I came across the following question in an old qualifying exam:
Find a Galois extension of $\mathbb{Q}$ with Galois group $\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$.
One very simple way to approach this problem is to find two distinct polynomials of degree $3$ with Galois groups $A_3 \cong \mathbb{Z}/3\mathbb{Z}$, and show that the field extensions they generate do not coincide. Then the Galois group must be $\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$ by Lang Th. VI.1.14.
There are an abundance of such polynomials; e.g. if $a = k + k^2 + 7$ for some integer then, $X^3 -aX + a$ does the trick (see Conrad). Unfortunately, I have no clue on how to show that two such field extensions do not coincide, except for possibly explicitly finding the roots of the two polynomials, and then trying to derive a contradiction trying to express a root of one polynomial in terms of the roots of the other. However, doing this would make me very sad, thus my question is:
How do I show (elegantly) that two field extension of $\mathbb{Q}$ of degree $3$ do not coincide?
Best Answer
If I were answering the exam question, I would have approached the question a little more computationally and specifically. The extensions $\Bbb Q(\zeta_7)\supset\Bbb Q$ and $\Bbb Q(\zeta_9)\supset\Bbb Q$ are both of degree six, with cyclic Galois group, so each has a subfield cubic over $\Bbb Q$. Since $\Bbb Q(\zeta_7)$ is ramified only at $7$ and $\Bbb Q(\zeta_9)$ is ramified only at $3$, the cubic extensions also are ramified only at $7$ and $3$, respectively. So they’re different.