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.

  • 2
    $\begingroup$ 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 $\bf{F}_{2^{12}}\supseteq L$. $\endgroup$ Commented Oct 1, 2016 at 12:13
  • $\begingroup$ @JyrkiLahtonen, sorry but why in this problem it suffices to prove that $x^{p^n}-x$ splits? $\endgroup$
    – RFZ
    Commented Apr 27, 2019 at 2:35
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    $\begingroup$ @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. $\endgroup$ Commented Apr 27, 2019 at 2:51
  • $\begingroup$ @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? $\endgroup$
    – RFZ
    Commented Apr 27, 2019 at 18:17
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    $\begingroup$ @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. $\endgroup$ Commented Apr 28, 2019 at 20:50

1 Answer 1


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.

  • $\begingroup$ Would you mind explaining why $q^{2}$ cannot divide $a$ can imply there is a prime number $w\neq q$ which can divide $a$? $\endgroup$ Commented Apr 24, 2018 at 21:24
  • $\begingroup$ 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? $\endgroup$
    – RFZ
    Commented Apr 27, 2019 at 17:20

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