When is $X_1^{a_1} \cdots X_n^{a_n}-1$ irreducible?

Solution 1:

Thanks to darij grinberg for the following approach. Here are the details written out:

(1): Let $R = F[X_1, ... , X_n], S = \{ (X_1 \cdots X_n)^t : t \geq 0 \}$. Then $f$ is irreducible in $R$ if and only if it is irreducible in $S^{-1}R = F[X_1, X_1^{-1}, ... , X_n, X_n^{-1}]$.

Since $R$ is a Noetherian unique factorization domain, so is $S^{-1}R$. If $f$ is irreducible in $R$, then it generates a prime ideal $Rf$, and the extension of this ideal to $S^{-1}R$ remains prime, because $Rf \cap S$ is empty.

Conversely, if $f$ is irreducible in $S^{-1}R$, then it generates a prime ideal $\mathfrak p$ of $S^{-1}R$, because $S^{-1}R$ is a unique factorization domain. For the inclusion map $R \rightarrow S^{-1}R$, $Rf$ is the contraction of its extension to $S^{-1}R$. This extension is $\mathfrak p$, so $Rf$ is a prime ideal of $R$, hence $f$ is irreducible.

(2): If $A = (\alpha_{ij}) \in \textrm{GL}_n(\mathbb{Z})$, then $A$ induces a bijection $\mathbb{Z}^n \rightarrow \mathbb{Z}^n$, which induces an $F$-linear automorphism $\phi$ of $S^{-1}R$, because $X_1^{a_1} \cdots X_n^{a_n} : a_i \in \mathbb{Z}$ is an $F$-basis for $S^{-1}R$. The only thing that isn't clear is that this is actually a ring homomorphism. To show this, it suffices to show that $\phi$ preserves multiplication of monomials. But this follows from the fact that $A(v+w) = Av+Aw$, where $v, w \in \mathbb{Z}^n$.

(3): If $a_1, ... , a_n$ are relatively prime integers, there is an $A \in \textrm{GL}_n(\mathbb{Z})$ which maps $(a_1, ... , a_n)$ to $(1, 0, ... , 0)$.

See this answer Information about Problem. Let $a_1,\cdots,a_n\in\mathbb{Z}$ with $\gcd(a_1,\cdots,a_n)=1$. Then there exists a $n\times n$ matrix $A$ ...

(4): The converse I mentioned in my question: since the GCD $d$ is equal to $1$, we can find an $A$ as in (3), hence a ring isomorphism $\phi$ which maps $f$ to $X_1 - 1$. Since $X_1 - 1$ is irreducible in $R$, it is so in $S^{-1}R$, hence $f$ is irreducible in $S^{-1}R$, hence in $R$.

Solution 2:

This is true, and can be seen easily using the Newton polytope of the polynomial and Minkowski sum of polytopes, look for this.

Try to prove that if $F=G*H$, then all the exponents $(b_1,...,b_n)$ of $G$ and $H$ must belong to the segment joining $(0,...,0)$ and $(a_1,...,a_n)$.

Hint: Prove that

1. Both $G$ and $H$ have nonzero constant term.
2. If $G$ has a monomial that does not belong to the segment $\overline{(0,\ldots,0)(a_1,\ldots,a_n)}$ then $F$ has at least a monomial that is not in the segment $\overline{(0,\ldots,0)(a_1,\ldots,a_n)}$.