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Suppose $M$ is a finitely generated free $\mathbb{Z}$-module of rank $n$, so $M=u_1\mathbb{Z} \oplus \cdots \oplus u_n\mathbb{Z}$. Suppose that $M'$ is a submodule of full rank, so it is also rank $n$, and let $\widetilde{M}=u_1\mathbb{Z} \oplus \cdots \oplus u_{n-1}\mathbb{Z}$.

Now, I'm pretty sure that $M'\cap\widetilde{M}$ is a submodule of $\tilde{M}$ that is full rank, i.e., of rank $n-1$. I'm not sure why that's true, though. Does this follow from some standard result, and/or is the proof real easy?

Thanks in advance.

G Tony Jacobs
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  • How can you be "pretty sure" and "not sure why that's true" at the same time? – Pedro Aug 14 '16 at 22:35
  • Call it a hunch? Is the answer to this particular psychology question terribly important? If my phrasing was unfortunate, feel free to ignore that part. – G Tony Jacobs Aug 14 '16 at 22:37
  • I'm sorry. It just seems a bit confusing to say that, since the reader cannot understand in what position you find yourself against the problem. If you say "I'm pretty sure it is true", I would guess you have a concrete reason that makes you think it is true, if you say "not sure why that's true", then I would think you're not really close to a solution. At any rate, you can tensor with $\mathbb Q$ and make this "a linear algebra exercise", and that might be the reason you think it is true. Rank of PIDs behaves much like dimensions of vector spaces. – Pedro Aug 14 '16 at 22:41
  • I guess the reason I'm pretty sure is because of some concrete examples, which probably shouldn't be enough to make me pretty sure, so maybe I'm just too trusting of my intuition. As to why I'm not entirely sure, it's probably because this is outside of my area (algebraic number theory), and I'm not familiar with tensor products, nor with short exact sequences that the answer below uses. I'm studying it now. If you can post an answer unpacking your statement about tensoring with $\mathbb{Q}$, I'll study it, too. :) – G Tony Jacobs Aug 14 '16 at 22:49
  • Yes, the analogy to linear algebra is part of what's behind my intuition. Good point. – G Tony Jacobs Aug 14 '16 at 22:50

1 Answers1

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There is a short exact sequence $$0\to M'\cap\tilde{M}\to M'\to M'/(M'\cap\tilde{M})\to0.$$ Note also that $M'/(M'\cap\tilde{M})$ is isomorphic to the image of $M'$ in $M/\tilde{M}\cong\mathbb{Z}$, so it free of rank at most $1$. It follows that the exact sequence splits, so the rank of $M'\cong M'\cap\tilde{M}\oplus M'/(M'\cap\tilde{M})$ is the sum of the ranks of $M'\cap\tilde{M}$ and $M'/(M'\cap\tilde{M})$. Since $M'$ has rank $n$ and $M'\cap\tilde{M}$ has rank at most $n-1$, the only way this can happen is if $M'\cap\tilde{M}$ has rank $n-1$ and $M'/(M'\cap\tilde{M})$ has rank $1$.

Eric Wofsey
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  • I see why the sequence is exact. What is the relationship between rank and exact sequences? That seems to be the missing piece that keeps me from fully understanding this answer. – G Tony Jacobs Aug 14 '16 at 22:51
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    I have edited to explain that explicitly. Any short exact sequence of free modules splits, so you just have a direct sum and the ranks add. (More generally, if you define the rank of a finitely generated module as the rank of its free part, this coincides with the dimension after tensoring with $\mathbb{Q}$, so you can tensor a short exact sequence with $\mathbb{Q}$ and get again that rank is additive in short exact sequences.) – Eric Wofsey Aug 14 '16 at 22:56