Cfrgtkky

If $dmid e$, does the ideal $(x^d-1, 1+x+cdots+x^{e-1})subsetmathbb{Z}[x]$ contain $1+x+cdots+x^{d-1}$?

For an integer $n$, let $chi_n(X) := 1+X+cdots+X^{n-1}$.

Certainly $chi_d(x)$ divides both $x^d-1$ and $chi_e(x)$. Moreover, the quotient $chi_e(x)/chi_d(x)$ is coprime to $(x^d-1)/chi_d(x)$.

Since the polynomials involved are all primitive, I feel like some form of Gauss's lemma should imply the result, but I'm not sure.

As a simple counterexample, let $d=1$ and $e=2$.

Note that when $d=1$, the expression
$$1+x+cdots+x^{d-1}$$
is presumably intended to be equal to $1$, not $2$.

Then in $mathbb{Z}[x]$, the ideal $(x-1,x+1)$ does not contain $1$.

Suppose instead that
$$a(x)(x-1)+b(x)(x+1)=1$$
for some $a,bin mathbb{Z}[x]$.

Now plug in $x=1$, and note that the $text{LHS}$ is divisible by $2$, while the $text{RHS}$ is equal to $1$.

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More generally, let $d,e$ be any positive integers such that $d < e$.

If we suppose that in $mathbb{Z}[x]$, the polynomial
$$1+x + cdots + x^{d-1}$$
is an element of the ideal
$$(x^d-1,1+x + cdots + x^{e-1})$$
we would have
$$a(x)(x^d-1)+b(x)(1+x + cdots + x^{e-1})=1+x + cdots + x^{d-1}$$
for some $a,bin mathbb{Z}[x]$.

But then if we plug in $x=1$, the $text{LHS}$ is divisible by $e$, while the $text{RHS}$ is equal to $d$.