If $\alpha,\beta\in C$ are algebraic over $Q$, show that $\alpha+\beta$ is a root of a quadratic polynomial in $Q[x]$ (with conditions).

Question

I need to prove (b) to (c): I have shown that $[Q(\alpha,\beta):Q(\alpha+\beta)][Q(\alpha+\beta):Q]=2$ But i cant proceed from there any suggestions?

what if b=-1?

Edit: any hints on how to proceed from c to a would also be appreciated

Answer 1

If $\alpha=a+b\beta$, then $\alpha+\beta=a+(b+1)\beta.$ Let $f$ be the polynomial of $\beta$, we have $g(c)=f(\frac{c-a}{b+1})$ to be the polynomial of $\alpha+\beta$.

Answer 2

By assumption we have $\alpha+\beta=a+b\beta+\beta=a+(b+1)\beta$. Since $\beta$ is algebraic over $\mathbb{Q}$ with degree 2. It exists a polynomial

$f\in\mathbb{Q}[X]$ with $f(\beta)=0$ and $\deg(f)=2$.

From this you can construct a polynomial $g$ with $g(a+(b+1)\beta)=0$ with $\deg(g)=2$.