# Given $n\in \mathbb N$, is there a free module with a basis of size $m$, $\forall m\geq n$?

Exercise IV.2.12 of Hungerford's Algebra asks to show the following:

If $F$ is a free module over a ring with identity such that $F$ has a basis of finite cardinality $n > 1$ and another basis of cardinality $n + 1$, then $F$ has a basis of cardinality $m$ for every $m > n\ (m \in \mathbb N)$.

This is easy to prove using induction and the fact that if $M$ is an $R$-module with a basis of size $n$, then $M\simeq\bigoplus_{k=1}^nR$ as $R$-modules.

My question is,

for each $n\geq 1$ is there a ring $R$ with identity and a free $R$-module $M$ such that $M$ has a basis of size $m$ for all $m\geq n$ and $M$ does not have a basis of size $k$ for all $k<n$?

In another exercise of the same section the case $n=1$ is established as follows:

Let $K$ be a ring with identity and $F$ a free $K$-module with an infinite denumerable basis $\{ e_1,e_2,\ldots\}$. Put $R =$ Hom$_K(F,F)$. Then the author shows that $R$ has a basis of size $2$ as an $R$-module; namely $\{f_1,f_2\}$, where $f_1(e_{2n})=e_n, f(e_{2n-1})=0,f_2(e_{2n})=0,f(e_{2n-1})=e_n$, and of course $R$ has a basis of size $1$; $\{1_R\}$.

But I don't know what to do when $n\geq 2$.

I thank beforehand any help.

• Does $R \times R$ solve the problem for $n=2$? – N. S. Mar 20 '13 at 0:11
• The case $n=2$ would be a ring $R$ such that $R\oplus R\simeq \bigoplus_{k=1}^nR$ for all $n\geq 2$ but $R\ncong R\oplus R$ – Camilo Arosemena-Serrato Mar 20 '13 at 0:14

I think I'm remembering right that their key feature that set them apart from the example you gave was that they could produce this $R^n\cong R^m$ behavior for prescribed $m,n\in \Bbb N$.