Clearly $\mathbb{Z}$ and $\mathbb{R}$ are $\mathbb{Z}$-module.

Then is $2\mathbb{Z}$ a submodule of $\mathbb{Z}$? What is a basis?

And what is a basis for $\mathbb{R}$?

  • $\begingroup$ $\;\Bbb Z -$ module = abelian group. This answers your first question. For the second one, think that $\;\Bbb Q\;$ is not a finitely generated abelian group, so imagine $\;\Bbb R\;$ ... $\endgroup$ – DonAntonio Jan 1 '14 at 3:54

To check that $2\mathbb{Z}$ is a submodule of $\mathbb{Z}$, you must check that it's an abelian subgroup of $\mathbb{Z}$ under addition and all elements $m \in 2\mathbb{Z}$, and for all elements $n \in \mathbb{Z}$, $nm \in 2\mathbb{Z}. $A basis for $2\mathbb{Z}$ is given by $2$, as this spans $2\mathbb{Z}$.

$\mathbb{R}$ does not admit a finite (even countable) basis.


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