# 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 $\alpha\in\mathbb Q(\beta)$

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

• Why is $\alpha=r_1+r_2\beta$ for some rationals $r_1$, $r_2$? – Christoph May 4 at 6:49
• If we assume α belongs to Q($\beta$). Elements of Q($\beta$) are of the form $r_1+r_2\beta$ – Nick Diaz May 4 at 6:50
• I don't think they are. Is $\beta^2$ of the form $r_1+r_2\beta$? – Christoph May 4 at 6:52
• I see Its a third degree polynomial so Elements of Q($\beta$) are of the form $r_1+r_2\beta+r_3\beta^2$ – Nick Diaz May 4 at 6:59
• 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. – Magdiragdag May 4 at 7:27