Question :

Let $\alpha$ be a root of $x^3+x+1$ and $\beta$ be a root of $x^3+x+3$. Show that it is not possible that $\alpha\in \mathbb Q(\beta)$.

My proof :

Given $\beta$ is a root of $x^3+x+3$.

Thus $\beta^3+\beta+3 = 0$. Then $(\beta^3+\beta+2 ) + 1 = 0$.

If $\alpha\in\mathbb Q(\beta)$, then $\alpha = r_1 + r_2\beta$ for some rationals $r_1$ and $r_2$.

Also $(r_1+r_2\beta)^3+r_1+r_2\beta+1 = 0$.

Therefore for some rationals $r_1$ and $r_2$ we have $(r_1+r_2\beta)^3+r_1 + r_2\beta = \beta^3+\beta+2$.

But equating and solving such $r_1$ and $r_2$ doesn't exist.

Thus, $\alpha$ doesn't belong to $\mathbb Q(\beta)$.

Do you think my proof is right?

If not correct me or provide a better easier proof

  • 2
    $\begingroup$ Why is $\alpha=r_1+r_2\beta$ for some rationals $r_1$, $r_2$? $\endgroup$ – Christoph May 4 at 6:49
  • $\begingroup$ If we assume α belongs to Q($\beta$). Elements of Q($\beta$) are of the form $r_1+r_2\beta$ $\endgroup$ – Nick Diaz May 4 at 6:50
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    $\begingroup$ I don't think they are. Is $\beta^2$ of the form $r_1+r_2\beta$? $\endgroup$ – Christoph May 4 at 6:52
  • 6
    $\begingroup$ I see Its a third degree polynomial so Elements of Q($\beta$) are of the form $r_1+r_2\beta+r_3\beta^2$ $\endgroup$ – Nick Diaz May 4 at 6:59
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    $\begingroup$ You say "By equating and solving ...". Even if you repair your argument by looking at polynomials expressions in $\beta$ of degree $2$ instead of sometimes $1$ and sometimes $3$, this is a daunting task. $\endgroup$ – Magdiragdag May 4 at 7:27

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