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Corollary 2.5 in this note states:

For any integer $k$, set $a = k^2 + k + 7$. The polynomial $x^3 − ax + a$ is irreducible over $\mathbb Q$ and has Galois group $A_3$.

I want to prove there're infinitely many non-isomorphic cubic Galois extension of $\mathbb Q$ by showing Galois extensions above are pairwise not isomorphic.

The general method would be:

Suppose $u$ is a root of $x^{3}-a_{1}x+a_1$ for $a_1=k_1^2+k_1+7$, $k_1 \in \mathbb Z$ and $v$ is a root of $x^{3}-a_{2}x+a_2$ for $a_2=k_2^2+k_2+7$, $k_2 \in \mathbb Z$.

Then suppose on the contrary that there exists field isomorphism $\varphi: \mathbb Q(u) \to \mathbb Q(v)$ with $\varphi(u) = a v^2 + b v + c$, $a, b, c \in \mathbb Q$, and $(a v^2 + b v + c)^{3}-a_{1}(a v^2 + b v + c) + a_1 = 0$, where $v^{3} - a_{2}v + a_2 = 0$.

This method works perfectly in showing $\mathbb Q(\sqrt{m}) \cong \mathbb Q(\sqrt {n}) \iff m = n$ for square-free $m$ and $n$, but in this case, it doesn't help much.

Another method would be directly using Cardano formula.

We know $a$ and $b$ have different minimal polynomials doesn't guarantee field extensions containing them are not isomorphic, but in this case, they have similar minimal polynomials $x^3-ax+a$ where $a=k^2+k+7$, $k \in \mathbb Z$, does this help?

And are there other thoughts and understandings?

Thanks for your time and effort.

Andrews
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1 Answers1

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There are infinitely many cubic Galois extensions for the polynomial $X^3-a\in \Bbb Q[X]$, since isomorphic splitting fields $\Bbb Q(\sqrt[3]{a},\omega)$ have the same discriminant and we have infinitely many different discriminants as $a\in \Bbb Q$ varies. This would show your title claim a little bit easier than using $X^3-aX+a$.

References:

Discriminant of $\Bbb Q(\sqrt[3]{2})$

Dietrich Burde
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  • Thanks for your answer. There still remain two questions: 1. Are splitting fields of $x^3−a x+a$ for $a=k^2+k+7, k∈Z$ not isomorphic? (It might be hard to answer in general) 2. Do we have a general conclusion that two fields have same discriminant can imply or be implied by they're isomorphic? – Andrews Mar 25 '20 at 13:59
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  • In general, two isomorphic number fields have the same discriminant. The converse might not be true, see this post. 1. So it is enough for us to produce infinitely many different discriminants.
    • – Dietrich Burde Mar 25 '20 at 14:19