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Let $u,v$ be algebraic over the field $K$, such that $[K(u):K]=n$ and $[K(v):K]=m$. Show that if $\gcd(m,n)=1$, then $[K(u,v):K]=mn$.

Once $[K(u,v):K]=[K(u,v):K(u)][K(u):K]=n[K(u,v):K(u)]$, I'm trying to show that $[K(u,v):K(u)]=m$.

user26857
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Andre Gomes
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    What you need to be showing is that $[K(u,v):K]$ is divisible by both $n$ and $m$. Use transitivity of degrees, and the fact that $K(u)$ and $K(v)$ are both intermediate fields. – anon Aug 09 '15 at 20:37

1 Answers1

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Note $r=[K(u,v):K(u)]$ and $s=[K(u,v):K(v)]$. As $K \subset K(v)$ we have $s \le n$.

We also have $rn=sm$ and if $\gcd(m,n)=1$, $n$ divides $s$ (by Euclid lemma). So finally $n=s$ and also $m=r$.

mathcounterexamples.net
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  • Why is $[K(u,v):K(v)] \le [K(u):K]$? – Oliver G Sep 24 '16 at 19:50
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    Let $p \in K[X]$ be the minimal polynomial of $u$ over $K$. $p$ is also a polynomial of $K(v)[X]$ which vanishes at $u$. Therefore the minimal polynomial of $u$ over $K(v)$ divides $p$ and its degree is less or equal to the one of $p$. – mathcounterexamples.net Sep 24 '16 at 19:53