## Tag Archives: projective module

### 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.

### A brief review of projective modules

Projective modules were introduced by Henri Cartan and Samuel Eilenberg in 1956.

In general one can define a $R$-module $P$ to be projective by any one of the following:
(1) Any short exact sequence $0\rightarrow N\rightarrow M\rightarrow P\rightarrow 0$ splits;
(2) $P$ is a direct summand of a free module, which $\exists\, Q$ such that $P\oplus Q$ is a free $R$-module;
(3) Given an epimorphism $\displaystyle f:M\to N$ and $p:P\to N$, then $\exists\, g:P\to M$ such that $p=fg$, that is

(4) hom$(P,\textendash)$ is an exact functor.

These four definitions show that projective module has very rich properties and is related to other types of modules. (2) means projective module is a generalization of free modules. (3) is called the lifting property of projective module, it is also a generalization of free module, because if $P$ is a free module, there exists a unique $g:P\to M$ such that the following diagram commutes

In (3) we don’t require $g$ to be unique, so $P$ does not have the universal property, consequently not necessarily a free object in category sense. Dual to definition (3), injective module is introduced
1)Given a monomorphism $f:M\to N$ and a module homomorphism $q:M\to Q$, then $\exists\, g:N\to Q$ such that the following diagram commutes

2) hom$(\textendash, Q)$ is an exact (contravariant) functor;
3) Any short exact sequence $0\rightarrow Q\rightarrow M\rightarrow N\rightarrow 0$ splits.

Since we know free modules are all projective ones, two natural questions emerge naturally.
(1)whether there is a projective module but not free;
(2)when the projective module is free.
For the first question, there is an easy answer. Let $R=\mathbb{Z}/6\mathbb{Z}$ regarded as a module for $R$-module, then $R$ is a free module. By the Chinese remainder theorem, $R\cong \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/3\mathbb{Z}$. Thus $\mathbb{Z}/2\mathbb{Z}$ is a direct summand of free $R$-module. $\mathbb{Z}/2\mathbb{Z}$ is projective from the definition, but not free $R$-module because actually it is a torsion $R$-module. Carefully investigating this example, the ring $R$ is decomposable, $R\cong R_1\oplus R_2$, then it is easy to find a non-free projective module, in particular $R_1$. What if we require the $R$ is not decomposable, say $R$ is an integral domain in particular? Is it still possible to find a non-free projective module? Our main result of this paper is devoted to this purpose. It turns out to be much more difficult to give such examples. Before finding an example, we should have a rough understanding on which kind of ring projective modules are always free, which is related to the answer of question (2).

The study of (2) can be summarized as following:
$\mathbf{Fact 1}$ If $R$ is a PID (the commutative principal ideal domains), all projective modules are free.
This is because any submodule of free module over $R$ is free. Projective is direct summand of free module, hence a submodule of free module.
$\mathbf{Fact 2}$ $R$ is a local ring, any finitely generated projective module over $R$ is free.
This result is proved by Irving Kaplansky in 1958. It is easy to prove this result for finitely generated projective modules, but the general case is difficult.
The following theorem is a well known result of H. Bass.
$\mathbf{Fact 3}$ If $R$ is a connected (i.e.without nontrivial idempotents) Noetherian ring. Then any infinitely generated projective $R$-module is free.
The following theorem is known as Serre’s Conjecture, which was solved by Quillen, Daniel and Suslin, Andrei A. independently in 1976.
$\mathbf{Fact 4}$ If $R=D[x_1,x_2\cdots,x_n]$, where $D$ is a PID, then every projective module over $R$ is free.
Note that Bass’s theorem is a special case of Serre’s Conjecture.