Deducing if $\gcd(\deg(m_K(x)),\deg(m_K(y)))=1$ then $[K(x,y):K]=\deg(m_K(x))\times \deg(m_K(y))$.

Question

I've shown that -

If $x,y\in L$ are algebraic over $K$,then $[K(x,y):K]\le \deg(m_K(x))\times \deg(m_K(y))$.

How can we deduce from the above result that if $$\gcd(\deg(m_K(x)), \deg(m_K(y)))=1,$$then the equality holds.

Answer 1

Note that both $K (x)$ and $K (y) $ are intermediate fields of $K .$ Then by Tower rule, we can get the following two equations: $[K(x,y):K]=[K(x,y):K(x)][K(x):K]$ and $[K(x,y):K]=[K(x,y):K(y)][K(y):K].$ Observe that $[K(x):K]=\deg(m_K(x))$ and $[K (y)=K]=\deg(m_K(y)).$ Since $[K(x,y):K]$ is a multiple of both $\deg(m_K(x))$ and $\deg(m_K(y)),$ $\text {lcm}(\deg(m_K(x)), \deg(m_K(y)))$ divides $[K(x,y):K]$. As $\text {gcd}(\deg(m_K(x)), \deg(m_K(y)))=1$, we have $\text {lcm}(\deg(m_K(x)), \deg(m_K(y)))=\deg(m_K(x))\times \deg(m_K(y)).$

Hence $$\deg(m_K(x))\times \deg(m_K(y))\leq [K(x,y):K].$$ Since you've already shown $[K(x,y):K]\leq\deg(m_K(x))\times \deg(m_K(y)),$ the equality follows.