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Context.

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)?

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  • $\begingroup$ 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
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    $\begingroup$ @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
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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.

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    $\begingroup$ That's exactly what I was looking for, thanks :) $\endgroup$ – E. Joseph May 22 '18 at 22:18
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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: $P(x) = x^n+3$, $\ \ p=3$, $a=9$. $$ Q(x)=1/2(\ (x+9i)^n+(x-9i)^n)+3 $$

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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.

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  • $\begingroup$ 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

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