Can every algebraic number field be constructed by adjoining a root of a trinomial (with rational coefficients) to $\mathbb{Q}$?
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1$\begingroup$ The same question can be asked about finite fields. In $\mathbf F_p[x]$ there is no irreducible trinomial of degree $d$ if the pair $(p,d)$ is $(2,8)$, $(3,49)$, or $(5,35)$. $\endgroup$KConrad– KConrad2025-10-12 13:27:48 +00:00Commented Oct 12 at 13:27
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2 Answers
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The answer is no. Take some totally real number field $K$ such that $K/\mathbb Q$ is Galois and has degree $n>4$ (e.g. the totally real subfield of $\mathbb Q(\zeta_{11})$). If $K=\mathbb Q(\alpha)$ and $\alpha$ is a root of some $f\in\mathbb Q[x]$, then all $n$ conjugates of $\alpha$ must also be roots of $f$. However, by Descartes' rule of signs, a trinomial can have at most $4$ distinct nonzero real roots, so $\alpha$ is not a root of a rational trinomial.
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3$\begingroup$ You've (negatively) answered the question of whether every number field — or even just every Galois number field — is the splitting field of a trinomial. Now I wonder about the question of whether every Galois number field is the decomposition field of a trinomial (which is another way to interpret the title question). Do you happen to have a quick answer to that? Otherwise I might ask it as a separate question. $\endgroup$Gro-Tsen– Gro-Tsen2025-09-30 10:49:54 +00:00Commented Sep 30 at 10:49
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1$\begingroup$ @Gro-Tsen: By sciencedirect.com/science/article/pii/S002186939998033X , there iare very few possibilites for the Galois groups of irreducible trinomials. Even if, not all the cases are covered in this paper, I seriously doubt that you can obtain any Galois extension of Q (say) as the decomposition field of a trinomial (irreducible or not). If I were you, I would search a counterexample amongst the multiquadratic field extensions of Q. $\endgroup$GreginGre– GreginGre2025-09-30 11:23:58 +00:00Commented Sep 30 at 11:23
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3$\begingroup$ @Gro-Tsen My example answers both questions the same way - since $K$ is Galois, if $f$ has any root in $K$, then all its conjugates are in $K$. Since $f$ has at most $4$ real roots, this shows any such root has degree at most $4$. Since $[K:\Bbb Q]=5$, the roots must be rational. So $K$ can be neither a rupture or a splitting field of a trinomial, irreducible or not. $\endgroup$Wojowu– Wojowu2025-09-30 13:46:48 +00:00Commented Sep 30 at 13:46
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2$\begingroup$ The trinomial $x^5-4x^3+x$ has $5$ distinct real roots. $\endgroup$mr_e_man– mr_e_man2025-09-30 17:55:58 +00:00Commented Sep 30 at 17:55
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2$\begingroup$ I think you can also make a non-archimedean version of this argument, though a concrete example will be a bit messier -- the Newton polygon of a trinomial has at most three slopes, which constrains the ways in which a prime can split/ramify in an extension cut out by a trinomial. $\endgroup$Alison Miller– Alison Miller2025-10-02 00:06:07 +00:00Commented Oct 2 at 0:06
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The answer of @wojowu leaves open the case whether there exist field extensions $K/\mathbb{Q}$ of degree $4$ that can't be generated by a root of a trinomial. Although the question has been answered very satisfactorily by @wojowu, I still thought it would be fun to add this case here.
I claim that if $f$ is a degree $4$ polynomial all of whose roots are real, with Galois group $S_4$ or $A_4$, and if $\beta$ is a zero of $f$, then $K=\mathbb{Q}(\beta)$ can't be generated by the zero of a trinomial.
To see this, suppose $\alpha \in K$ is a zero of a trinomial $g$ (which we may assume does not have zero as a root). Then since $\alpha$ is totally real, the number of real roots of $g$ must be exactly $4$. Assuming $g$ to be monic, this forces $g$ to be of the form $g = x^n + ax^m + b$, where $a<0$ and $b>0$. Moreover we must also have $m,n$ both even, since if $m$ would be odd, the trinomial $g(-x)$ would not have a sign change from the second term to the constant term, and if $n$ would be odd but not $m$, we would not have a sign change between the first and the second term. So in fact $g$ is of the form $g=g_0(x^2)$.
If $\alpha$ is a root of $g$, it must be a root of some irreducible factor $h$ of $g$. Now it is a relatively easy exercise that the (monic) irreducible factors of a polynomial of the form $g_0(x^2)$ are either of the form $h_0(x^2)$, or they come in pairs $h_0(x),h_0(-x)$. But since $\alpha$ has degree $4$ over $\mathbb{Q}$, and $g$ only has four (non-zero) real roots, $\alpha$ can be a zero of only one irreducible factor, which therefore must have the form $h=h_0(x^2)$. However, this means that $\mathbb{Q}(\alpha^2)$ is of degree $2$. But this is a contradiction, since degree $4$ extensions whose Galois closure has group $S_4$ or $A_4$ do not have quadratic subextensions. QED
