An example of non-free projective module over integral domain

An ideal $I$ of an integral domain $R$ is a free $R$ -module if and only if it is generated by one element. Indeed, given any two non-zero elements $x,y\in I$ , $x\cdot y+(-y)\cdot x=0$ . So any two elements are linear dependent. Hence, if $I$ were free module, it must generated by one element, or $I$ is a principle ideal. Our strategy is to find an ideal $I$ which is not principle and show that $I$ is projective.

Let $R=\mathbb{Z}[-21]=\{a+b\sqrt{-21}|a,b\in\mathbb{Z}\}$ . Obviously $R$ is an integral domain. Let $I$ be the ideal generated by 5 and $2+\sqrt{-21}$ . $I=\{5a+(2+\sqrt{-21})b|a,b\in R\}$ .

$\mathbf{Claim:}$ $I$ is not a principal ideal.
$\mathbf{Proof:}$
Define $\phi:R\to \mathbb Z$ by $\phi(a+b\sqrt{-21})=a^2+21b^2$ , then it is straightforward to verify $\phi(r_1r_2)=\phi(r_1)\phi(r_2)$ , for all $r_1,r_2\in R$ . If $r\in R$ is a unit, then $\phi(r)=1$ , which means $a=\pm 1$ and $b=0$ . So the only units in $R$ are $\pm 1$ .

Now suppose $I$ is a principle ideal, then $\exists\,u$ such that $I=Ru$ . Then $\exists\, r\in R$ such that $5=ru$ , which means $25=\phi(r)\phi(u)$ . Since $x^2+21y^2=5$ has no solution in $\mathbb Z$ , $\phi(r)$ or $\phi(u)$ can not be 5. So we must have $\phi(r)=25$ and $\phi(u)=1$ or $\phi(r)=1$ and $\phi(u)=25$ . The second case is impossible, because $\phi(r)=1$ means $r=\pm 1$ . Then $I=5\cdot R$ . We can obtain $2+\sqrt{-21}=5(a+b\sqrt{-21})$ , this is impossible.

The only case we left is $\phi(u)=1$ , which means $u=\pm 1$ . Then $I=R$ . We have $R\cong \mathbb Z[x]/(x^2+21)$ , $I$ corresponds to the ideal $(5,2+x)/(x^2+21)$ . If $I=R$ , then $(5,2+x)=\mathbb Z[x]$ . This fact leads to $1=f(x)\cdot 5+g(x)\cdot (2+x)$ , for some $f(x)$ , $g(x)\in \mathbb Z[x]$ . If we set $x=-2$ , then $1=f(-2)\cdot 5$ in $\mathbb Z$ , this is impossible.

In conclusion, $I$ can not be principle. $\text{Q.E.D. }\hfill\square$

$\mathbf{Claim:}$ $I$ is a projective module.

$\mathbf{Proof:}$ Define $\psi:R^2\to I$ by $\psi(a,b)=5a+(2+\sqrt{-21})b$ . If we can find $i:I\to R^2$ such that $\psi i=id_I$ , then $I$ is a direct summand of $R^2$ , which is a free $R$ -module. Consequently, $I$ is a projective module.

Let $x=5a+(2+\sqrt{-21})b\in I$ , $a,b\in R$ . There exist unique $u,v\in R$ such that
$(3-4\sqrt{-21})x=5u$
$(16+2\sqrt{-21})x=5v$
A straightforward calculation shows that, $u=3a+18b-(4a+b)\sqrt{-21}$ , $v=16a-2b+(2a+4b)\sqrt{-21}$ . Define $i(x)=(u,v)\in R^2$ . It is easy to verify that $i$ is an $R$ -module homomorphism. Moreover,
$\psi(u,v)=u\cdot 5+v\cdot(2+\sqrt{-21})$
$=5(3a+18b-(4a+b)\sqrt{-21})+[16a-2b+(2a+4b)\sqrt{-21}](2+\sqrt{-21})= 5a+(2+\sqrt{-21})b.$