Exercise 11.1.10

Proof. (a) With the following Sage instructions:

sage: R.<x> = PolynomialRing(GF(3))
sage: f = -1-x-x^2-x^3-x^4-x^5-x^6-x^7+x^8+x^9+x^10
sage: gcd(f,x^(3^3) -x)
x^3 + 2*x^2 + x + 1
sage: gcd(f,x^(3^7) -x)
x^7 + 2*x^6 + 2*x^5 + x^4 + 2*x^3 + x^2 + 2
sage: gcd(f,x^3-x)
1

we know that $f$ has 3 roots in ${𝔽}_{3^3}$, 7 roots in ${𝔽}_{3^7}$ (and no root in ${𝔽}_3$).

(The degree of $x^{3^{21}}-x$ is $10,460,353,203$. Too big for my computer!) (b) We take all the finite field ${𝔽}_{p^{\nu }}$ as subfields of an algebraic closure $\mathrm{\Omega }$ of ${𝔽}_p$, so by Corollary 11.1.8,

(If you don’t like algebraic closure, you can replace $\mathrm{\Omega }$ by ${𝔽}_{3^{21}}$ which contains all the fields considered in this exercise.)

Note that

Indeed, ${𝔽}_{3^3}\cap {𝔽}_{3^7}$ is a finite subfield of $\mathrm{\Omega }$, so is of the form ${𝔽}_{3^n}$, where $n\mid 3$ and $n\mid 7$, thus $n=1$.

Moreover $f$ has not root in ${𝔽}_3$, since $\gcd <!--\; FUNCTION\; APPLICATION\; -->(f,x^3-x)=1$. So the sets of the roots of $f$ in ${𝔽}_{3^3}$ and ${𝔽}_{3^7}$ are disjointed. Therefore $f$ has 10 distinct roots in any field which contains ${𝔽}_{3^3}$ and ${𝔽}_{3^7}$. In particular $f$ has 10 distinct roots in ${𝔽}_{3^{21}}$, and $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=10$, so $f$ splits completely in ${𝔽}_{3^{21}}$.

Name ${\alpha }_1,{\alpha }_2,{\alpha }_3$ the three roots of $f$ in ${𝔽}_{3^3}$, and ${\alpha }_4,\dots ,{\alpha }_{10}$ the seven roots of $f$ in ${𝔽}_{3^7}$.

Since 7 is prime, there is no strictly intermediate field between ${𝔽}_3$ and ${𝔽}_{3^7}$, and since ${\alpha }_i\notin {𝔽}_3$, then ${𝔽}_3({\alpha }_4,\dots ,{\alpha }_7)={𝔽}_{3^7}$, and similarly ${𝔽}_3({\alpha }_1,{\alpha }_2,{\alpha }_3)={𝔽}_{3^3}$.

Let $K={𝔽}_3({\alpha }_1,\dots ,{\alpha }_{10})$. Then $K={𝔽}_{3^r}(r\in \mathbb{N})$, is a finite subfield of $\mathrm{\Omega }$, which contains ${𝔽}_{3^3}$ and ${𝔽}_{3^7}$, so $3\mid r$ and $7\mid r$, with $\gcd <!--\; FUNCTION\; APPLICATION\; -->(3,7)=1$, thus $21\mid r$, therefore $K={𝔽}_{3^r}\supset {𝔽}_{3^{21}}$, so ${𝔽}_{3^{21}}\subset {𝔽}_3({\alpha }_1,\dots ,{\alpha }_{10})$. Moreover ${\alpha }_1,\dots ,{\alpha }_{10}$ are in ${𝔽}_{3^{21}}$, so

(In other words $𝔽({\alpha }_1,\dots ,{\alpha }_{10})$ is the compositum in $\mathrm{\Omega }$ of ${𝔽}_3({\alpha }_1,{\alpha }_2,{\alpha }_3)={𝔽}_{3^3}$ and ${𝔽}_3({\alpha }_4,\dots ,{\alpha }_7)={𝔽}_{3^7}$, and the compositum of ${𝔽}_{3^3}$ and ${𝔽}_{3^7}$ in $\mathrm{\Omega }$ is ${𝔽}_{3^{21}}$, so ${𝔽}_3({\alpha }_1,\dots ,{\alpha }_{10})={𝔽}_{3^{21}}$.)

Therefore, ${𝔽}_{3^{21}}$ is the splitting field of $f$ over ${𝔽}_3$. □