It is a fact that every non-trivial ring \(R\) contain at least one maximal
ideal. Showing this in general requires some fancy set-theoretic notions. This
exercise develops a simple proof in the case where \(R\) is countable (see \(\S\)
A3).
(a) Show that if \(R\) is non-trivial but finite, then it contains a maximal
ideal.
(b) Assume that \(R\) is countably infinite, and let \(a_{1}, a_{2}, a_{3},
\ldots\) be an enumeration of the elements of \(R\). Define a sequence of ideals
\(I_{0}, I_{1}, I_{2}, \ldots,\) as follows. Set
\(I_{0}:=\left\\{0_{R}\right\\},\) and for each \(i \geq 0,\) define
$$
I_{i+1}:=\left\\{\begin{array}{ll}
I_{i}+a_{i} R & \text { if } I_{i}+a_{i} R \subsetneq R \\
I_{i} & \text { otherwise }
\end{array}\right.
$$
Finally, set \(I:=\bigcup_{i=0}^{\infty} I_{i},\) which by Exercise 7.37 is an
ideal of \(R\). Show that \(I\) is a maximal ideal of \(R\). Hint: first, show that
\(I \subsetneq R\) by assuming that \(1_{R} \in I\) and deriving a contradiction;
then, show that \(I\) is maximal by assuming that for some \(i=1,2, \ldots,\) we
have \(I \subsetneq I+a_{i} R \subsetneq R,\) and deriving a contradiction.