# 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?

• I guess the polynomials have different discrimant.
– lhf
Jan 9, 2017 at 16:05
• @lhf: The discriminant of the field can be a proper divisor of the discriminant of a defining polynomial. That approach works if the discriminants are relatively prime, though, since $\mathbb{Q}$ is the only number field with discriminant $1$.
– user14972
Jan 9, 2017 at 16:09
• @Ihf: I've considered using this, but I don't see how. Let $\mathbb{Q} \subseteq K_1, K_2$ be two field extensions of degree three; consider the minimal polynomials $f_1$ and $f_2$ of non-rational elements $x_1 \in K_1$ and $x_2 \in K_2$ respectively, then their discriminants coinciding only tells that there are three elements in each field (the differences of the roots of $f_1$ and $f_2$) such that their respective products coincide in $\mathbb{Q}$ (up to sign). Jan 9, 2017 at 16:14
• I don't think the discriminant of the polynomial will be useful in this case because it is always a square. In fact, the discriminant of a cubic Galois extension of $\mathbf Q$ is the square of the conductor and there are many such extensions with the same conductor! However, for a cubic given by $x^3 - ax + a$ as in your question, the conductor is the product of the primes congruent to $1$ mod $3$ that divide $p$. You might have to multiply this product by $9$ but either way you can use this to construct nonisomorphic cubic extensions. Jan 9, 2017 at 17:39

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.

• This is definitely the best approach and is generalizable to arbitrary products of arbitrary cyclic groups. Learning this in my number theory course was super cool. Jan 11, 2017 at 15:29
• Excuse me, but what does "ramified" mean?
– Labo
Mar 24, 2017 at 22:17
• Let $K\supset k$ be a pair of number fields, with $[K:k]=n$. Let $\mathfrak o$ and $\mathfrak O$ be the rings of integers of $k$ and $K$ respectively. Now let $\mathfrak p$ be a prime of $\mathfrak o$. Then $\mathfrak p\mathfrak O$ is writable as $\mathfrak P_i^{e_1}\cdots\mathfrak P_g^{e_g}$, where the $\mathfrak P_i$ are primes of $\mathfrak O$. The $e_i$ are called the ramification indices, and if any of them is $>1$, you say that there is ramification, or that $\mathfrak P_i$ is ramified over $\mathfrak p$, or that $\mathfrak p$ is ramified. Mar 25, 2017 at 2:35

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.

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 $Q(\sqrt  a)$, with $a \in Q^{*}$, they even start in an obvious way ./.