Ring Theory – Ideals of ?[X]

commutative-algebraidealsring-theory

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.

Best Answer

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.

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