Vtyjkyuk

Let $mathbb F_3$ be the field with 3 elements and $overlinemathbb F_3 $ its algebraic closure. Let $K$ be the splitting field of $f(x)=x^21-1$ (I guess over $mathbb F_3$). Find the number of zeros of $f$ in $overlinemathbb F_3 $ and the number of elements in $K$.

There are 3 questions below in different parts of the text, they are in italics.

First of all, I don't understand why would anyone ask about the number of zeros of $f$ in the closure, since it is always equal to the degree of the polynomial. What am I supposed to answer here?

For the second question. Since $mathbb F_3$ has characteristic 3, $x^21-1=(x^7-1)^3$. The set of roots of $f$ is the set of roots of $g=x^7-1$. So the splitting field of $f$ is the same as that of $g$. Now let $a$ be a root of $g$ in the algebraic closure of $mathbb F_3$. Consider the extension $mathbb F_3(a)$. Now

1. either $a=1$ in $mathbb F_3(a)$ or


2. the order of $a$ in the multiplicative group of the extension $mathbb F_3(a)$ is $7$.

Note that $|mathbb F_3(a)|=3^n$ since $mathbb F_3(a)$ is a vector space over $mathbb F_3$. In the second case above, $3^n-1$ must be divisible by $7$, whence $nge 6$. What to do with the first case?

Consider $F=mathbb F_3^6$. There is an element $a$ of order $7$ in the multiplicative group of that field. Then $a^2,dots,a^6$ also have order $7$. So they all are roots of $g$ (or equivalently $f$). Thus $F$ contains all roots of $f$. How to show it is the minimal extension containing the roots of $f$?

Since in $mathbbF_3$ we have $x^21-1=(x^7-1)^3$, $x^21-1$ has seven roots in $overlinemathbbF_3$, all of them with multiplicity $3$. Indeed $x^7-1 = (x-1)Phi_7(x)$ and the degree of the splitting field $mathbbK$ of $Phi_7(x)$ over $mathbbF_3$ is given by the least $d$ such that $7mid(3^d-1)$, i.e. $d=6$, since $3$ is a generator for $mathbbZ/(7mathbbZ)^*$.

$$ mathbbK=mathbbF_3[x]/(Phi_7(x)) simeq mathbbF_3^6$$
is a finite field with $3^6=729$ elements.

As you've said the number of zeroes equals the degree of the polynomial. However I suspect that the question asks for the number of distinct zeroes of maybe the number of zeroes in F.

For the second part note that $1$ is one of the roots. Also there are not other roots in $mathbbF_3$. So let $a$ be such a root in $overlinemathbbF_3$. This means that you don't have to split the problem in two cases, as you will have to adjoin $a not = 1$ to get the splitting field.

Also your idea, which was mentioned in the link works fine, as the order of $3$ modulo $7$ is $6$. Another way (if you're familair with Galois Theory) is to notice that the maps defined by $a to a^3^i$ for $i=0,1,cdots 5$ are distinct automorphisms of $mathbbF_3(a)$. This would mean that the Galois group of the extension $mathbbF_3 subset mathbbF_3(a)$ has order at least $6$ and thus the extension degree is at least $6$. On the other side the extension has degree at most $6$, as $a$ is a root of $x^6+x^5+x^4+x^3+x^2+x+1$. This can help you conclude that the extension degree is $6$ and the field has $3^6$ elements.

Question 1.

The question is asking for the number of distinct roots.

Question 2.

In the case that $a=1$, $mathbbF_3(a)$ is not the splitting field of $g$ over $mathbbF_3$.

(1) The question is unclear, the point being maybe the fact that there are some multiple roots, and the question wants then without multiplicity. The question is answered if we write $x^21-1 = (x^7-1)^3$, and the last polynomial $f=x^7-1$ has distinct roots. (The question may be seen as a help, as a hint.)

(2) The root $1$ already lives in the prime field. There is nothing to be done. Let $a$ be a root of $g=x^6+x^5+x^4+x^3+x^2+x+1$. This polynomial is irreducible.

sage: F = GF(3)
sage: R.<x> = PolynomialRing(F)
sage: f = x^21 - 1
sage: f.factor()
(x + 2)^3 * (x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)^3
sage: g = (x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)
sage: g.factor()
x^6 + x^5 + x^4 + x^3 + x^2 + x + 1

The above sage code confirms this. Else, $a$ would be a root of a polynomial of degree $k$ among $ 2,3,4,5$, so would satisfy $a^3^k=a$. We get a contradiction as in the OP.

(3) The third question is clear, a lives in $Bbb F_3^6$, but in no subfield. And there is the following splitting of $g$ over this bigger field:

sage: K.<a> = GF(3^6, modulus=g)
sage: g.base_extend(K).factor()
(x + 2*a) * (x + 2*a^2) * (x + 2*a^3) * (x + 2*a^4) 
* (x + a^5 + a^4 + a^3 + a^2 + a + 1) * (x + 2*a^5)

In other words, as expected,
$$g=prod_1le kle 6(x-a^k) .$$

An other argumentation would be given by the observation that $x^7-1$ divides $x^3^6-x$, the last polynomial being the product of all $(x-s)$, $s$ in $Bbb F_3^6$. (Because $7$ divides $3^6-1=728$, or Fermat.)