Isomorphism on a collection of R, $\mathbf{Z}$-isomorphism between ideal and $\mathbf{Z}[5i]$

tough (*) question from a previous examination paper:

6.(*)
a) Let $R=\mathbf{C}[q]/q^{2}\mathbf{C}[q]$ and let $m,n \in \mathbf{N}$ where $R^{m},R^{n}$ are R-isomorphic. Show that m is equal to n.
b) Now let $R=\mathbf{Z}[5i]$ and $I=2R+(1+5i)R=2\mathbf{Z}+(1+5i)\mathbf{Z}$. Show that R and I are $\mathbf{Z}$-isomorphic but not R-isomorphic.

For a) I thought about showing that $\dim(R^{m})=\dim(R^{n})$, but I don't see how to do that (nor if it is a correct assumption). I lack an idea for b). Since solutions are not available, help is greatly appreciated.

edit: this theorem is probably relevant for the solution of the first one: $R^{m}$ isomorphic to a module M $\Leftrightarrow$ M has a basis with m elements.

Is there a way to use a) to solve b)?

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Thanks for your efforts. – PumaDAce Dec 1 '11 at 14:25

a) Notice that since $\mathbf{C}$ is a subring of $R$, any $R$-module is a $\mathbf{C}$-module (or rather $\mathbf{C}$-vector space). Since an isomorphism of $R$-modules is also an isomorphism of $\mathbf{C}$-vector spaces, it is enough to show that if $R^n$ and $R^m$ are isomorphic as $\mathbf{C}$-vector spaces, then $n=m$. But if they are, then $2n = dim_\mathbf{C} R^n = dim_\mathbf{C} R^m = 2m$, thus $n=m$

b) as $\mathbf{Z}$-modules, $R$ and $I$ are both isomorphic to $\mathbf{Z}^2$ (you can easily exhibit a basis for them as $\mathbf{Z}$-modules) Showing that $I$ is not isomorphic to $R$ as an $R$-module is a bit trickier: if there is an isomorphism $f:R \to I$, then it is determined by the image of $1$, and that isomorphism would simply be multiplication by $f(1)$. In other word, you have to show that $I$ is not of the form $aR$ where $a \in I$, i.e. is not a principal ideal of $R$.

If $I = aR$, then since $2 \in R$, $2$ has to be a multiple of $a$, which forces $a$ to be $2$ or $-2$. But $2R$ doesn't contain $1+5i$, so $I$ is not $aR$.

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Got it. Thank you so much mercio! – PumaDAce Dec 6 '11 at 15:02

I guess this all depends upon what you know. Call a ring $R$ an IBN-ring if the $R$-isomorphism of $R^m\cong R^n$ implies $m=n$. A common theorem states that commutative unital rings are IBN-rings. The basic idea being that by Krull's theorem you can find some maximal ideal $\mathfrak{m}$ of $R$. You can then show that $R^m/\mathfrak{m}R^m$ an $R^n/\mathfrak{m}r^n$ can both be given $R/\mathfrak{m}$-module structures in such a way that the image of the usual bases of $R^m$ and $R^n$ under the canonical epimorphisms are bases for $R^m/\mathfrak{m}R^m$ and $R^n/\mathfrak{m}r^n$ respectively. Moreover, you can prove that the $R$-isomorphism $R^n\cong R^m$ descends to a $R/\mathfrak{m}$-isomorphism $R^m/\mathfrak{m}R^m\cong R^n/\mathfrak{m}R^n$. But, note that $R/\mathfrak{m}$ is a field, and so by first principles $m=\dim_{R/\mathfrak{m}}R^m/\mathfrak{m}=\dim_{R/\mathfrak{m}}R^n/\mathfrak{m}R^n=n$. If you were doing a qual though, I feel like you could just state this theorem, no?

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Thanks for your idea, but the only thing that came up in th lectures I can think of as relevant for the solution of this problem, is that: $M$ isomorphic to $R^{m} \Leftrightarrow M$ has a basis with m elements – PumaDAce Dec 1 '11 at 14:24