Not all quadratic extensions over $\mathbb{Q}$ are contained in the compositum of all the splitting fields of irreducible cubics in $\mathbb{Q}[X]$

Question

Question: Let $F$ be the composite of all the splitting fields of irreducible cubics over $\mathbb{Q}$. Prove that $F$ does not contain all quadratic extensions of $\mathbb{Q}$.

(This is exercise 16 in section 14.9 in the book "Abstract Algebra" by Dummit and Foote, 3rd edition.)

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My progress:

Any quadratic extension over $\mathbb{Q}$ is of the form $\mathbb{Q}[\sqrt D]$, where $D$ is a square-free integer. Thus, if $Q[\sqrt{\pm p}]$ was contained in $F$ for every prime $p$, then, by taking compositums, any quadratic extension of $\mathbb{Q}$ would be contained in $F$. Hence, the question is equivalent to finding a prime $p$ and $\epsilon \in \{\pm 1\}$ such that $\sqrt{\epsilon p} \notin F$.

Another relevant fact is that if $D$ is the discriminant of a cubic polynomial $f$, then $\sqrt D$ lies in the splitting field of $f$.

Consider the polynomial $f(X) = X^3 - 3X - 4 \in \mathbb{Q}[X]$. Its discriminant turns out to be $-18^2$, which can be found using the formula for the discriminant of a depressed cubic. Also, it is irreducible over $\mathbb{Q}$ as it has no rational roots, which can be concluded using the rational root theorem. Hence, the splitting field of $f$ over $\mathbb{Q}$ contains $\sqrt{-18^2} = 18i$, and hence, it contains $i$. Thus, $i \in F$.

Hence, the original question is now equivalent to finding a prime $p$ such that $\sqrt{p} \notin F$.

(Consider the polynomial $g(X)=X^3-3$, which is irreducible over $\mathbb{Q}$ by Eisenstein's criterion. It's discriminant is $-3^5$, and hence $9\sqrt{3}i\in F \implies \sqrt 3 \in F$. So, $3$ is one prime that won't work as an example for the previous statement. By similar computation, maybe more primes can be ruled out.)

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Please let me know if the above conclusions that I have made are indeed correct.

I could not make any further progress. Any help would be greatly appreciated.

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Edit: Turns out the question is wrong, and $F$ does contain every quadratic extension over $\mathbb{Q}$. Thanks to Peter Wu for pointing it out and providing the solution to the corrected question; as pointed out in the comments, the question is corrected in the errata (https://site.uvm.edu/ddummit/files/2021/04/errata_3rd_edition_4_19_2021.pdf), page 14.

Answer 1

I strongly suspected that it is a wrong problem.

The discriminant of $x^3-ax+b=0$ is $4a^3-27b^2$, for which we consider its class in $\mathbb{Q^{\times}}/\mathbb{Q^{\times2}}$, which can be simplified to $a^3-3(3/2*b)^2$. Let $a=(3/2*b)=n+3$ for $n\in \mathbb{Z}$. Then $n$ and discriminant is in same class. So $\sqrt{n}\in F$ if $x^3-(n+3)x+\frac{2}{3}(n+3)$ is irreducible.

Then we prove that only for a finite number of $n$, $x^3-(n+3)x+\frac{2}{3}(n+3)$ is reducible. Firstly there must be a rational root and it must be an integer. So $\frac{2}{3}(n+3)$ is integral. Let $\frac{1}{3}(n+3)=m$ the cubic polynomial equation is $x^3-3mx+2m$.

Consider $x^3-3mx+2m=0$, then $m=\frac{x^3}{3x-2}$, which means only a finite choice of $x$ can make $m\in \mathbb{Z}$. So for all but finitely many $n$, $\sqrt{n}\in F$. Then for any number $k$ we can find an $l$ s.t. $\sqrt{l},\,\sqrt{kl}\in F$. Thus $\sqrt{k}\in F$.

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On why root of $x^3-(n+3)x+\frac{2}{3}(n+3)$ must be integer, we introduce the evaluation: fix prime $p$, evaluation of an nonzero integer $n$ is defined as the maximal integer $m$ such that $p^m|n$, denoted as $v_p(n)$.

And for $q=a/b\in \mathbb{Q}^{\times}$, $v_p(q)=v_p(a)-v_p(b)$, and $v_p(0)=+\infty$. Then we have if $v_p(a)\not=v_p(b),\,v_p(a+b)=min(v_p(a),v_p(b))$.

Then if $x$ is a rational root and exist a prime $p$ s.t. $v_p(x)<0$, let $v_p(x)=-a$. Thus $v_p(x^3)=-3a,\,v_p(-(n+3)x)\ge -a$ and $v_p(\frac{2}{3}(n+3))\ge -1$, in which $v_p(x^3)=-3a$ is minimal, so the sum has valuation $-3a$ and can not be 0.