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A subgroup of $\mathbb{Z}$ has a generator, the g.c.d of its elements.

Is there such a result for a subgroup of $\mathbb{Z}^n$ ? (= Can we always find m generators ?)

Context: I'm reading a book on Galois's theory and it says: if $M = \mathbb{Z}[e^{i\pi/n}]$, M is a finite $\mathbb{Z}$ module and so $M \cap \mathbb{Q}$ is a finite $\mathbb{Z}$ module too (and hence equals $\mathbb{Z}$). I can prove the part "finite module / ring on integers $\implies$ equals $\mathbb{Z}$" but Im stuck for the other part.

Thomas
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1 Answers1

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Let $\zeta = e^{2\pi i /n}$. Then $\zeta$ is a root of the polynomial $\frac{x^n-1}{x-1} = x^n + x^{n-1} + \cdots + x + 1$. This shows that $\zeta$ is integral over $\mathbb{Z}$, which is equivalent to $\mathbb{Z}[\zeta]$ being a finitely generated $\mathbb{Z}$-module (see the section equivalent definitions in the above link). Thus $\mathbb{Z}[\zeta]$ is an integral extension of $\mathbb{Z}$, so $\mathbb{Z}[\zeta] \cap \mathbb{Q}$ is also an integral extension of $\mathbb{Z}$. But since $\mathbb{Z}$ is integrally closed, then every element of $\mathbb{Q}$ that is integral over $\mathbb{Z}$ already belongs to $\mathbb{Z}$. Since $\mathbb{Z}[\zeta] \cap \mathbb{Q} \subseteq \mathbb{Q}$ and every element of $\mathbb{Z}[\zeta] \cap \mathbb{Q}$ is integral over $\mathbb{Z}$, then we must have $\mathbb{Z}[\zeta] \cap \mathbb{Q} \subseteq \mathbb{Z}$.

Viktor Vaughn
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  • Thanks. I wonder why it is phrased like this in the book Out of curiosity do you have an answer to the "subgroup of $Z^n$ question? – Thomas Nov 27 '17 at 03:28
  • Not all subgroups of $\mathbb{Z}^n$ are cyclic for $n \geq 2$, so it's not possible to find a single generator in general. For instance, consider the subgroup of $\mathbb{Z}^2$ generated by $(2,0)$ and $(0,3)$. You can find a "nice" set of generators by computing the Smith normal form of a matrix: see this post for example. – Viktor Vaughn Nov 27 '17 at 03:36
  • (sorry I meant a base of generators, editing the question) – Thomas Nov 27 '17 at 04:00