Is $ K(t)/K(t^3)$ Galois?

124 Views Asked by At

Let $K(t)$ be a field of rational function over a field $K$. I want to know whether $K(t)/K(t^3)$ is Galois or not when $K=\Bbb R,\Bbb Q,\Bbb F_p$.

I know Galois theory in the subfield of $\Bbb C$, but I'm not familiar with a field of rational function, so I need your help. I think $[K(t):K(t^3)]=3$(but I don't know how to prove this) ,I cannot find minimal polynomial of this extension so I cannot go any further (I cannot find conjugate element). I would be appreciated if you could help me checking whether $K(t)/K(t^3)$ is Galois or not.

3

There are 3 best solutions below

0
On BEST ANSWER

You can think about the extension $K(t)/K(t^3)$ as adjoining a cube root ($t$) to the element $t^3$ in $K(t^3)$. The minimal polynomial of $t$ over $K(t^3)$ is $x^3 - t^3$.

If $K$ has characteristic $3$, then $x^3 - t^3$ factors in $K(t)$ as $(x-t)^3$. Thus the extension $K(t)/K(t^3)$ is not separable, hence not Galois.

If $K$ does not have characteristic $3$, then $x^3 - t^3$ factors in the algebraic closure of $K(t)$ as $(x-t)(x-\xi t)(x- \xi^2 t)$ where $\xi$ is a primitive cube root of unity. So the question of whether this splitting already occurs in $K(t)$ depends on whether $\xi\in K$ or not.

Putting these observations together, $K(t)/K(t^3)$ is Galois if and only if $K$ has characteristic not equal to $3$ and $K$ contains all three cube roots of unity. Of course, you may want to prove the assertions above a bit more carefully.

0
On

Hint $t$ is a root of the Polynomial $X^3-t^3 \in K(t^3)[X]$. Can this be factored in $K(t^3)[X]$?

Hint 2 Pay extra attention to the case $K=\mathbb F_3$.

0
On

What about

$$P(X)=X^3-t^3$$

as a minimal polynomial.