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

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.

• Note that $K(x)$ and $K(y)$ are both intermediate subfields. What are their degrees over $K$, and what can you conclude about $[K(x,y):K]$ from that information? – arctic tern Feb 20 at 16:57

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.