Distinguishing field extensions of $\mathbb{Q}$ of degree $3$.
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.
Many pairs fields can be distinguished by their discriminant. And knowing the discriminant of a defining polynomial instead of that of the field is often good enough.
Recall that if $f$ is an irreducible polynomial over $\mathbb{Q}$ that defines a number field $K$, then there exists some integer $s$ such that
$$ \operatorname{disc}(f) = s^2 \operatorname{disc}(K) $$
Therefore, if you have two irreducible polynomials $f$ and $g$ such that the ratio of their discriminants is not a square, then they cannot define the same field.
Although, as pointed out in the comments, cubics with galois group $A_3$ are precisely those whose discriminant is a square, so for this approach to work you have to do more than look at the discriminant of a single defining polynomial; e.g.
- Obtain the actual field discriminant
- Find polynomials defining the two fields whose discriminants are relatively prime (exploiting the fact that $\operatorname{disc}(K) \neq 1$ for every number field $K \neq \mathbb{Q}$).
Another simple test is to factor an integer. If $K$ and $L$ are two fields such that $p\mathcal{O}_K$ and $p\mathcal{O}_L$ have factorizations into ideals that don't have the same pattern of degrees and ramification indices, then they aren't the same field.
For example, if the discriminants of $f$ and $g$ are cubics with galois group $A_3$ and both have discriminant relatively prime to $p$, and $f$ has three roots in $\mathbf{F}_p$ but $g$ has no roots in $\mathbf{F}_p$, then the fields defined by $f$ and $g$ are distinct.
In that case, in the field defined by $f$, $(p)$ factors into a product of three ideals of degree $1$, but in the field defined by $g$, $(p)$ is a prime ideal.