# Ideals of $\mathbb{Z}[X]$

Is it possible to classify all ideals of $\mathbb{Z}[X]$? By this I mean a preferably short enumerable list which contains every ideal exactly once, preferably specified by generators. The prime ideals are well-known, but I'm interested in all the ideals. I couldn't find any literature about this. I don't even know the number of generators we need (Edit: Hurkyl has pointed out that any number appears!).

An ideal of $\mathbb{Z}[X]$ restricts to some ideal of $\mathbb{Z}$, say $n \mathbb{Z}$. For $n>0$ the ideal corresponds to some ideal in $\mathbb{Z}/n [X]$, and by CRT we may assume that $n$ is a prime power, say $n=p^k$. For $k=1$ we have the PID $\mathbb{F}_p[X]$, whose ideal structure is well-understood. What happens for $k=2$? The case $n=0$ seems to be even more complicated.

Since $\mathbb{Z}[X]$ is noetherian, every ideal is an intersection of primary ideals, and we should classify them first? What about other special classes of ideals, for example radical ones?

If the question is too naive in this generality, what about special cases, or slight weakenings? For example, is it possible to classify all (reduced, finite, coprimary, ...) commutative rings which are generated by a single element? This comes down to the classification of the ideals of $\mathbb{Z}[X]$ modulo isomorphic quotient rings.

• The ideal $(2,x)^n$ requires $n+1$ generators.
– user14972
Feb 11, 2013 at 12:59
• Also, an ideal requiring $2$ generators may still satisfy $I \cap \mathbb{Z} = 0$, such as $I = (x^2, 2x)$
– user14972
Feb 11, 2013 at 13:04
• To address the radical ideal blurb: If you know the prime ideals you know the radical ideals: they are just all possible intersections of prime ideals. Feb 11, 2013 at 14:59
• @rschwieb: Oh, yes! Is it also possible to give generators? I will think about this ... Feb 11, 2013 at 15:01
• @MartinBrandenburg Beyond my ken :) Glad you asked this question so that I can slowly learn commutative algebra too... Feb 11, 2013 at 15:04

First some simple facts:

The ring $\mathbb{Z}[X]$ has dimension $2$ and it is a unique factorization domain with unit group $\{\pm 1\}$. We immediately conclude:

• Every ideal has height $0$, $1$, or $2$.
• The only ideal of height $0$ is $(0)$.
• If the primary decomposition of an ideal contains only height $1$ ideals, then:
• The ideal is principal
• These ideals are in one-to-one correspondence with nonzero polynomials with positive leading coefficient

Now let $I$ be a nonprincipal ideal. Suppose $I \cap \mathbb{Z} = (0)$. Then $I \otimes \mathbb{Q}$ is a proper, nonzero ideal of $\mathbb{Q}[X]$, generated by a primitive polynomial $g(x)$ (i.e. integer coefficients of gcd 1). It follows that $I = g(x) J$ where $J \cap \mathbb{Z} \neq (0)$.

Alternatively, $g(x)$ can be obtained a the gcd of all elements of $I$. (the $g(x)$ so constructed would have content $n$ rather than $1$ if $I$ is divisible by $(n)$). But the above is the method of proof I thought of to show the cofactor contains an integer.

Taking a different tactic, essentially by the Euclidean algorithm any ideal has a basis of the form

$$I = \langle g_i(x) \mid 1 \leq i \leq k \rangle$$

where

• $g_i(x) = c_i x^{n_i} + f_i(x)$
• where $c_i \in \mathbb{Z}$, $c_i > 0$, $\deg f_i(x) < n_i$
• $n_i < n_{i+1}$
• $c_{i+1} | c_i$ and $c_i \neq c_{i+1}$

For example, the ideal

$$\langle 8, 4x + 4, 2x^2 + 2 \rangle$$

In fact, I'm pretty sure this works out to a normal form by the algorithm:

• For each $d$, pick (if any) the smallest polynomial of degree $d$ in $I$
• Throw out any polynomial whose leading term is divisible by a picked polynomial of lesser degree

"Smallest", here, is determined by first comparing the coefficients on $x^d$, and if those are equal comparing the coefficients on $x^{d-1}$, and so forth. Integers are compared by absolute value first, and if equal, $n$ is considered smaller than $-n$.

e.g. the polynomial $4x + 4$ is considered smaller than $8x$ and than $4x-4$.

