Is this example right (ideals of $\mathbb{Z}[x]$)?

Question

I encountered the following problem:

Let $I_{0}=\{f(x)\in \mathbb{Z}[x] \ | \ f(0)=0\}$. For any positive integer, show that there exists a sequence of ideals such that $I_0\subsetneq I_1\subsetneq I_2\subset\cdots\subsetneq I_n \subsetneq \mathbb{Z}[x]$.

The author (Gallian) porposes $\langle x\rangle \subset \langle x,2^n\rangle \subset \langle x,2^{n-1}\rangle\subset\cdots\subset \langle x,2\rangle$, but I thought in this one:

Define $I_j=\{f(x)\in \mathbb{Z}[x] \ | \ f(j)=0\}$ for $j\in \{0,1,...,n\}$. These ideals satisfies the problem, aren't they?

Thank you very much.

Answer 1

It is certainly true that for all $j \in \Bbb Z$ the set $I_j = \{f(x) \in \Bbb Z[x] \mid f(j) = 0 \}$ is an ideal: $p(x), q(x) \in I_j$ implies $p(j) - q(j) = p(0) - q(0) = 0 - 0 = 0$, so $p(x) - q(x) \in I_j$, and for any $g(x) \in \Bbb Z[x]$ we have $g(j)p(j) = g(j) (0) = 0$, so $g(x)p(x) \in I_j$; $I_j$ thus satisfies the defining requirements to be an ideal in $\Bbb Z[x]$. The problem which arises, however, is that these ideals per se do not form either an increasing or decreasing sequence: we have in general neither $I_j \subset I_k$ nor $I_k \subset I_j$ if $j \ne k$, as may be quite simply seen by choosing $f(x) = x - j \in I_j$; then $f(x) \notin I_k$ since $f(k) = k - j \ne 0$. A possible remedy for this obstacle is to consider, for any finite $\varnothing \ne A \subset \Bbb Z$ the ideal $I_A = \{ f(x) \in \Bbb Z[x] \mid f(a) = 0, a \in A \}$. $I_A$ may easily be seen to be an ideal by a simple extension of the argument presented here for $I_j$; furthermore if $\varnothing \ne B \subset A$, then $I_A \subset I_B$ is also easily seen. Taking the special case of $A_n = \{0, 1, 2, \ldots, n \} \subset \Bbb Z$, we then have, since $A_j \subset A_k$ for $0 \le j \le k$, that $I_{A_k} \subset I_{A_j}$; we may thus form a chain of ideals $I_n \subset I_{n - 1} \subset I_{n - 2} \subset \ldots \subset I_0$; other increasing sequences of ideals may easily be constructed by considering other finite (and probably, with some refinements, infinite) $\varnothing \ne A \subset \Bbb Z$ and subsets $\varnothing \ne B \subset A$; presumably the general notion is clear from what I have written here; e.g., if you want in increasing sequence of ideals you need an decreasing sequence of subsets, but they don't have to be taken from $\{0, 1, 2, \ldots, n \}$ for any $n$. Perhaps our colleague in this question, Karl Kronenfeld, was thinking along similar lines when he suggested a modification of the idea presented by the OP.

So, I hate to be the bearer of evil tidings but in fact the $I_j$ don't satisfy the requirements of the problem.

Since $\{ j \}$ is a singleton, we can't use subsets of $\{j\}$ to form larger ideals than $I_j$ as above; we need resort to some other construction. The example given in the problem is one such. My initial hunch is that there are others.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

Answer 2

If you make that choice of $I_j$, then you do not have the containments $I_j\subset I_{j+1}$. Indeed, $x-j$ is an element of the first but not the second.

There is a slight modification of your idea that does work though. Feel free to ask in the comments to this answer for hints.

Answer 3

Ideals of $\mathbb{Z}[x]$ containing $I_0=(x)$ correspond 1:1 to ideals of $\mathbb{Z}[x]/(x) = \mathbb{Z}$. Thus we only have to find arbitrary long chains of ideals of $\mathbb{Z}$, which amounts to find arbitrary long "divided by" chains of integers. For example, $(a^n)$ is such a sequence when $a>1$. The preimage in $\mathbb{Z}[x]$ is $(x,a^n)$.