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Let $p$ be a prime and $K = \Bbb Q(α)$, where $α ^3 = p$. Find the minimal polynomial of $α + α ^2$ over $\Bbb Q$.

my attempt : I was taking $α + α ^2$= $α(1 + α )$=$0$ and getting $α =0$ and $α= -1$ as I don't know how to proceed further

Please, Help me.

Ri-Li
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    If $\alpha$ satisfies $\alpha^3-p=0$, find a polynomial satisfied by $x = \alpha+\alpha^2$. Then factor that. – GEdgar Feb 25 '18 at 12:50
  • im not getting @GEdgar –  Feb 25 '18 at 12:52
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    Just take powers of your expression and look for relations. If $\beta=\alpha+\alpha^2$ then $\beta^2=\alpha^2+2\alpha^3+\alpha^4=\alpha^2+2p+\alpha p$ and so on. – lulu Feb 25 '18 at 12:52
  • For general find-the-minimal-polynomial problems, I address one technique here (start reading at "Alternative method" halfway down the post). The downsides of this method include the Galois group not being easily computable or having a mess of radicals to deal with when computing the product of linear factors (they do simplify!). If the Galois group is well-understood and if you're content with a CAS like Mathematica doing some multiplication for you, then it's perfect. – Kaj Hansen Feb 25 '18 at 13:12
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    This post may be useful. – Viktor Vaughn Feb 25 '18 at 15:34
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    thanks a lots @Quasicoherent –  Feb 25 '18 at 15:55

3 Answers3

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$$A=\pmatrix{0&1&0\\0&0&1\\p&0&0}$$ has $\alpha$ as an eigenvalue (for eigenvector $\pmatrix{1\\\alpha\\\alpha^2}$). Then $B=A+A^2$ has $\alpha+\alpha^2$ as an eigenvalue, so that $f(\alpha)=0$ where $f$ is the characteristic polynomial of $B$.

Angina Seng
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A fully automated way to approach such questions is to use resultants or Gröbner bases. For example in this case the resultant of $x^3-p$ and $y - (x^2+x)$ with respect to $x$ turns out to be $y^3 - 3 p y - p^2 - p$ which happens to be irreducible. The same polynomial appears in the Gröbner basis of the ideal $(x^3-p, y-(x^2+x))$ with respect to the lexicographical order with $x>y>p$.

WimC
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Hint: start with $x=\alpha + \alpha^2\Rightarrow x-\alpha^2=\alpha$ then try to cube it and then try to make the coefficients of the polynomial in $\Bbb Q$.

Adding details:

After taking cube $x-\alpha^2=\alpha$ we get $x^3-3\alpha^2x^2+3\alpha^4 x-\alpha^6=\alpha^3=b \Rightarrow x^3-3\alpha^2x^2+3p\alpha x-p^2=b$

Now observe that $x^3-p=0\Rightarrow \big(\frac x {\alpha}\big)^3-1=0\Rightarrow \alpha^2+x\alpha+x^2=0$. Now you take from here.

Ri-Li
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