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Let $R=\mathbb{Z}[\sqrt{-2}]$

1) Is it true that for every free $R-$module $M$ of rank=$n$ that $M$ is a free $\mathbb{Z}-$module of rank=$2n$?

2) Find two non-isomorphic $R$-modules with $19$ elements each.

For (1) I tried to begin with a basis $\{m_1,...,m_n\}$ of $M$ as a $R-$module and construct a basis $\{m_1',...,m_{2n}'\}$ of it as a $\mathbb{Z}-$module.

For (2) I have no idea.

Any hints??

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  • $\begingroup$ The first question is very simple : given the basis $[r_i]$ over $\mathbb Z [\sqrt{-2}]$, why don't you show that it is a module generated by $[r_i, \sqrt{-2}r_i]$ over$\mathbb Z$? $\endgroup$ – астон вілла олоф мэллбэрг Aug 31 '18 at 6:23
  • $\begingroup$ @астон вілла олоф мэллбэр Aaa I see! Thank you! $\endgroup$ – Γιάννης Παπαβασιλείου Aug 31 '18 at 6:24
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Let $M = [m_1,...,m_n]$. Then, every element of $M$ can be written uniquely as $\sum a_im_i$ where $a_i \in \mathbb Z[\sqrt -2] = b_i + c_i \sqrt{-2}$.

Clearly, $M$ is then generated by $[m_i, m_i\sqrt{-2}]$ over $\mathbb Z$.The question is : is this set linearly independent over $\mathbb Z$?

Well, let $\sum c_im_i + \sum d_i m_i\sqrt{-2} = 0$, where $c_i,d_i$ are integers. Then, $\sum(c_i + d_i \sqrt {-2}) m_i = 0$, so by the fact that $m_i$ are linearly independent over $\mathbb Z[\sqrt{-2}]$, we get that $c_i,d_i = 0$ for all $i$, as $c_i + d_i \sqrt{-2} = 0$ for all $i$.


The second question is already answered above , but I wish to point out something important, and be more elaborate. This will result in an expanded answer.

The point is, $\frac{R}{\langle \alpha\rangle} \not \equiv \frac{R}{\langle\beta\rangle}$ as $R$- modules is important, because they are indeed isomorphic as additive groups (modules are abelian groups under addition) , having the same cardinality $19$ which is prime, so both are cyclic and we may map a generator to another generator of the additive abelian group.

What is the difference? An $R$-module homomorphism $\phi : \frac{R}{\langle \alpha\rangle} \to \frac{R}{\langle \beta\rangle}$, in addition to the fact that it is additive, also must satisfy $f(rx) = rf(x)$ for all $r \in \mathbb Z[\sqrt{-2}]$ and $x \in \frac{R}{\langle \alpha\rangle}$.


This very key fact makes a great difference. For example, let us consider $f(1 + \langle \alpha\rangle)$. This element satisfies the fact that $(\alpha)(1 + \langle\alpha\rangle) = 0 + \langle \alpha \rangle$. Hence, it follows that $\alpha \times f(1 + \langle \alpha \rangle) = 0 + \langle \beta\rangle$. That is, if $f(1 + \langle \alpha\rangle) = c + \langle \beta\rangle$, then $\beta | \alpha c$, as $\alpha c -0 \in \langle \beta\rangle$.

The catch? Well, $\beta$ is co-prime to $\alpha$, because $a$ and $b$ have the same norm which is $19$ , a prime number, and they are not associates, so no element of non-unit norm can divide both. So $\beta | c$ (note that we are in a unique factorization domain), and consequently, we get $f(1+\langle\alpha\rangle) = 0$. If one generator maps to zero, the whole group maps to zero. Consequently, every module homomorphism between the two modules is the zero map, and therefore none is an isomorphism.

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  • $\begingroup$ Thank you again. Your answers really help me! $\endgroup$ – Γιάννης Παπαβασιλείου Aug 31 '18 at 7:11
  • $\begingroup$ You are welcome. I thought that the fact that the homomorphism was of modules and not merely of additive groups merited attention. $\endgroup$ – астон вілла олоф мэллбэрг Aug 31 '18 at 7:12
  • $\begingroup$ I have a question about (1): You say $M$ is then generated by $[m_i,m_i\sqrt{-2}]$ over $\mathbb{Z}$. But how do I know that $m_i\sqrt{-2}\in M$? $\endgroup$ – Γιάννης Παπαβασιλείου Aug 31 '18 at 7:14
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    $\begingroup$ The point is, that $M$ is generated, as a $\mathbb Z[\sqrt{-2}]$ module, by the elements $[m_i]$. This is what we assume in the starting. So every element of $M$ is a linear combination of the form $\sum a_jm_j$ where $a_j \in \mathbb Z[\sqrt{-2}]$. In particular, the element $\sqrt{-2}$ is an element of $\mathbb Z[\sqrt{-2}]$, and therefore $\sqrt{-2}m_i$ is a $\mathbb Z[\sqrt{-2}]$ - linear combination of the generators of $M$, like so : $\sqrt{-2}m_i = 0m_1 + 0m_2 + ... + \sqrt{-2}m_i + 0m_{i+1} + ... + 0m_n$ , hence also an element of $M$. $\endgroup$ – астон вілла олоф мэллбэрг Aug 31 '18 at 7:17
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    $\begingroup$ You are welcome. Note that $M$ is a module, but over which ring it is a module, that also matters. The first question seeks to question this concept. $\endgroup$ – астон вілла олоф мэллбэрг Aug 31 '18 at 7:20
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For $2$ look for modules $R/I$ where $I$ is an ideal of $R$. As $R$ is a principal ideal domain, $I=\left<\alpha\right>$ for some $\alpha=a+b\sqrt{-2}$. Then $|R/\left<\alpha\right>|=|\alpha|^2=a^2+2b^2$. If we have $\alpha=\eta\beta$ where $\eta$ is a unit in $R$. Then $\left<\alpha\right> =\left<\beta\right>$. But the only units in $R$ are $\pm 1$. Taking $\alpha=1+3\sqrt{-2}$ and $\beta=-1+3\sqrt{-2}$, then $|R/\left<\alpha\right>|=|R/\left<\beta\right>|=19$. Can you prove $R/\left<\alpha\right>\not\equiv R/\left<\beta\right>$ as $R$-modules?

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