Problem 1.23 in Fulton's Algebraic curves

Question

In Fulton's book "Algebraic curves - an introduction to Algebraic Geometry" (freely available from the author's web page http://www.math.lsa.umich.edu/~wfulton/CurveBook.pdf ) problem 1.23 says: Give an example of a collection $\mathscr S$ of ideals in a Noetherian ring such that no maximal member of $\mathscr S$ is a maximal ideal.

But a lemma just before this problem says:

Lemma. Let $\mathscr S$ be any non-empty collection of ideals in a Noetherian ring $R$.Then $\mathscr S$ has a maximal member, i.e. there is an ideal $I$ in $\mathscr S$ that is not contained in any other ideal of $\mathscr S$ .

Does this mean that the only solution for the exercise is ${\mathscr S}=\emptyset$? If not, what am I missing?

Answer 1

DonAntonio's answer is nice, but unless I am missing something there seems to be a nice "trivial" example as well.

Let $\mathcal S = \{ I \}$ where $I$ is not maximal. Then the maximal element in $\mathcal S$ is $I$, but $I$ is not maximal (by definition).

Answer 2

For example, in $\;R:=\Bbb Z\;$ , take

$$\mathcal L:=\left\{\,I\le \Bbb Z\;|\;\forall\,x\in I\,,\,\,4\,\mid\,x\,\right\}$$

Find a (the) maximal element in $\;\mathcal L\;$ , and now show it is not a maximal ideal of $\;R=\Bbb Z\;$ .

Answer 3

You may consider a noetherian U.F.D. of dimension $>1$, e.g. $k[X,Y]$, where $k$ is a field, and take for $\mathscr S$ the set of of ideals generated by the irreducible polynomials, which are prime ideals of height $1$, and therefore are not maximal since $ k[X,Y]$ has dimension $2$ and is catenary.

Answer 4

In general you can just take collections of non-maximal ideals, the non-closed points from the spectrum. Take some irreducible affine curve $C$ and consider the generic point $\eta \in C$. This is basically the same example as Rubens, but in the context of curves.