If $d\mid e$, does the ideal $(x^d-1, 1+x+\cdots+x^{e-1})\subset\mathbb{Z}[x]$ 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 $a,b\in \mathbb{Z}[x]$.
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 $a,b\in \mathbb{Z}[x]$.
But then if we plug in $x=1$, the $\text{LHS}$ is divisible by $e$, while the $\text{RHS}$ is equal to $d$.