Unfortunately, not every choice of $g_i$'s is admissible. For example, the ideal

$$\langle 4, 2x+1 \rangle$$

actually has normal form

$$\langle 1 \rangle$$

I conjecture a sequence of reduced $g_i$'s (meaning we reject $\langle 8, 4x-4 \rangle$ because we can reduce $4x-4$ to $4x+4$ by a monomial multiple of $8$) satisfying the above properties is the normal form of an ideal if and only if $c_i \mid f_i(x)$.

Alas, I don't know enough about the theory of Groebner bases over Euclidean rings to say for sure.

• Thank you. Could you explain a little bit more the paragraph beginning with "Now let $I$ be a height $2$ ..."? I don't really understand your arguments. Feb 11, 2013 at 15:00
• @Martin: If $I$ contains no integers, then we can invert every integer without killing $I$, making it an ideal of $\mathbb{Q}[X]$.
– user14972
Feb 11, 2013 at 15:52
• How do you define $J$, and how do you prove $I=g(x) J$? Feb 11, 2013 at 16:54
• @Martin: If $f \in I$, then there exists $q \in \mathbb{Q}$ and $h \in \mathbb{Q}[X]$ such that $fq = gh$. By adjusting $q$, we may assume $h \in \mathbb{Z}[X]$. If $g$ has content $1$, then $f = gh/q$ implies $g | f$ (and that $h/q \in \mathbb{Z}[X]$).
– user14972
Feb 11, 2013 at 17:00
• Dear Hurkyl, I think a height $1$ ideal is not necessarily principal, e.g. $(6, 2X)$; and a height $2$ ideal contains necessarily a non-zero integer (see my answer). Your example $I=g(X)J$ is height $1$.
– user18119
Feb 11, 2013 at 22:35

There are generalizations of Grobner/standard ideal basis algorithms that work over coefficient rings more general than fields, e.g. certain PIDs (which, in particular, apply to $$\rm\,\Bbb Z[x]).$$ There is, by now, much literature on such topics. Perhaps the easiest way to chase it down is to start with the references listed in Buchberger's survey [1] on the history of the Grobner basis algorithm.

See in particular the work mentioned in section "Generalizations of the CPC approach for polynomial rings", p. 19. If you chase citations of the work mentioned there, it will quickly lead you to the most recent work on said generalizations.

[1] B. Buchberger. History and basic features of the critical pair / completion procedure.
Journal of Symbolic Computation, 3(1/2) (1987) 3-38

The earliest such classification (of canonical generating sets of ideals in $$\mathbb{Z}[X]$$) I could find is from 1952, due to Szekeres. I summarize the main points:

Theorem: Let $$I \subseteq \mathbb{Z}[X]$$ be an ideal. Let $$f \in \mathbb{Z}[X]$$ be the polynomial with positive leading coefficient defining the smallest principal ideal containing $$I$$. Then there exists $$m \in \mathbb{N}$$, and $$q_k, b_{k,i} \in \mathbb{Z}$$ satisfying $$0 \le b_{k,i} < q_k, \qquad \forall 1 \le k \le m, \; 0 \le i < k \tag{*}\label{*}$$ such that recursively defining $$g_k \in \mathbb{Z}[X]$$ via \begin{align*} g_0 &= \prod_{i=1}^m q_i \\ q_kg_k &= Xg_{k-1} + \sum_{i=0}^{k-1} b_{k,i} g_i, \qquad k > 0 \end{align*} % g_k = \Big( \prod_{j > k}^m q_j \prod_{j=0}^{k-1} (x + b_{j+1,j}) \Big) + \sum_{i=1}^{k-1} b_{ki} \Bigg[ \prod_{j > i, j \ne k}^m q_j \prod_{j=0}^{i-1} (x + b_{ij}) \Bigg] one has $$I = (fg_0, fg_1, \ldots, fg_m)$$. Conversely, for any choice of $$f, q_k, b_{k,i}$$ satisfying $$\eqref{*}$$, there is an ideal generated by $$fg_0, \ldots, fg_m$$.

