Let $M$ be a rank $k$ submodule of $\mathbb{Z}^n$. When is its image in $\mathbb{F}_p^n$ a rank $k$ subspace (where $p$ is any prime)? Can this situation be characterized in terms of a flatness condition?
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1What is $p$? Please update your question – unseen_rider Mar 23 '17 at 17:51
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The image of $M$ in $\mathbb{F}_p^n$ is the image of the map $f:M\otimes\mathbb{F}_p\to \mathbb{Z}^n\otimes\mathbb{F}_p$ induced by the inclusion $M\to\mathbb{Z}^n$, and the image is $k$-dimensional iff $f$ is injective. This map is part of a long exact sequence $$\operatorname{Tor}(\mathbb{Z}^n,\mathbb{F}_p)\to\operatorname{Tor}(\mathbb{Z}^n/M,\mathbb{F}_p)\to M\otimes\mathbb{F}_p\stackrel{f}\to \mathbb{Z}^n\otimes\mathbb{F}_p.$$ Since $\operatorname{Tor}(\mathbb{Z}^n,\mathbb{F}_p)=0$, $\ker(f)$ is isomorphic to $\operatorname{Tor}(\mathbb{Z}^n/M,\mathbb{F}_p)$. Thus $f$ is injective iff $\operatorname{Tor}(\mathbb{Z}^n/M,\mathbb{F}_p)=0$: that is, iff $\mathbb{Z}^n/M$ has no $p$-torsion.
Eric Wofsey
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If we wanted the image of $M$ to have rank $k$ in $\mathbb{F}_p^n$ for all $p$, would this be the same as asking that $\mathbb{Z}^n/M$ is flat as a $\mathbb{Z}$ module? – Madeline Brandt Mar 23 '17 at 20:29
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