Prove that if $a$ and $b$ are algebraic over field $\mathbb F$ and $b \neq 0$, then $\frac ab$ is algebraic over $\mathbb F$.
I know that algebraic means that there are polynomials $f$, $g$ such that $f(a)=0$ and $g(b)=0$, but I don't know how to continue, I have never solved problems like this..
I have seen a similar question: Prove: if $a$ and $b$ are algebraic, then $a + b$, $a - b$ and ab are also algebraic but I did not understand the answer using field extensions.
Thanks in advance!
If you are looking for an approach without field extensions:
As you already mentioned, there are monic polynomials $f,\;g$ such that $f(a)=0$ and $g(b)=0.$ Wolog $g(0)\neq 0$. If it was, the whole polynomial could be divided by the variable and would still fulfill $g(b)=0.$
Let $A$ be the companion matrix of $f$ and $B$ the companion matrix of $g.$ $B$ is invertible because $g(0)\neq 0.$ One of the eigenvalues of $A$ is $a,$ and one of the eigenvalues of $B$ is $b,$ which means that one of the eigenvalues of $B^{-1}$ is $b^{-1}.$
Now set $C=A\otimes B^{-1}$ (Kronecker product). Then one of the eigenvalues of $C$ is $ab^{-1},$ which means that $ab^{-1}$ is one of the roots of the characteristic polynomial $\chi_C$. This makes $ab^{-1}$ algebraic over $\mathbb{F},$ because all of the elements of $C$ are in $\mathbb{F}.$ Therefore, all coefficients of $\chi_C$ are in $\mathbb{F},$ too.