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Suppose $f$ is the irreducible polynomial $x^4+x+1$ in $\mathbb{Q}[X]$ and we will call $E_f$ the splitting field of $f$ (which is a subfield of $\mathbb{C}$). Suppose $\alpha$ is a complex root of $f$ and suppose $x^2+ax+b$ is the minimal polynomial of $\alpha$ over $\mathbb{R}$. Then I have the following questions which I cannot solve:

-Show that $a$ and $b$ are in $E_f$

-Find the minimal polynomial $g$ of $b$ over $\mathbb{Q}$ and show that $b$ is not $0$

-Show that $g(1/b)=0$

-Find the minimal polynomial of $b+1/b$

I just don't know how to start with these questions. Any help is appreciated, thank you!

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closed as off-topic by user26857, John B, Watson, S.C.B., choco_addicted Apr 30 '16 at 12:59

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user26857, John B, Watson, S.C.B., choco_addicted
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  • $\begingroup$ For the first one: recall that the minimal polynomial over $\Bbb R$ of $\xi\in\Bbb C\setminus \Bbb R$ is always $(x-\xi)(x-\overline\xi)$. Are the roots of $x^4+x+1$ closed under complex conjugation? $\endgroup$ – user228113 Apr 18 '16 at 9:55
  • $\begingroup$ Yes, they are closed I believe $\endgroup$ – user2720690 Apr 18 '16 at 10:26
  • $\begingroup$ Thank you, I have the solution for the first one, it was quite simple actually. Any help for the other ones? $\endgroup$ – user2720690 Apr 18 '16 at 12:32
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This is a very nice problem: I struggled with it for some time. I’ll go part way, and you should be able to finish it off.

Certainly $b$ is nonzero, ’cause it’s the constant term of an irreducible quadratic. Now, since the roots of $x^2+ax+b$ are among the roots of $f$, we have $(x^2+ax+b)\big|f(x)$, so we may write $f(x)=(x^2+ax+b)(x^2+a'x+b')$, a real factorization of $f$. From the form of $f$, we get $a+a'=0$ and $bb'=1$, so that we may rewrite: $$ x^4+x+1=(x^2+ax+b)(x^2-ax+1/b)\,. $$ Expand this out and compare coefficients to get a pair of equations in $a$ and $b$, solve the easy one for $a$ and make a substitution, and get a sextic equation for $b$, coefficients happening to come from the set $\{0,\pm1\}$.

I think you ought to be able to finish it off now, but I’ll give more hints if you can’t.

EDIT, a week later:
You have asked for a method for finding the polynomial for $b+1/b$, when $b^6-b^4-b^3-b^2+1=0$. We also know that $b^3-b-1-1/b+1/b^3=0$, so let’s write $\beta=b+1/b$ and first compute $\beta^3$: \begin{align} \beta^3&=b^3+3b+3/b+1/b^3\\ &=4b+1+4/b\qquad\text{(by subtracting zero)}\\ &=4\beta+1\,, \end{align} so that $\beta^3-4\beta-1=0$, which is $\Bbb Q$-irreducible by the rational root test.

FURTHER EDIT, a few days later yet:
You asked for irreducibility (over $\Bbb Q$) of $x^6-x^4-x^3-x^2+1$. Here’s my argument, which I’m still a little worried about.

Over $\Bbb F_2$, we have $x^6+x^4+x^3+x^2+1=(x^2+x+1)(x^4+x^3+x^2+x+1)$, both factors being irreducible. (The second factor has for its roots the primitive fifth roots of unity, and they show up only in the field with $16$ elements.)

Over $\Bbb F_3$, we have $x^6-x^4-x^3-x^2+1=(x^3+x^2+x-1)(x^3-x^2-x-1)$, and both factors are $\Bbb F_3$-irreducible ’cause they don’t have roots in $\Bbb F_3$.

Now, what kind of factorization can $h=x^6-x^4-x^3-x^2+1$ have over $\Bbb Z$? No linear factors, we know, so either (1) three quadratics, (2) two cubics, or (3) a quadratic and a quartic. But if there were a quadratic irreducible factor of $h$, there would be a quadratic factor over $\Bbb F_3$, and there isn’t. So possibilities (1) and (3) are excluded. In possibility (2), take one of those $\Bbb Z$-irreducible cubic factors, call it $g$, and look at it in characteristic two. There, $g$ still divides $h$, so can’t remain irreducible, and therefore has a root modulo $2$, but $h$ itself doesn’t have such a root, so (2) is also excluded. (There must be a better argument!) Conclusion? It follows that $h$ has no nontrivial $\Bbb Z$-factorization.

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  • $\begingroup$ Thank you for your answer, I have the polynomial $x^6-x^4-x^3-x^2+1$ for $b$. But I still have no idea how I can find the polynomial of $b+1/b$ $\endgroup$ – user2720690 Apr 26 '16 at 13:07
  • $\begingroup$ And I think I still need to prove that that polynomial is irreducible, am I correct? $\endgroup$ – user2720690 Apr 26 '16 at 13:48
  • $\begingroup$ Is $x^6-x^4-x^3-x^2+1=0$ irreducible? I am not sure and I cannot prove it. I am actually thinking it is not because in the next question I have to prove $E_g=E_f$($E_g$ is the splitting field of $x^6-x^4-x^3-x^2+1=0$). And is $E_g$ Galois over $Q$? Thanks a lot already for your answers Lubin, I really appreciate your help. Galois-theory is so beautiful, I love exercises like this. $\endgroup$ – user2720690 Apr 29 '16 at 18:09
  • $\begingroup$ I'm away from home right now, it may take a while. $\endgroup$ – Lubin Apr 29 '16 at 22:21

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