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I was thinking about how to show whether or not a polynomial irreducible over $\mathbb Q$ could be irreducible over $\mathbb Q(i)$

I am aware of $x^4 + 1$ being irreducible over $\mathbb Q$ but not $\mathbb Q(i)$. However, it also has strictly complex roots. I was wondering if whenever there are only two non real roots and at least one real root, that the polynomial is irreducible over both $\mathbb Q$ and $\mathbb Q(i)$.

Are there any more examples of polynomials irreducible over $\mathbb Q$ but not $\mathbb Q(i)$ that have only two complex roots and at least one real root? I've realised that my original reasoning is actually flawed but I still suspect that maybe there isn't such a polynomial.

Also, a related question:

I am aware of many tests for irreducibility over $\mathbb Q$, but what I can I do other than check the roots, to see that a polynomial is or isn't irreducible over $\mathbb Q(i).$ I am aware that Eisenstein's criterion may be generalised to integral domains, but what if the constant of the polynomial is $2$? In that case Eisenstein wouldn't apply, so is there anything else that I could do?

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  • $\begingroup$ $x^2+1$ is a slightly easier example than yours, and the first I thought of before reading your entire question. Still has the same problems, though. $\endgroup$ – Arthur Jan 2 '18 at 13:13
  • $\begingroup$ What do you make of $x^4-3$ and $x^4-5$? $\endgroup$ – Mark Bennet Jan 2 '18 at 13:19
  • $\begingroup$ @MarkBennet: Why, yes! The day after New Year's Day can be hard sometimes… $\endgroup$ – Bernard Jan 2 '18 at 13:29
  • $\begingroup$ Newton's polygon gives an occasionally useful extension of Eisenstein's criterion. The condition is that if the polygon of a degree $n$ polynomial consists of a single line segment with slope $-a/n,\gcd(a,n)=1$, then the polynomial is irreducible. Consider, for example, $p(x)=x^5-4x+2$ over $\Bbb{Q}(i)$ and the valuation related to the prime $p=1+i$. The constant term is divisible by $p^2$, the linear with $p^4$, and the leading with $p^0$. The polygon thus reduces to a line with slope $-2/5$ and we are done. $\endgroup$ – Jyrki Lahtonen Jan 2 '18 at 15:53
  • $\begingroup$ (cont'd) Compare with this segment from WP on Eisenstein. $\endgroup$ – Jyrki Lahtonen Jan 2 '18 at 15:55
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Let $f$ be irreducible over $\Bbb Q$ but not over $\Bbb Q(i)$. Then over $\Bbb Q(i)$ we have the factorisation $f=g\overline g$ where the coefficients of $\overline g$ are the complex conjugates of those of $g$. The zeroes of $\overline g$ the complex conjugates of those of $g$. A real zero of $f$ is then both a zero of $g$ and of $\overline g$ and so a double zero of $f$, which we'd assumed irreducible...

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  • $\begingroup$ Does this imply that $f$ is of even degree, since $g, \bar g$ would have the same degree? I'm also not really sure what your response implies. What is the importance of this double root? $\endgroup$ – user366818 Jan 2 '18 at 17:51
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    $\begingroup$ @user366818 If $f$ is irreducible over $\Bbb Q$ but not over $\Bbb Q(i)$ that it has even degree. A polynomial over $\Bbb Q$ with a multiple root cannot be irreducible. $\endgroup$ – Lord Shark the Unknown Jan 3 '18 at 8:41

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