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Is it possible to embed $\mathbb Z^n$ inside $ \mathbb Z^m$ as a $ \mathbb Z$-module for $m < n$ ?

I think it's not possible. It might be a easy problem for some of you, but I really don't have any clean way to show this. Any ideas?

user26857
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Arpit Kansal
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    Related: http://math.stackexchange.com/questions/254347/prove-that-the-group-isomorphism-mathbbzm-cong-mathbbzn-implies-that/254357#254357 – user26857 Aug 28 '15 at 11:32

2 Answers2

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Let $m < n$. If $\mathbb{Z}^n$ embeds in $\mathbb{Z}^m$, then the $n$ linearly independent basis elements of $\mathbb{Z}^n$ map to $n$ linearly independent elements of $\mathbb{Z}^m$. The fact that they are linearly independent over $\mathbb{Z}$ implies they are linearly independent over $\mathbb{Q}$, in $\mathbb{Q}^m$. But as $\mathbb{Q}^m$ is a vector space of dimension $m < n$, there do not exist $n$ linearly independent elements in it, contradiction.

Jyrki Lahtonen
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Caleb Stanford
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  • What exactly do you mean by linearly independent where we arn't talking about vector spaces ? – Belgi Aug 28 '15 at 10:04
  • @Belgi It's linear independence in a module over $\mathbb{Z}$. Anyway, it means exactly the same as it does in the vector space case: $v_1, \ldots, v_n$ are linearly independent over $\mathbb{Z}$ if for any integers $k_1, \ldots, k_n$, if $k_1 v_1 + \cdots + k_n v_n = 0$, then $k_i = 0$ for all $i$. – Caleb Stanford Aug 28 '15 at 10:07
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If $0\to\mathbb Z^n\to\mathbb Z^m$ is an exact sequence of $\mathbb Z$-modules, then $0\to S^{-1}\mathbb Z^n\to S^{-1}\mathbb Z^m$ is an exact sequence of $S^{-1}\mathbb Z$-modules, where $S=\mathbb Z-\{0\}$, hence $0\to\mathbb Q^n\to\mathbb Q^m$ is an exact sequence of $\mathbb Q$-vector spaces, so $n\le m$.

user26857
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  • Sorry..what do you mean by $S^{-1} \mathbb Z^n$ ? – Arpit Kansal Aug 28 '15 at 12:10
  • @ArpitKansal It's localization. – Caleb Stanford Aug 28 '15 at 16:15
  • Anyway, +1 for this answer, I think it's the right way of looking at it, if you know about modules and exact sequences and localization. – Caleb Stanford Aug 28 '15 at 16:17
  • @6005 unfortunately I don't know about localization,can you say something about it in short? – Arpit Kansal Aug 28 '15 at 16:18
  • @6005 I see.Fortunately I know about modules and exact sequences but I've never heard about localization in any of my courses.Regards, – Arpit Kansal Aug 28 '15 at 16:26
  • @ArpitKansal oh, sorry! then here is a better explanation. For a set $S$ that is a subset of a ring $R$ and is closed under multiplication, you can form the ring of fractions: $S^{-1} R$, where formal fractions look like $r / s$, where $r \in R$ and $s \in S$. You then make an equivalence relation on fractions and define addition and multiplication in the obvious way (think of the construction of rational numbers as an example). It then turns out if you have an $R$-module $M$, you can make the module $S^{-1} M$ which is the set of formal fractions $\frac{m}{s}$, $m \in M$, $s \in S$. – Caleb Stanford Aug 28 '15 at 16:32
  • $S^{-1} M$ is an $S^{-1} R$-module. You define addition and scalar multiplication in again the most obvious way. – Caleb Stanford Aug 28 '15 at 16:33
  • By "formal fraction" I really mean ordered pair, I just use fraction notation to make it easier to understand. – Caleb Stanford Aug 28 '15 at 16:33
  • @6005 Thanks a lot.I atleast understand the construction of $S^{-1}M$ but I think I'll be more comfortable with it when I'll use it in some course.Best Regards, – Arpit Kansal Aug 28 '15 at 16:37