Section 4.6 Kronecker’s Theorem
Proof.
\begin{equation*}
h=qf+r,\quad\text{i.e.,}\quad h+(f)=r+(f)\in K.
\end{equation*}
Then \((f)+(r)=F[X]\text{.}\) Indeed, if \(d=\gcd(r,f)\) then \(d\mid r\) and \(d\mid f\text{.}\) As \(f\) is irreducible, \(d\) is a unit or \(d=f\text{.}\) Since \(d\mid r\) and \(\deg(r)\lt\deg(f)\text{,}\) we must have \(d\) is a unit. Hence \((r)+(f)=F[X]\text{.}\) Thus there exists \(x,y\in F[X]\) such that \(rx+fy=1\text{.}\) Thus
\begin{equation*}
rx\equiv 1\mod f
\end{equation*}
i.e., \(r\) is invertible in \(K\text{.}\) This shows that \(K\) is a field. We now show that \(f\text{,}\) when considered as a polynomial over \(K\text{,}\) has a root in \(K\text{.}\) An element \(r\in F[X]\) when considered as an element in \(K\) is denoted by \(\overline{r}\text{.}\) Suppose that \(f(X)=a_0+a_1X+\cdots+a_nX^n\in F[X]\text{.}\) Consider the polynomial \(\overline{f}(t)=\overline{a_0}+\overline{a_1}t+\cdots+\overline{a_n}t^n\in K[X]\text{.}\) We show that \(\overline{X}\) is a root of \(\overline{f}\text{.}\) Indeed,
\begin{align*}
\overline{f}(\overline{X})\amp=\overline{a_0}+\overline{a_1}\overline{X}+\cdots+\overline{a_n}\overline{X}^n\\
\amp= (a_0+a_1X+\cdots+a_nX^n)+(f)\\
\amp=0\in K
\end{align*}
Corollary 4.6.2. (Prime if and only if maximal).
A nonzero ideal of \(F[X]\) is prime if and only if it is maximal.Proof.
Suppose that \(I\) is a nonzero prime ideal. Then there exists an irreducible polynomial \(f\in F[X]\) such that \(I=(f)\) (see Proposition 4.5.9). By Theorem 4.6.1, \(F[X]/I\) is a field. By correspondence theorem, every ideal of \(F[X]/I\) is obtained by an ideal of \(F[X]\) containing \(I\text{.}\) As \(F[X]/I\) is a field there are no proper ideals of \(F[X]\) that contains \(I\text{.}\) Hence \(I\) is maximal.
Conversely, assume that \(I\) is maximal. Suppose that \(I=(f)\) (see Theorem 4.5.5). If \(f\) is reducible then there exists \(g,h\in F[X]\) such that \(f=gh\text{.}\) In this case \(I\subseteq (g)\) and \(I\subseteq (h)\text{.}\) The maximality of \(I\) implies that either \(I=(g)\) or \((g)=F[X]\) (similarly for \(h\)). If \(I=(g)\) then \(f\mid g\) and there exists \(k\in F[X]\) such that \(g=fk\text{.}\) Thus, \(f=gh=fkh\text{.}\) Since \(F[X]\) is an integral domain, \(kh=1\text{.}\) This shows that \(h\) is a constant polynomial. If \((g)=F[X]\) then \(g\) is a unit. Therefore, we obtain that \(f\) is irreducible. By Proposition 4.5.9 we get the desired result.
Observation 4.6.3.
We keep the notations of Theorem 4.6.1. We have a ring monomorphism given by composition:
\begin{equation*}
F\xrightarrow{i}F[X]\xrightarrow{\pi}F[X]/(f)
\end{equation*}
given by
\begin{equation*}
i(\alpha)=\alpha 1\quad\text{and}\quad\pi(g(X))=g(X)+(f)=g(\overline{X}).
\end{equation*}
Thus \(F\simeq \pi\circ i(F)\subset K\text{.}\)
