# $K(A)\cong \mathbb Z$ for a PID $A$

In Atiyah and Macdonald, chapter 7, exercise 26, iii), it's required to show the Grothendieck group $K(A)\cong \mathbb Z$ for a PID $A$. By ii) of this problem, it's easy to show that $K(A)$ is generated by the element $\gamma(A)$, but I cannot prove it has infinite order. Can anyone give a hint?

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I think it would help if you reproduced the problem statement and showed some of your work. – Brian Fitzpatrick Feb 13 '14 at 8:38
No one without the book in his hand can even know what you are talking about. – Mariano Suárez-Alvarez Feb 13 '14 at 8:42

Suppose that $K$ is the fraction field of $A$.
If $P$ is a finitely generated projective $A$-module, then $P\otimes_AK$ is a finite dimensional $K$-vector space, so $\dim_K P\otimes_AK$ is a non-negative integer. You should be able to show that there is a (unique!) homomorphism of abelian groups $d:K(A)\to\mathbb Z$ such that for each f.g. projective $A$-module $P$ we have $d(\gamma(P))=\dim_K P\otimes_AK$.
Now $d(\gamma(A))=1$, which has infinite order in the abelian group $\mathbb Z$, so $\gamma(A)$ has to have infinite order in $K(A)$.
Not really. ${}{}$ – Mariano Suárez-Alvarez Feb 13 '14 at 8:51