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!
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.