Degree of splitting field of $x^6+tx^3+t$ in $\mathbb{F}_3(t)[x]$

Question

I'm trying to find the degree splitting field of $x^6+tx^3+t$ in $\mathbb{F}_3(t)[x]$. After substituting $z= x^3$, and using the qudratic formula, and thensubstituting $x$ back in, I get the roots

$$\frac{\sqrt[3]{-t\pm\sqrt{t^2-4 t}}}{\sqrt[3]{2}},$$ but how do I find the minimal polynomial for this? No matter what power I take, I can't seem to get rid of the radicals. Is there another way to think about square/cube roots when you are in a finite field? Not sure how to proceed.

Answer 1

As @jyre says, the polynomial $X^6 + tX^3 + t$ is irreducible, so you can make a first field extension of degree $6$. $\Bbb F_3(t) \to k = \Bbb F_3(t)[x]/(x^6+tx^3+t)$.

Over this field $k$, $x$ is a root of $X^6 + tX^3 + t$ so the polynomial factors a bit. It turns out $x$ is a multiple root :

$X^6 + tX^3 + t = X^6 + tX^3 +t - (x^6 + tx^3 + t) \\= X^6-x^6 + t(X^3-x^3) = (X^3-x^3)(X^3+x^3+t) \\= (X-x)^3(X^3+x^3+t)$

Now you need to check if $X^3+x^3+t$ is further reducible over $k$.

But putting $t = -x^6/(x^3+1)$, $k$ is isomorphic to $\Bbb F_3(x)$, and now the remaining cubic factor is $X^3 + x^3-x^6/(x^3+1) = X^3+x^3/(x^3+1) = (X+x/(x+1))^3$

Therefore, your polynomial splits in $k$ as $(X-x)^3(X+x/(x+1))^3$

And so the splitting field is $k$, and the degree is $6$.

Answer 2

Here’s a slightly different take on the problem, using the method recently recommended by Dietrich Burde somewhere. Calling $f(X)=X^6+tX^3+t$, we have $F(X)=g(X^3)$, $g(X)=X^2+tX+t$, quadratic, so that one root generates a Galois extension of the base field $k=\Bbb F_3(t)$. The roots are $$ \zeta=\frac{-t\pm\sqrt{t^2-4t}}2=t\pm\sqrt{t^2-t}\,, $$ in particular, the two roots $\zeta_1,\zeta_2$ are related by $\zeta\mapsto-t-\zeta$, an involution.

The roots $\xi_i$ of $f$ satisfy $\xi_i^3=\zeta_i$, so there are only two of them. Just as $g$ factors as $g(X)=(X-\zeta)(X+t+\zeta)$, so $f$ factors as $f(X)=(X^3-\xi^3)(X^3+t+\xi^3)=(X-\xi)^3(X+t^{1/3}+\xi)^3$, if only $t^{1/3}\in k(\xi)$, but it is, since $t=-\frac{\xi^3}{1+\xi^3}$, so $t^{1/3}=-\frac{\xi}{1+\xi}$.

Thus all (both) roots of $f$ come when you adjoin one $\xi$, and $k(\xi)$ is the splitting field of $f$ over $k$, degree six.