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Here is the question I want to answer:

Let $R = \mathbb Z[\sqrt{-5}]$

Show that $I = (2, 1 + \sqrt{-5})$ is not a free $R-$module.

Hint: It helps to view $R$ as $\mathbb Z[x]/(x^2 + 5).$

My questions are::

1- How can I use the hint?

I have found here How to show that an $\mathbb{Z}[\sqrt{-5}]$-module is not free? how to show that it is not free as it says in the answer in that link the following:

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.

2- for showing this "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 $1+\sqrt{-5}$.", should I use the norm function?

Could someone push me in the right direction to answer those questions please?

Dylan C. Beck
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  • I have removed "unique factorization domains" from the tags. As you can see, there are two distinct factorizations of $6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5}).$ Once you establish that $2, 3, 1 \pm \sqrt{-5}$ are irreducible in $R,$ then we are done. – Dylan C. Beck Feb 06 '21 at 21:44
  • @Carlo we are done on what? –  Feb 06 '21 at 23:01
  • I apologize for the confusion. The quadratic integer ring $R = \mathbb Z[\sqrt{-5}]$ is not a unique factorization domain. One need only show that $2, 3,$ and $1 \pm \sqrt{-5}$ are irreducible in order to see this because $6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5}).$ – Dylan C. Beck Feb 07 '21 at 04:02
  • The question is asking to show that $I$ is not principle not that $\mathbb Z [\sqrt {-5}] $ is a unique factorization domain. So why you are telling me to prove that $2,3$ and $1 \pm \sqrt{-5} $ are irreducible? –  Feb 09 '21 at 02:39
  • You had tagged "unique factorization domain." – Dylan C. Beck Feb 09 '21 at 06:06
  • Also, if you can show that $1 \pm \sqrt{-5}$ and $2$ are irreducible, then by definition, their only divisors are themselves and the units of $R.$ – Dylan C. Beck Feb 09 '21 at 06:08
  • And how this proves that $I$ is principal?@Carlo –  Feb 09 '21 at 07:53
  • It doesn't; it proves that $I$ is not principal, actually. – Dylan C. Beck Feb 09 '21 at 18:15

1 Answers1

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Hint: Any two elements of $R$ are linearly dependent. (There is an obvious relation. Can you see this?)
From this, can you conclude that an ideal is free only if it is principal?

Edit: To see the first part, for distinct $a, b \in R$, note that $$(-b)\cdot a + a \cdot b = 0$$ and not both $a$ and $-b$ are $0$.

The second part now follows because any $R$-basis of $I \subset R$ must now either be $\varnothing$ or a singleton.

Aryaman Maithani
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  • How is that? that contradicts the presentation matrix that I knew for $R$ here in this question https://math.stackexchange.com/questions/4013135/understanding-the-solution-of-14-5-1-in-artin-and-comparing-2-solutions-of-it I have 2 rows of the matrix that are linearly independent. –  Feb 06 '21 at 16:42
  • Do you have exact answers, specifically to my questions in my post? –  Feb 06 '21 at 16:43
  • Two rows of a matrix with values in $R$ can be linearly independent. I'm saying that if ${a, b} \subset R$, then the set is not linearly independent over $R$. ($a \neq b$) – Aryaman Maithani Feb 06 '21 at 16:44
  • How can I prove that? –  Feb 06 '21 at 16:49
  • In $\Bbb Z$, do you see any $\Bbb Z$-linear combination of $2$ and $3$ which is $0$? – Aryaman Maithani Feb 06 '21 at 16:50
  • Yes $-3.2 + 3 .2 = 0$ –  Feb 06 '21 at 16:51
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    I would personally write that as $(-3)\cdot2 + 2\cdot3 = 0$. Do you see how that can generalise to $a, b$ in a general $R$? – Aryaman Maithani Feb 06 '21 at 16:52
  • Also, I think that an ideal $I$ is free iff it is principle is not a generally correct statement .... so why it is correct here in our case? –  Feb 06 '21 at 16:53
  • The "iff" is not true in general. The "only if" is but the "if" would require the ring to be an integral domain. – Aryaman Maithani Feb 06 '21 at 16:54
  • So you are saying that this direction $\implies$ is true but the other one is not? –  Feb 06 '21 at 16:57
  • I'm saying that: $I$ is free $\implies$ $I$ is principal. – Aryaman Maithani Feb 06 '21 at 16:58
  • Yeah this is generally correct? I think this is exactly what I am saying in my previous comment. –  Feb 06 '21 at 16:58
  • Yes, by the first part of the hint. – Aryaman Maithani Feb 06 '21 at 16:59
  • So we want to show that $I$ is not a principal ideal ..... is that correct? ..... will we show that using the norm function? –  Feb 06 '21 at 17:01
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  • You want to show that a free ideal must be principal. 2. Then, you want to show that this given $I$ is not principal. Using the norm is one way to do that, yes.
    • – Aryaman Maithani Feb 06 '21 at 17:03
  • But how I will use the hint given to me above in the question? –  Feb 06 '21 at 17:10