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I do not know how to start with the following exercise:

Let $R := \mathbb{Z}[\sqrt{-5}]$. Show that the left $R$-module $M:=\big\langle 2, 1+\sqrt{-5} \big\rangle$ (so the $R$-module generated by $2$ and $1+\sqrt{-5}$) is not free.

I do not see how I could prove this. (I know only the basic definition of a module.) All I recognise is that $\mathbb{Z}[\sqrt{-5}]$ is an ideal and two elements of an ideal in a commutative ring must be linearly dependent, but I do not know if this could help here. Could you give me a hint?

3nondatur
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1 Answers1

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It is generated by two elements so if it were free, it would be free of rank $1$ or $2$.

If it were free of rank $1$, it would be generated by a single element: you can then try to prove that no element of $R$ divides both $2$ and $1+\sqrt{-5}$.

If it were free of rank $2$, then $2$ and $1+\sqrt{-5}$ would have to be free generators (can you see why ?). You can then try to prove that this is not the case, that is, that $2$ and $1+\sqrt{-5}$ are not linearly independent.

Maxime Ramzi
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