For $a,b$ algebraic over field $K$ it holds $[K(a,b):K] \leq [K(a):K]\cdot [K(b):K]$

Question

For $a,b$ algebraic over field $K$ I need to show that it holds that $$[K(a,b):K] \leq [K(a):K]\cdot [K(b):K]$$

Now since $a,b$ algebraic $[K(a,b):K]< \infty,[K(a):K]:=n < \infty,[K(b):K]:=m < \infty$ and with the degree formula we get

$$ [K(a,b):K] =[K(a,b):K(a)]\cdot [K(a):K] = x\cdot m $$ and also $$ [K(a,b):K] =[K(a,b):K(b)]\cdot [K(b):K] = y\cdot n $$ Where do I go from here?

Answer 1

Let $p(x)$ be the minimal polynomial of $a$ over $K$, and let $q(x)$ be the minimal polynomial of $b$ over $K$. Let $n$ and $m$ be the degrees of $p(x)$ and $q(x)$ respectively. Hence we have that

$$[K(a):K]=n\quad\text{and}\quad[K(b):K]=m.$$

Let $f(x)$ be the minimal polynomial of $b$ over $K(a)$, and let $r$ be the degree of $f$. So we have that

$$[K(a,b):K(a)]=r.$$

Now note that $q(x)$ is a polynomial with coefficients in $K(a)$ with $q(b)=0$. It follows that $f(x)$ divides $q(x)$. Hence $r\le m$. Thus

$$[K(a,b):K]=[K(a,b):K(a)]\cdot[K(a):K]=rn\le mn=[K(b):K]\cdot[K(a):K].$$