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The title is pretty much self explanatory; Can an irreducible polynomial of $\mathbb{Q}[x]$ have more than two real roots? And if so, what is an example of a polynomial with more than two real root? All the polynomials I've seen, had maximum two real roots, but from what I've learned of Galois Theory it seems to indicate that there could be more than two.

Eric Wofsey
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slowpoke
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    A cubic polynomial has three real roots if and only if its discriminant is positive. The discriminant of $x^3+px+q$ is $-4p^3 -27q^2$. Pick $q$ prime and $p$ a large negative multiple of $q$ so that the polynomial is irreducible by Eisenstein, and you're done. – Arturo Magidin Sep 29 '21 at 17:20
  • But the polynomial $(x^2-2)(x^2-3)$ isn't irreducible, which is why it doesn't work just to patch polynomials together. – slowpoke Sep 29 '21 at 17:21
  • @slowpoke The product of two rational polynomials is not irreducible over $\mathbb Q$, of course. For an irreducible quartic, take the polynomial with roots $\pm\sqrt{2}\pm\sqrt{3}$ for example. – dxiv Sep 29 '21 at 17:25
  • @ArturoMagidin Thank you! That sure seem to work. – slowpoke Sep 29 '21 at 17:25
  • @dxiv: that doesn't work. A polynomial with roots $\pm\sqrt{2}$ would be a multiple of $x^2-2$, so it won't be irreducible. – Arturo Magidin Sep 29 '21 at 17:26
  • @ArturoMagidin I meant the four roots $(\pm\sqrt{2}\pm\sqrt{3})$, not the two pairs of square roots. In other words, the minimal polynomial of $\sqrt{2}+\sqrt{3}$. – dxiv Sep 29 '21 at 17:27
  • @dxiv: Ah, missed the absence of a comma. – Arturo Magidin Sep 29 '21 at 17:28

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A cubic polynomial has three real roots if and only if its discriminant is positive. If it is a depressed cubic, $$f(x) = x^3+px+q$$ then the discriminants is $$\Delta(f) = -4p^3 - 27q^2.$$ Now, if there is a prime $\ell$ such that $\ell$ divides both $p$ and $q$, and $\ell^2$ does not divide $q$, then by Eisenstein's Criterion we would have that $f(x)$ is irreducible over $\mathbb{Q}$. So all we need is to pick appropriate $p$ and $q$ to get this done. For example, $$ f(x) = x^3 - 9x + 3$$ will do, since it is Eisenstein at $3$, hence irreducible, and $$\Delta(f) = 4(9)^3 - 27(9) = 2673\gt 0.$$

Arturo Magidin
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  • $F\cap\mathbb{R}$ is not a Galois extension of $\mathbb{Q}$ in general (for instance, consider if $F$ is the splitting field of an irreduciple polynomial with a real root but not all real roots). $[F:F\cap\mathbb{R}]$ is the order of the corresponding subgroup of the Galois gruop, not its index. – Eric Wofsey Sep 29 '21 at 22:44
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    If you just want to find an irreducible cubic, make it monic then use the rational roots test. You can just shove a massive square term in to make it cross the axis enough times. $x^3-10000x^2+1$ should do the trick. – David A. Craven Sep 29 '21 at 22:50
  • @EricWofsey: Thanks for pointing it out; I shouldn't add stuff when I'm rushing out to go teach. – Arturo Magidin Sep 30 '21 at 01:09
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Have you heard about totally real number fields? If $K$ is a totally real number field of degree $n$, then there is an irreducible polynomial $f$ over $\mathbb Q$ of degree $n$ such that all roots are real and generate $K$. This post https://math.stackexchange.com/a/2645829 implies that for every $n \in \mathbb N$ you find such a totally real number field of degree $n$. So for each $n$ you find an irreducible polynomial $f$ of $\mathbb Q$ with $n$ real roots.

principal-ideal-domain
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