 1
    No. For example $x^2 + 1$ doesn't have a root in $\mathbb Z/3\mathbb Z$. – Mathmo123 Jul 03 '22 at 04:43
  • 2
    $\Bbb{Z}/p\Bbb{Z}$ is most definitely not algebraically closed. Two simple ways of seeing that are: 1) Have you ever heard of quadratic non-residues? 2) The polynomial $x^p-x-1$ has no zeros in $\Bbb{Z}_p$ because $a^p-a=0$ for all $a\in\Bbb{Z}_p$. – Jyrki Lahtonen Jul 03 '22 at 04:44
  • 1
    The title question has been answered here. I'm inclined to close this as a duplicate. But I want to hear other opinions given that my dupehammer privilege would not leave any room for differing views. – Jyrki Lahtonen Jul 03 '22 at 04:46
  • @JyrkiLahtonen I've got my answer so it doesn't make much difference to me if the question is closed, but I would suggest that it may have value for the site since the comments and answer here make a simpler presentation of the topic also addressed in the question you link. – mjc Jul 03 '22 at 04:58
  • @QiaochuYuan Already linked by Jyrki. I suspect the answer is 'yes', but I've found the answers here easier to follow. – mjc Jul 03 '22 at 05:07
  • 1
    That comment was automatically generated; it does that whenever you vote to close as a duplicate. – Qiaochu Yuan Jul 03 '22 at 05:41

2 Answers2

3

No finite field can be algebraically closed. If a field has $k$ elements, the polynomial $x^{k + 1} - 1$ has $k + 1$ distinct roots in the algebraic closure. So the field is clearly not algebraically closed.

Mark Saving
  • 31,855
  • 1
    Sorry if I am missing something obvious, but why must the roots be distinct and not instead have some algebraic multiplicity. I agree that it must not be algebraically closed, but isn't Fermat's little theorem the more obvious way of concluding it (at least for prime order fields) – Fishbane Jul 03 '22 at 06:05
  • 1
    Take the derivative of $x^{k + 1} - 1$ and you get $(k + 1) x^k$. Since $k$ is the number of elements of the field, we have that $k + 1 = 1$ in the field. So the derivative is $x^k$, which is comprime with $x^{k + 1} - 1$. This shows there are no multiple roots. – Mark Saving Jul 03 '22 at 06:16
  • @Frisbane But admittedly, an easier way to do it is to enumerate the elements of the field as $a_1, a_2, \ldots, a_n$, and note the polynomial $1 + \prod\limits_{i = 1}^n (x - a_i)$ has no root in the field. I just didn’t think of this method as quickly. – Mark Saving Jul 03 '22 at 06:18
  • Thanks for explaining. I agree that the method of enumerating the elements to get an unsatisfiable polynomial is nicer, I also wouldn't have thought of it until I saw Cpc's answer. – Fishbane Jul 03 '22 at 06:20
3

Before @Jyrki uses his dupehammer, let me just give a proof which is easily found on the internet: An algebraically closed field is infinite. Proof:

Let $|\Bbb F|=k$. So $F=\{a_1,\dots,a_k\}$. Consider $p\in \Bbb F[x]$ given by $p(x)=(x-a_1)\cdots(x-a_k)+1$. $p$ has no root in $\Bbb F$.

That $\Bbb Z_p$ is not closed algebraically can also be seen by considering quadratic nonresidues, for $p>2$. Thus we have polynomials $x^2-a$ with no root for each odd $p$. There are $(p-1)/2$ such $a$.

For instance, in $\Bbb Z_3$, $2$ is not a quadratic residue.

calc ll
  • 8,427
  • 2
    No problem. I'm more worried about high volume veteran answerers feasting on dupes, trying to pick low-hanging fruits. Relatively new users such as yourself, have less of an idea how thoroughly we have covered the basics. Some people think that I worry too much :-) – Jyrki Lahtonen Jul 03 '22 at 09:26