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Let $E$ be an extension of $F$, and let $a, b \in E$ be algebraic over $F$. Suppose that the extensions $F(a)$ and $F(b)$ of $F$ are of degrees $m$ and $n$, respectively, where $(m,n)=1$. Show that $[F(a,b):F]=mn$.

Since $[F(a,b):F]=[F(a,b):F(a)][F(a):F]$ and $[F(a):F]=n$ we have $n|[F(a,b):F]$ with the same argument we prove that $m|[F(a,b):F]$, then $mn|[F(a,b):F]$ and $mn \le [F(a,b):F]$.

My problem is with the converse, I need help.

Thank you

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up vote 3 down vote accepted

You should be able to prove the degree of $F(a,b)$ over $F(a)$ is at most the degree of $F(b)$ over $F$, and what you want follows from that.

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Note that $F(a,b)=(F(a))(b)$. take a basis for $F(a)/F$ and for $F(b)/F$. What is a sapning set for $(F(a))(b)/F$ ?

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what is sapning set? – user42912 Nov 1 '12 at 16:52
@user42912 - Do you remember what a basis is ? (I may be miss-spelling, probably am) – Belgi Nov 1 '12 at 18:35
The term is "spanning." – anon Nov 15 '12 at 15:08

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