If $E/F$ is a field extension with $a,b\in E$ of "degree" $n,m$ then $ab$ and $a+b$ have "degree" less then $mn$

Question

Let $E/F$ be a field extension. Let $a,b\in E$ be algebric in $F$. Assume that $$\deg(\text{irr}(a,F))=n,\deg(\text{irr}(b,F))=m$$ Show that $$\deg(\text{irr}(t,F))\leq mn$$ where $t=ab,a+b$.

I'm not sure how to approach this: If we set $$f,g=\text{irr}(a,b;F)$$ then $(fg)(ab)$ or $(fg)(a+b)$ does not zeroed.

Answer 1

$F(a)$ has degree $n$ and the degree of $F(a,b)$ is $[F(a,b):F(a)][F(a):F]$ the degree of $[F(a,b):F(a)]$ is inferior to $m$ since a basis of the $F$ field $F(b)$ is a system of generators of the $F(a)$ field $F(a,b)$; this implies that $[F(a,b):F]$ is inferior to $mn$ since $a+b, ab\in F(a,b)$ this implies the result.

Answer 2

If you write $\{a_1,\dots, a_n\}$ the roots of $\text{irr}(a,F)$ (on some algebraic closure of $E$) with $a_1=a$, and analogously $\{b_1,\dots, b_m\}$ the roots of $\text{irr}(b,F)$, then the polynomials $$ \prod_{1\le i \le n, 1\le j \le m} (X-(a_i+b_j)) $$ and $$ \prod_{1\le i \le n, 1\le j \le m} (X-(a_ib_j)) $$ have both coefficients in $F$ by Newton's result on symmetric polynomials, degree $nm$ and roots $a+b$ and $ab$ respectively. So the respective irreducible polynomials divide them, and have degree $\le nm$.

Answer 3

Note that $$ [F(a,b):F(a)]=\deg\text{irr}(b,F(a)) $$ Due to the fact that $\text{irr}(b,F)\in F[x]\subseteq F(a)[x]$ is zeroes by $b$ we get that $$ \text{irr}(b,F(a)) | \text{irr}(b,F) $$ and thus $$ [F(a,b):F(a)]=\deg(\text{irr}(b,F(a))\leq\deg(\text{irr}(b,F)=m $$ From the building property $$[F(a,b),F]=[F(a,b:F(a)]\cdot [F(a):F]\leq mn\Rightarrow \\ \deg\text{irr}(t,F)=[F(t):F]\leq[F(t):F]\leq mn$$ where $t=ab,a+b$.