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I'm trying to understand some differences in vector spaces over a division ring or field and over an arbitrary ring. I came across this post, and have found it very enlightening. I have a question regarding one of the bullet points in the second answer.

A maximal linearly independent subset needs not be a basis: consider $2\mathbb{Z}$ in $\mathbb{Z}$ as a $\mathbb{Z}$-module

I understand on the surface that we can find such an example, but I don't understand why this is one. Specifically, I don't understand what makes $2\mathbb{Z}$ a linearly independent set in this case. Can't all elements of $2\mathbb{Z}$ be written as linear combinations of 2, making them dependent? I'm definitely missing something, so if someone could enlighten me as to why this example works, I'd be very appreciative. Thanks!

GabeT
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    I think what they mean is consider ${2}$ as a maximal linearly independent subset, which generates $2\mathbb Z \subsetneq \mathbb Z$ and therefore is not a basis. – Moisés Nov 19 '19 at 16:34
  • I see! Thank you for the clarification! – GabeT Nov 19 '19 at 16:45

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