Algebraic Closure of Fp [Lang, Algebra, Chapter 6, Problem 22]

[caffeinemachine]

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.

 

[jvicens]

Could someone help me? I would suggest approaching the problem in a systematic and logical way. First, we need to understand the field $K$ and its properties. It is clear from the definition that $K$ is a field extension of $\mathbf{F}_p$ obtained by adjoining all primitive $\ell$-th roots of unity for primes $\ell \neq p$. This means that $K$ contains all the $\ell$-th roots of unity for $\ell \neq p$, which implies that it is a Galois extension of $\mathbf{F}_p$. Furthermore, since $\mathbf{F}_p$ is a finite field, we know that $K$ is also a finite field.

Next, we need to understand the polynomial $x^{q^n}-1$ and its roots in $K$. Since $K$ contains all the primitive $\ell$-th roots of unity for $\ell \neq p$, it follows that $K$ also contains all the $q^n$-th roots of unity. This means that all the roots of $x^{q^n}-1$ are contained in $K$. Therefore, $x^{q^n}-1$ splits in $K$.

Now, we can use induction to show that $x^{p^n}-x$ splits in $K$ for all $n$. The base case $n=1$ is already established above. For the induction step, assume that $x^{p^n}-x$ splits in $K$. We want to show that $x^{p^{n+1}}-x$ also splits in $K$. Note that $x^{p^{n+1}}-x = (x^{p^n}-x)^p$. By our assumption, $x^{p^n}-x$ splits in $K$, so its roots are contained in $K$. Since $K$ is a field, this means that $(x^{p^n}-x)^p$ also splits in $K$. Therefore, $x^{p^{n+1}}-x$ splits in $K$, completing the induction step.

In conclusion, we have shown that $x^{p^n}-x$ splits in $K$ for all $n$, which implies that $K$ is algebraically closed.