Remarks:

1. $$f$$ can always be obtained as the gcd of any generating set of $$I$$ (up to sign), so it suffices to consider the case $$f = 1$$, i.e. $$I$$ has height $$2$$, in which case $$I \cap \mathbb{Z} \ne 0$$ (by expressing $$1$$ as a $$\mathbb{Q}[X]$$-linear combination of generators and clearing denominators). This is the setting that Szekeres works in.
2. It follows from the recursive definition that for all $$k$$, $$\deg g_k = k$$ and the leading coefficient $$a_k$$ of $$g_k$$ satisfies $$a_{k-1} = q_ka_k$$. In particular $$a_k \mid a_{k-1}$$, and $$a_m = 1$$, i.e. $$g_m$$ is monic.
3. (Assuming $$f = 1$$) The $$g_k$$ are canonical in the following sense: for each $$k$$, $$a_k$$ is the least positive integer occurring as the leading coefficient of a degree $$k$$ polynomial in $$I$$ (this property is called minimal by Szekeres, but I avoid this to prevent confusion with the minimality in (5)).
4. The conditions in (2) alone are not sufficient for a canonical enumeration in the sense of (3). E.g. $$(4, 2X+2, X^2+X+1)$$ does not occur as a valid $$(g_0, g_1, g_2)$$, as the ideal would contain $$2(X^2+X+1) - X(2X+2) = 2$$ (so $$4$$ is not the canonical generator of degree $$0$$).
5. The generating set $$fg_0, \ldots, fg_m$$ need not be a minimal generating set, but by (3), any non-minimality is "obvious", i.e. only occurs when $$q_k = 1$$. Thus the generating set can be made minimal by removing the $$fg_k$$ with $$q_k = 1$$. E.g. the ideal in (4) is $$(2, X^2+X+1)$$, which has $$f = 1$$, $$m = 2$$, $$q_1 = 1$$, $$q_2 = 2$$, $$b_{1,0} = 0$$, $$b_{2,0} = b_{2,1} = 1$$, and $$g_0 = 2$$, $$g_1 = 2X$$ (which is redundant), and $$g_2 = X^2 + X + 1$$.
• For a non-paywall link to Szekeres' paper, see here Jan 4 at 16:15

Say $$I$$ is an ideal. If $$I=0$$, we stop.
If $$I\neq 0$$, we look at the set of the elements of $$I\setminus \{0\}$$ with minimal degrees, in this set we pick an element $$f_1$$ with minimal positive leading coefficient. If $$I=(f_1)$$, we stop.
If not, consider the set $$I\setminus (f_1)$$. Similar as above, pick an element $$f_2$$ with minimal positive leading coefficient in the subset of the elements with minimal degree. If $$I=(f_1,f_2)$$, we stop. Repeat this process. With finite steps, we can stop at a step $$I=(f_1,f_2,\ldots,f_n)$$.
We can see that the degree of $$f_{i+1}$$ is strictly greater than the degree of $$f_i$$ for $$i=1,2,\ldots, n-1$$.
Say the leading coefficient of $$f_i$$ is $$a_i$$. Then $$a_{i+1}| a_i$$ and $$a_{i}$$ is strictly greater than $$a_{i+1}$$. So we can see the length of the chain can be bounded by $$a_1$$. For example, if $$a_1=1$$, then $$I=(f_1)$$, if $$a_1$$ is prime, then $$I$$ is generated by one or two elements. If $$a_1=p^2$$, $$p$$ is prime, then $$I$$ can be generated by 3 elements.