Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do you construct, for each $n\geq 1$, an ideal in $\mathbb Z[x]$ of the form $(a_1,a_2,\dots,a_n)$ with $a_i\in \mathbb Z[x]$ such that it is impossible to have $(b_1,b_2,\dots,b_m)=(a_1,a_2,\dots,a_n)$ with $m<n$ and $b_j\in\mathbb Z[x]$?

share|improve this question
    
What if you consider the ideal $I = (p, f)$ where $p$ is a prime number and $f$ is an irreducible polynomial mod $p$? –  user38268 Feb 17 '12 at 19:55
    
I did a body & title edit. If my edit gets reviewed, then it is worth noting that I changed the title from "Construction of a special ideal" to ``... with least number of generators'' to reflect (my interpretation of) the condition "impossible .. with $m < n$". –  user2468 Feb 17 '12 at 19:57
    
Without any extra requirement on the ideal, any principal ideal will work. –  Brandon Carter Feb 17 '12 at 19:57
    
@BenjaminLim: So something like $(2,x)$? what about for some arbitrary $n$? –  rad Feb 17 '12 at 20:00
    
@BrandonCarter: Are you saying that any principal ideal cannot b generated by any fewer elements? –  rad Feb 17 '12 at 20:01
show 16 more comments

1 Answer 1

up vote 5 down vote accepted

This is just an elucidation of the hint given in this MO post.

Let $p$ be a prime. The claim is that the ideal $I_n=(p^n, p^{n-1}x, p^{n-2}x^2,\ldots, px^{n-1}, x^{n})$ cannot be generated by fewer than $n+1$ elements.

Claim. $I_n=(p,x)^n$.

Proof of claim. Induction on $n$: if $n=1$, then $I_1=(p,x)$ and we are done.

Assume the result holds for $n$, and let $I_{n+1}=(p^{n+1},p^{n}x,\ldots,px^{n},x^{n+1})$. Then $$I_{n+1} = I_n(p,x) = (p^n,p^{n-1}x,\ldots,px^{n-1},x^n)(p,x);$$ indeed, elements of the right hand side are sums of the form $$\sum_{i=1}^n a_ib_i$$ where $a_i\in (p^n,p^{n-1}x,\ldots,x^n)$ and $b_i\in (p,x)$ for each $i$. If $a_i = k_0p^n + k_1p^{n-1}x + \cdots + k_{n-1}px^{n-1} + x^ng(x)$ and $b_i = \ell_0p + xh(x)$, then the constant term of $a_ib_i$ is congruent to $0$ modulo $p^{n+1}$; the linear term is congruent to $0$ modulo $p^n$; the quadratic term is congruent to $0$ modulo $p^{n-1}$; etc. so $a_ib_i\in I_{n+1}$ for each $i$, hence the sum is in $I_{n+1}$.

Conversely, any element of $I_{n+1}$ can be written as $$b_0p^{n+1}+b_1p^nx + \cdots + b_npx^{n} + x^{n+1}g(x).$$ Since $b_ip^{n+1-i}x^i = (b_ip^{n-i}x^i)p \in (p^n,p^{n-1}x,\ldots,x^n)(p,x)$, each of the first $n+1$ summands lie in the product; and $x^{n+1}g(x) = (x^ng(x))x$, which also lies in the product. Thus, $I_{n+1}$ is contained in the product, giving equality.

By the Induction Hypothesis, $$I_{n+1} = (p^n,p^{n-1}x,\ldots,px^{n-1},x^n)(p,x)= I_n(p,x) = (p,x)^n(p,x) = (p,x)^{n+1}.$$ This proves the claim. $\Box$

Now consider $I_{n}/I_{n+1} = (p,x)^n/(p,x)^{n+1}$. This makes sense, since in general $IJ\subseteq I\cap J\subseteq I$, so $I_{n+1}$ is contained in $I$.

Consider the generators: $p^n\in I_n$ maps to an element of order dividing $p$ in the quotient (since $p^{n+1} = pp^n\in (p,x)^{n+1}$); $p^nx$ likewise maps to an element of order $p$; and so on; every generator of $I_n$ is of order dividing $p$ in the quotient. That means that, as an abelian group, $I_n/I_{n+1}$ is of exponent $p$.

Since we have a finitely generated abelian group of exponent $p$, it is isomorphic, as an abelian group, to a direct sum of copies of the cyclic group of order $p$. The number of cyclic summands is equal to the dimension of $I_n/I_{n+1}$ as a vector space over $\mathbb{F}_p$, the field with $p$ elements.

I claim that the images of the generating set of $I_n$ are linearly independent in $I_n/I_{n+1}$. Indeed, suppose that $\alpha_0,\ldots,\alpha_n\in\mathbb{F}_p$ are such that $$\alpha_0[p^n] + \alpha_1[p^{n-1}x] + \cdots + \alpha_n[x^n] = [0],$$ where $[r]$ denotes the image of $r\in I_n$ in the quotient. Replacing $\alpha_i$ with an integer, $0\leq \alpha_i\lt p$, this amounts to saying that $$\alpha_0p^n + \alpha_1p^{n-1}x + \cdots + \alpha_nx^n \in I_{n+1}.$$ In order for this polynomial to lie in $I_{n+1}$, we need $p^{n+1}|\alpha_0p^n$, $p^n|\alpha_1p^{n-1},\ldots,p|\alpha_n$. Since $\alpha\neq 0$ implies $(p,\alpha_i)=1$ by construction, this is only possible if $\alpha_i=0$ for each $i$. Thus, the images of the generators of $I_n$ are linearly independent in $I_n/I_{n+1}$.

Therefore, $\dim_{\mathbb{F}_p}(I_n/I_{n+1})\geq n+1$. Since the dimension is less than or equal to the size of any generating set for $I_n$, it follows that if $I_n$ can be generated by $m$ elements, then $m\geq n+1$; since we know $I_n$ can be generated by $n+1$ elements, it follows that the minimum size of a generating set for $I_n$ is $n+1$, as desired.

share|improve this answer
    
Thank you very much! –  rad Feb 18 '12 at 23:27
    
I wonder if this can be modified to work for $(x^n,2x^{n-1},6x^{n-2},30x^{n-3},\dots,(2)(3)(5)\cdots(p_n))$, where the coefficient of $x^{n-k}$ is the product of the first $k$ primes. –  Gerry Myerson Feb 18 '12 at 23:27
    
@Gerry: I think it would. –  Arturo Magidin Feb 19 '12 at 22:32
    
Where is the claim that $I_n = (p,x)^n$ used in the subsequent argument? I thought the facts $I_{n+1} \subseteq I_n$ and $I_n / I_{n+1}$ having exponent $p$ may be shown directly without using the claim? –  hwhm Dec 21 '12 at 15:42
    
The claim is trivial and follows from the definition of the ideal product $I_1 \cdot \dotsc \cdot I_n$ as the smallest ideal containing all products $a_1 \cdot \dotsc \cdot a_n$ with $a_i \in I_i$, neither induction nor sums are needed. What neads more reasoning is the sentence "In order for this polynomial to lie in $I_{n+1}$, we need ...". –  Martin Brandenburg Feb 11 '13 at 16:46
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.