Let $K$ be a field extension of the field $F$ and $a,b \in K$ be algebraic over $F$.If a has degree $m$ over $F$ and $b \neq 0$ has degree $n$ over $F$, then show that the elements $a + b, ab, a − b , ab^{−1}$ each has at most degree $mn$ over $F$.[Hint: first show $a + b, ab, a − b , ab^{−1}$ are algebraic over $F$]
We know that $a + b, ab, a − b , ab^{−1}$ are algebraic over $F$ which comes from showing that $[F(a,b)]:F] \leq[F(a):F][F(b):F]=mn = \text{finite}$
and as all these elements $a + b, ab, a − b , ab^{−1} \in F(a,b)$ and as every element of finite extension is algebraic so these are algebraic.
Now $F(a,b)$ is a vector space over $F$ with a basis of atmost mn elements. From this how to show that degree of minimal polynomial of $a + b, ab, a − b , ab^{−1}$ is atmost mn?
It's direct. Note that $F(a+b)\subset F(a,b)$, $F(ab)\subset F(a,b)$, $F(a-b)\subset F(a,b)$ and $F(ab^{-1})\subset F(a,b)$. So each one is a subextension of the original extension, so the degree of each one must divide $mn$, , so it coud be any divisor of $mn$, from $1$ (trivial subextension) to $mn$ (that implies that the element is a primitive element of the extension).