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.

This means \psi i(x)=x. Thus I is a projective module. \text{Q.E.D. }\hfill\square

As discussed in the beginning of this section, since I is not principle, we get a non-free projective module.

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