Past exam of the options I course of the University of Oxford / 2012 / A (Introduction to Fields)

The degree of a field extension is the dimension of the larger field, regarded as a vector space over the smaller field:The scalar multiplication is the multiplication of $K$ restricted to scalars in $F$. A finite extension has finite dimension; otherwise the degree is infinite.

The notation $K=F(\delta )$ means that $K$ is the smallest subfield containing both $F$ and $\delta$, namely the simple field extension generated by $\delta$. Its elements can be expressed as rational functions in $\delta$ with coefficients in $F$ and nonzero denominator.

Write $d={\delta }^2\in F$. The element $\delta$ is a root of $X^2-d$, so its minimal polynomial over $F$ has degree at most two. Since $\delta \notin F$, that degree cannot be one. The degree of the simple field extension equals the degree of the minimal polynomial, givingIn particular $1,\delta$ form an $F$-basis, so every element is uniquely $a+b\delta$. This argument also works in characteristic two; separability is not needed for this part.

Choose $\alpha \in K\setminus F$. Since $[K:F]=2$, the linearly independent elements $1,\alpha$ form a basis of $K$ over $F$. Thus there are $a,b\in F$ with ${\alpha }^2=a+b\alpha$. The assumption on the characteristic of a field allows division by two. SetCompleting the square givesAlso $\delta \notin F$, since otherwise $\alpha =\delta +b/2$ would lie in $F$. Because $\alpha \in F(\delta )$ and $\delta \in F(\alpha )$, these generated subfields agree. The basis representation shows $F(\alpha )=K$, henceThe square-root generator is obtained by translating an arbitrary generator to remove the linear term of its quadratic relation.