Vtyjkyuk

Let $R=mathbbZ[sqrt-2]$

1) Is it true that for every free $R-$module $M$ of rank=$n$ that $M$ is a free $mathbbZ-$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 $mathbbZ-$module.

For (2) I have no idea.

Any hints??

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 $[m_i, m_isqrt-2]$ over $mathbb Z$.The question is : is this set linearly independent over $mathbb Z$?

Well, let $sum c_im_i + sum d_i m_isqrt-2 = 0$, 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, $fracRlangle alpharangle not equiv fracRlanglebetarangle$ 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 $phi : fracRlangle alpharangle to fracRlangle betarangle$, 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 $x in fracRlangle alpharangle$.

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

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 $f(1+langlealpharangle) = 0$. 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.

For $2$ look for modules $R/I$ where $I$ is an ideal of $R$. As $R$
is a principal ideal domain, $I=left<alpharight>$ for some $alpha=a+bsqrt-2$. Then $|R/left<alpharight>|=|alpha|^2=a^2+2b^2$. If we
have $alpha=etabeta$ where $eta$ is a unit in $R$. Then $left<alpharight>
=left<betaright>$. But the only units in $R$ are $pm 1$. Taking $alpha=1+3sqrt-2$ and $beta=-1+3sqrt-2$, then
$|R/left<alpharight>|=|R/left<betaright>|=19$. Can you prove
$R/left<alpharight>notequiv R/left<betaright>$ as $R$-modules?