Distinguishing field extensions of $\mathbb{Q}$ of degree $3$. 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? 

 A: 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.
A: 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.
A: In your question at the end, you do not specify wheteher your cubic fields are Galois above $Q$, but the whole context of your post suggests that they are. In this case your problem admits a purely Galois theoretic solution (arithmetic will come into play only over $Q$).
Let $p$ be an odd (for simplification) prime and $k$ be a field of characteristic $\neq p $. Denote by $G_k$ the absolute Galois group of $k$. Any cyclic extension of $k$ of degree $p$ (say a $C_p$- extension for short) is the fixed field of a subgroup of index $p$ of $G_k$, and any such subgroup is the kernel of a non trivial homomorphism $G_k \to C_p$, so that the $C_p$-extensions of $k$ are classified by the non trivial elements of the group $Hom(G_k, C_p)$. To progress further, introduce the field $K = k(\mu_p)$, where $\mu_p$ is the group of $p$-th roots of unity, and consider the restriction map $Hom(G_k, C_p) \to Hom(G_K, C_p)$. For any Galois module $X$, $\Delta=Gal(K/k)$ acts on $Hom (G_K , X)$ in the usual way, more precisely via  $(\delta, f) \in  \Delta \times Hom(G_K, X) \to f^{\delta}$ given by $f^{\delta}(x)= \delta (f(\delta^{-1}(x))$ for $x \in X$ . Note that this action is the only one which is functorial; here $\Delta$ acts trivially on $C_p$. Since $\Delta$ has order prime to $p$, it is classically known that $Hom(G_k, C_p)\cong Hom(G_K, C_p)^{\Delta}$, where the superscript $(.)^{\Delta}$ denotes the invariants under $\Delta$. Since $G_K$ acts trivially on $\mu_p$, one has $Hom(G_K, C_p)\cong Hom(G_K,\mu_p)$ as groups, but not as $\Delta$-modules because of the canonical action of $\Delta$ defined above. Actually, one can check easily that $Hom(G_K, C_p)\cong Hom(G_K,\mu_p)(-1)$, where $(.)(-1)$ denotes the so called "Tate twist", which means that the Galois action on $Hom(G_K,\mu_p)$ has been replaced by a new action defined by $\delta^{new} (f)= \kappa(\delta)^{-1}.\delta^{old}(f)$, where $\kappa$ is the mod $p$ character defined by $\delta(\zeta)=\zeta^{\kappa(\delta)}$ for $\zeta \in \mu_p$ . Consequently, $Hom(G_k, C_p)$ can be identied with the elements $f\in Hom(G_K,\mu_p)$ such that $\delta (f)=\kappa(\delta).f$ (these can be viewed as "eigenvectors" corresponding to the "eigenvalues" $\kappa(\delta)$.)
Now, by Kummer theory, $Hom(G_K,\mu_p)\cong (K^{*}/K^{*})^p$ , which means more concretely that any $C_p$-extension of $K$ is of the form $K(\sqrt [p]a)$, where $a$ represents a class $\bar a \in (K^{*}/K^{*})^p$, and two such extensions $K(\sqrt [p]a)$ and $K(\sqrt [p]b)$ coincide iff $\bar a = \bar b^j$ for a certain $j\neq 0$  mod $p$. Summarizing : two $C_p$-extensions $L$ and $L'$ of $k$ coincide iff the composite fields $L.K = K(\sqrt [p]a)$ and $L'.K = K(\sqrt [p]b)$ coincide, which means that $\delta(\bar a)=\kappa(\delta).\bar a$ and $\bar a = \bar b^j$, $j\neq 0$  mod $p$. This is an "elegant" theoretical way to distinguish between two $C_p$-extensions of $k$, but of course, in concrete examples, it requires some computations - although in your paticular case $k=Q$ and $p=3$, these do not seem especially complicated. For pure cubic fields, i.e. of the form $Q(\sqrt [3] a)$, with $a \in Q^{*}$, they  even start in an obvious way ./. 
