Let \(E/F\) be an extension of fields and \(X\) be a subset of \(E\text{.}\)
The ring generated by \(X\) is the intersection of all subrings of \(E\) containing \(X\text{,}\) denoted by \(F[X]\text{.}\) If \(X=\{a_1,a_2,\cdots,a_n\}\) then, we write \(F[X]=F[a_1,a_2,\cdots,a_n]\text{.}\)
The field generated by \(X\) is the intersection of all subfields of \(E\) containing \(X\text{,}\) denoted by \(F(X)\text{.}\) If \(X=\{a_1,a_2,\cdots,a_n\}\) then, we write \(F(X)=F(a_1,a_2,\cdots,a_n)\text{.}\)
Consider the evaluation \(F\)-algebra homomorphism \({\mathrm ev}_a\colon F[x]\to E\) given by \(p(x)\mapsto p(a)\text{.}\) The image of this homomorphism is a subring of \(E\) containing \(a\text{.}\) By the closure of addition and multiplication the image of \({\mathrm ev}_a\) is contained in every subring of \(E\) containing \(F\) and \(a\text{.}\) Thus, the image of \({\mathrm ev}_a\) is the ring generated by \(a\text{,}\) i.e., \(F[a]\text{.}\)
Keep the notations of the above Proposition 26. If the kernel of the evaluation homomorphism \(\ker(\ev_a)\neq (0)\) (note that \(\ker(\ev_a)\neq F[x]\) as ring homomorphisms maps unity to unity and that \(F[x]\) is not a zero ring) then, \(F[x]/\ker(\ev_a)\) is a field. Therefore, we ge that
\begin{equation*}
F[a]=F(a).
\end{equation*}
The \(\ker(\ev_a)\) is nonzero (resp. zero) if and only if \(a\) is a root of some nonzero polynomial in \(F[x]\text{,}\) i.e., \(a\) is algebraic (resp., transcendental) over \(F\text{.}\)
