Past exam of the options I course of the University of Oxford / 2006 / B / B1

A field extension $K/F$ is finite when $K$, regarded as a vector space over $F$, has finite dimension. Its degree of a field extension isThus a finite basis $u_1,\dots ,u_m$ means that every $k\in K$ is uniquely expressible as ${\sum }_i{c}_i{u}_i$ with coefficients $c_i\in F$.

Since $K$ is an $F$-vector subspace of the finite-dimensional vector space $L$, it has a finite $F$-basis $u_1,\dots ,u_m$. Any finite $F$-spanning set for $L$ also spans $L$ over $K$, so choose a finite $K$-basis $v_1,\dots ,v_r$ of $L$.

Every $\ell \in L$ can be expanded first in the $v_j$ and then in the $u_i$, showing that the products $u_i{v}_j$ span $L$ over $F$. If ${\sum }_{i,j}{c}_{ij}{u}_i{v}_j=0$ with $c_{ij}\in F$, independence of the $v_j$ over $K$ gives ${\sum }_i{c}_{ij}{u}_i=0$ for every $j$; independence of the $u_i$ then gives every $c_{ij}=0$. The products form an $F$-basis, proving the tower law for field extensions:

Put $E=F(a,b)$. The tower law for field extensions shows that both $m$ and $n$ divide $[E:F]$. Since $m,n$ are coprime, their product divides $[E:F]$, so $[E:F]\ge mn$.

The minimal polynomial of $b$ over $F$ has degree $n$ and is still an annihilating polynomial over $F(a)$. The minimal polynomial over $F(a)$ divides it, soThe upper and lower bounds agree:Coprimality is the step forcing the lower bound to equal the product; without it the two generated fields can overlap substantially.