Past exam of the options I course of the University of Oxford / 2004 / B

The inclusion $F\subseteq K$ makes $K$ a vector space over $F$ by its own addition and multiplication by elements of $F$. The extension is finite when this vector space has a finite basis. Its degree of a field extension isBeing a finite field extension concerns vector-space dimension, not the cardinality of the underlying field.

Since $K\subseteq L$ and $L/F$ is finite, an $F$-linearly independent set in $K$ remains independent in $L$, so $K/F$ is finite. A finite $F$-spanning set of $L$ also spans it over $K$, so $L/K$ is finite.

Take an $F$-basis $a_1,\dots ,a_m$ of $K$ and a $K$-basis $b_1,\dots ,b_n$ of $L$. Expanding coefficients twice shows that all products $a_i{b}_j$ span $L$ over $F$. If ${\sum }_{i,j}{c}_{ij}{a}_i{b}_j=0$ with $c_{ij}\in F$, independence of the $b_j$ forces ${\sum }_i{c}_{ij}{a}_i=0$ for each $j$, and then independence of the $a_i$ forces every coefficient zero. Thus these products form an $F$-basis, proving

The polynomial $x^2-2$ is an irreducible polynomial over $\mathbb{Q}$, so $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$. This intermediate field is real, so it does not contain $i$. Hence $x^2+1$ has no root in it and adjoining $i$ has degree two. The tower law givesA basis is $1,\sqrt{2},i,i\sqrt{2}$. Square-bracket adjoining notation in this algebraic example gives the same field as round-bracket adjoining notation.