For simplicity, we consider commutative rings $R$ with unity $1$ different from $0$ and module to be finitely generated.

If $P$ is a projective $R$-module, it is well-known that there exists a generating set $u_1,\cdots, u_r$ for $P$ over $R$ and $f_1,f_2,\cdots, f_r$ in $\mbox{Hom}_R(P,R)$ such that every $x\in P$ can be written as $$x=f_1(x)u_1 + f_2(x)u_2 +\cdots + f_r(x)u_r.$$

If $P$ would have been free on $\{u_1,\cdots,u_r\}$ then the expression for each $x$ would have been unique, am I right? In this case, $\{f_1,\cdots, f_r\}$ is a dual basis of $\{u_1,\cdots,u_r\}$ in the sense $$f_i(u_j)=\delta_{ij}.\hskip8mm(*)$$

Q.1 Is there example of finitely generated module over $R$ such that

(1) $P$ is projective but not free.

(2) The basis $\{f_1,\cdots, f_r\}$ (or any such basis) does not satisfy $(*)$?

I would be happy if some simple example could be given from ring of algebraic integers from a finite extension of $\mathbb{Q}$; since I have heard (without proof) that these rings of integers are Dedekind domains, and in these domains, all the ideals are projective.

Cohn in his book Basic Algebra says that for projective module, there always exists dual basis $\{f_1,\cdots, f_r\}$ which not necessarily satisfy $(*)$. I wanted to see example in this regard.

  • 1
    $\begingroup$ I have not made the computations but you can start from the following example: set $R=\mathbb{Z}[\omega]$n with $\omega=i\sqrt{5}$. Let $P$ be the $R$-submodule of $R^2$ generated by $(4,-1+\omega)$ and $(2+2\omega,-3)$. Finally let $Q$ be the $R$-submodule of $R^2$ generated by $(-2,1+\omega)$ and $(-3\omega,5+\omega)$. Then $P,Q$ are projective and non free, such that $R^2=P\oplus Q$. $\endgroup$ – GreginGre Mar 8 '18 at 8:05

For a really easy example, let $K$ and $L$ be two nonzero rings (fields, if you like) and $R=K\times L$. Note that as a module over itself, $R$ is the direct sum of the two ideals $I=K\times 0$ and $J=0\times L$, so these two ideals are projective modules. They obviously are not free, though, since they have nontrivial annihilator. Taking $M=I$, for instance, you can take $u_1=(1,0)$ and $f_1$ to be the inclusion $I\to R$ and then $u_1$ and $f_1$ satisfy $x=f_1(x)u_1$ for all $x\in I$. However, since $f_1(u_1)=u_1\neq 1$, they do not satisfy $(*)$. Moreover, it is impossible to satisfy $(*)$ with any other choice of generators, since every homomorphism $f:I\to R$ has image contained in $I$ (since the image must be annihilated by $(0,1)$), so it is impossible to have $f_i(u_i)=1$.

Let me remark more generally that any non-free projective module can serve as the example you are looking for (that is, your condition (1) automatically implies condition (2)). Note that given generators $u_1,\dots,u_r$ and $f_1,\dots,f_r$ as you describe, we have maps $U:R^r\to M$ and $F:M\to R^r$ given by the $u_i$ and $f_i$ which satisfy $U\circ F=1_M$. If $(*)$ is satisfied as well, that means that $F\circ U=1_{R^r}$. So if $(*)$ is satisfied, $F$ and $U$ are inverses and so $M\cong R^r$ is free (and the $f_i$ and $u_i$ are dual bases).

So if $M$ is any projective module that is not free, then $(*)$ can never be satisfied. So, for instance, you could take $M$ to be any non-principal ideal in a Dedekind domain. For an explicit example, you could have $R=\mathbb{Z}[\sqrt{-5}]$ and $M=(2,1+\sqrt{-5})$.

  • $\begingroup$ This last example is nice one (well-known non-UFD); I was thinking for it, but not justifying properly. Now I got some idea of example. Thanks for the help. $\endgroup$ – Beginner Mar 8 '18 at 9:28

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