I was trying to prove that for a given $n$, there exist a totally real number field of degree $n$.

I understood that it was equivalent to find a polynomial $P$ such that

(i) $P\in\mathbb Q[X]$ ;

(ii) $P$ is irreducible ;

(iii) $P$ only has real roots in $\mathbb C$ ;

(iv) $P$ has degree $n$.

I succeeded in my initial problem thanks to this and this two MSE questions.

But the construction is abstract, and I can not deduce from it an explicit polynomial satisfying the four conditions.

The question.

For a given $n$, do you know how can I construct an explicit polynomial $P$ satisfying (i), (ii), (iii) and (iv)?

  • $\begingroup$ If by "Number field" you simply mean an algebraic extension of $\;\Bbb Q\;$ of finite degree then $\;\Bbb Q(\sqrt[n]2)\;$ is such an example for any $\;n\in\Bbb N\;$ . But in this case condition (iii) isn't fulfilled...and I'm not sure whether this can be done. $\endgroup$ – DonAntonio May 22 '18 at 19:58
  • 1
    $\begingroup$ @DonAntonio That's not totally real. $\endgroup$ – Ravi May 22 '18 at 19:59
  • $\begingroup$ @Ravi What is "totally real" for you? $\endgroup$ – DonAntonio May 22 '18 at 20:00
  • $\begingroup$ @DonAntonio Every embedding $\iota: K\hookrightarrow \mathbb C$ lands in $\mathbb R$. $\endgroup$ – Ravi May 22 '18 at 20:00
  • $\begingroup$ @DonAntonio Totally real means that all the conjugates under the Galois group are subsets or $\mathbb R$. But this question is more about the polynomial. $\endgroup$ – E. Joseph May 22 '18 at 20:00

I am assuming that by "real roots in $\mathbb{C}$" you mean real roots (from $\mathbb{R}$). One possible family of polynomials that satisfy all these criteria (for $n\geq5$):

$$ f_n(x)=(x-1)(x-2)\cdots(x-n)+1. $$

Proof. Ad (i) and (iv) Trivial.

Ad (ii) This polynomial is well known to be irreducible, it follows nicely from irreducibility criterion of Pólya, see for example Prove that the polynomial $(x-1)(x-2)\cdots(x-n) + 1$, $n\ne 4$, is irreducible over $\mathbb Z$ and/or linked questions.

Ad (iii) (sketch) This is a bit technical but can be done (holds for $n\geq 4$). Idea is that we have $f_n(i)=1$ for $i=1\dots n$. Now if we look at $f_n(k/2)$ for $k=1,3,\dots,2n-1$, we get another $n$ points, out of which $\lceil n/2 \rceil$ are negative. So by examining sign changes, we can reach the conclusion that $f(x)$ has exactly $n$ real roots.

  • 1
    $\begingroup$ That's exactly what I was looking for, thanks :) $\endgroup$ – E. Joseph May 22 '18 at 22:18

Take an Eisenstein polynomial ( link ) $P(x)$ with integer coefficients for a prime $p$. Now consider $a$ integer such that $a+ Im(x_k)>0$ for all roots $x_k$ of $P(x)$ and moreover $a\equiv 0 \bmod{p^2}$. Then

  1. $P(x- i a) \in \mathbb{Z}[x]$ has all the roots in the upper half plane.

  2. $Re(P(x- i a))$ is an Eisenstein polynomial for $p$.

Denote by $Q(x)=Re(P(x-ia))$.

From 1. $Q(x)$ has all the roots real (and distinct) (see link )

From 2. $Q(x)$ is irreducible over $\mathbb{Q}$.

Example: $P(x) = x^n+3$, $\ \ p=3$, $a=9$. $$ Q(x)=1/2(\ (x+9i)^n+(x-9i)^n)+3 $$


Take a polynomial with large integer coefficients with $n$ real roots. Then perturb the coefficients slightly (add or subtract one from some of them) to make the polynomial reduce modulo $2$ to an irreducible polynomial modulo $2$. Your new polynomial is irreducible over $\Bbb Q$ and unless you have been unlucky or careless, will still have $n$ real roots.

  • $\begingroup$ Thanks, it convinces me that it exists. But do you think we can deduce a more explicit polynomial from this construction? $\endgroup$ – E. Joseph May 22 '18 at 20:09
  • $\begingroup$ For a more explicit example, take a prime $p\equiv1\pmod{2n}$. Take the Gaussian periods corresponding to the index $n$ subgroup of $\text{Gal}(\Bbb Q(\zeta_p)/\Bbb Q)$ en.wikipedia.org/wiki/Gaussian_period $\endgroup$ – Lord Shark the Unknown May 22 '18 at 20:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.