# Algebraic Closure of $\mathbf F_p$ [Lang, Algebra, Chapter 6, Problem 22]

Problem. Let $K$ be the field obtained from $\mathbf F_p$ by adjoining all primitive $\ell$-th roots of unity for primes $\ell\neq p$. Then $K$ is algebraically closed.

It suffices to show that the polynomial $x^{p^n}-x$ splits in $K$ for all $n$. In order to show this, it in turn suffices to show that the polynomial $x^{q^n}-1$ splits in $K$ for all primes $q\neq p$ and all $n$. This is because $x^{p^n}-1= x(x^{p^n-1}-1)$. Say $p^n-1=p_1^{k_1} \cdots p_m^{k_m}$, where $p_i$'s are distinct primes. Assuming each $f_i(x):=x^{p_1^{k_i}}-1$ splits in $K$, we deduce that $K$ has a primitive $p_i^{k_i}$-th root of unity for all $1\leq i\leq m$ since each $f_i$ is separable by the derivative test. If $\zeta_i$ denotes the primitive $p_i^{k_i}$-th root of unity in $K$, then we see that $\zeta_1\times \cdots\times \zeta_m$ is a primitive $p_q^{k_1}\times \cdots \times p_m^{k_m}$-th root of unity and we see that $x^{p^n-1}-1$ splits in $K$.

So the problem boils down to showing that $x^{q^n}-1$ splits in $K$ for all primes $q\neq p$ and all $n$.

I am stuck here.

• A cute question that I don't recall seeing before! The mechanism is surely to use suitable larger primes, or combinations of primes. For example in the case $p=2$ the first problem we have is getting a primitive ninth root of unity $\zeta$ to $K$. The extension $L=\bf{F}_2(\zeta)$ is of degree six. It is not itself generated by any root of unity of prime order, but we get $L$ by adjoining roots of order seven and three as those give cubic and quadratic extensions respectively. Alternatively we can use a root of unity of order $13$, because that generates the field $\bf{F}_{2^{12}}\supseteq L$. Commented Oct 1, 2016 at 12:13
• @JyrkiLahtonen, sorry but why in this problem it suffices to prove that $x^{p^n}-x$ splits?
– RFZ
Commented Apr 27, 2019 at 2:35
• @ZFR $x^{p^n}-x$ is the product of all the irredeucible polynomials of degree $d$, $d\mid n$. If $x^{p^n}-x$ splits, so do all its factors. If $x^{p^n}-x$ splits for all $n$, so do all the irreducible polynomials, and we are done. Commented Apr 27, 2019 at 2:51
• @JyrkiLahtonen, thank a lot for reply! But one moment confuses me: we consider irreducible polynomial $p(x)\in K[x]$. But $K$ is not finite field, right? So the above statement may not be true. I know that $x^{p^n}-x$ is the product of all irreducible monic polynomials with degree $d$, $d\mid n$. But I know that this result is true in finite field. However, in our case $K$ is not finite field. Could you clarify it, please?
– RFZ
Commented Apr 27, 2019 at 18:17
• @ZFR Hmm. I should have also explained the following. Assume that $p(x)$ is a polynomial over $K$. All the elements of $K$ are algebraic over the prime field $\Bbb{F}_p$. Therefore the zeros of $p(x)$ (possibly in some extension field of $K$) are also algebraic over $\Bbb{F}_p$. Therefore the zeros of $p(x)$ belong to some finite field. Therefore their minimal polynomials over $\Bbb{F}_p$ are factors of some $x^{p^n}-x$. So if all those polynomials split in $K$, $K$ must be algebraically closed. Commented Apr 28, 2019 at 20:50

The idea is that given any prime power $$q^k$$, we may take a prime $$w$$ such that $$w$$ divides $$p^{q^k} - 1$$ but does not divide $$p^{q^{k-1}} - 1$$, in other words, such that $$p$$ has order $$q^k$$ modulo $$w$$. First, assuming the existence of such a prime $$w$$, we observe that $$\mathbf F_p(\zeta_w)$$ is the finite field with $$p^{q^k}$$ elements, so it is the splitting field of $$X^{p^{q^k}} - X$$ over $$\mathbf F_p$$. Now, we proceed with the argument.

To see that such a prime $$w$$ exists, we use the polynomial identity

$$\frac{(1 + X)^q - 1}{X} = \sum_{k=0}^{q-1} C(q, k+1) X^{k}$$

and write

$$a = \frac{p^{q^k} - 1}{p^{q^{k-1}} - 1} = \sum_{j=0}^{q-1} C(q, j+1) (p^{q^{k-1}} - 1)^j$$

Clearly, we have $$a > q$$. On the other hand, if a prime $$v$$ divides both $$a$$ and the denominator, it must also divide $$q$$ by the sum on the right hand side, and since $$q$$ is prime we must have $$v = q$$. However, in that case $$q^2$$ cannot divide $$a$$, so $$a$$ has a prime factor $$w \neq q$$. Since $$w$$ cannot be a divisor of the denominator, it is the desired prime number.

• Would you mind explaining why $q^{2}$ cannot divide $a$ can imply there is a prime number $w\neq q$ which can divide $a$? Commented Apr 24, 2018 at 21:24
• Dear, Starfall! Let me ask you the following question: You've shown that $\mathbb{F}_p(\zeta_w)$ is the finite field with $p^{q^k}$ hence it is isomorphic to $\mathbb{F}_{p^{q^k}}$ and so it is the splitting field of $x^{p^{q^k}}-x$ but not $x^{q^k}-x$ as you wrote above, right? Hence it does not give full solution, right?
– RFZ
Commented Apr 27, 2019 at 17:20