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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.

For a simple field extension, the powers $1,\lambda ,\dots ,{\lambda }^{d-1}$ form a basis when the minimal polynomial has degree $d$: reduction modulo that polynomial spans the extension, and a relation of smaller degree would contradict minimality. Hence $[F(\lambda ):F]=d$. The field is $K$, whose dimension is $n$, giving $d=n$.

By the allowed finite multiplicative-subgroup theorem, the group $K^{\times }$ is cyclic. Let $z$ generate it. Every nonzero element of $K$ is a power of $z$, hence belongs to $F(z)$; zero belongs as well. Since the reverse inclusion is automatic,Thus a finite extension of a finite field is a simple field extension.

A simple field extension is a field extension generated by one element: $F(\alpha )$ is the smallest subfield of an ambient field containing $F$ and $\alpha$. Its elements are rational expressions in $\alpha$ with coefficients in $F$ and nonzero denominators. If $\alpha$ is algebraic, its minimal polynomial $m_{\alpha }$ identifies $F(\alpha )\cong F[t]/(m_{\alpha })$, so $[F(\alpha ):F]=\mathrm{deg}{m}_{\alpha }$. The classes of $1,t,\dots ,t^{d-1}$ give a basis when $d=\mathrm{deg}{m}_{\alpha }$.

In particular, ${\alpha }^2\in F$ but $\alpha \notin F$ implies $[F(\alpha ):F]=2$. Conversely, in characteristic different from two, any degree-two extension is of this form: for $\beta \notin F$, express ${\beta }^2=a+b\beta$ in the basis $1,\beta$, then $\alpha =\beta -b/2$ has ${\alpha }^2=a+b^2/4\in F$ and generates the same extension. The restriction on the characteristic makes division by two possible.