Homework 2 corrections

Question 2.

We should assume that the module is not simply stably free, but that it is also as good as possible in that regards. That is, $P\oplus R=R^2$.

Question 4. We do not have usually left exactness in the stated short exact sequence, so that part should be dropped.

Look at the K-book chapter 2 page 14 exercise 2.3 ( Excision for $K_0$ and how to make a non-unital ring unital)

In the exact sequence $K_0(R,I)\stackrel{{\pi }_{2*}}{\to }{K}_0(R)\to K_0(R/I)$

${\pi }_2:D_R(I)\to R$ is given by ${\pi }_2(x,y)=y$ and ${\pi }_{2*}[-]=[R\underset{D_R(I)}{\overset{{\pi }_2}{\otimes }}-]$

Question 1.

$(a)\Rightarrow (b)$ is fine.

but $(b)\Rightarrow (a)$ should be modified to:

if $R^n\stackrel{f}{\to }R\to 0$ is an exact sequence then $f$ is given by a unimodular row.

The simple example that makes statement of $(b)\Rightarrow (a)$ false is :

$0\to \mathbb{\mathbb{Z}}\stackrel{\times 2}{\to }\mathbb{\mathbb{Z}}$

Question 2.

One can define exterior powers over commutative rings + good properties. So if:

$P\oplus R^{n-1}\simeq R^n\Rightarrow P\simeq P{\otimes }_{R}R\simeq \underset{R}{\overset{1}{\wedge }}P{\otimes }_R\underset{R}{\overset{n-1}{\wedge }}{R}^{n-1}\simeq \underset{R}{\overset{n}{\wedge }}{R}^n\simeq R$

since rank of $P$ is one ( so constant ) over $R$ then $\underset{R}{\overset{k}{\wedge }}P=0\mathrm{\forall }k>1$. [ I have assumed in all this that $R$ is unital. ]

For reference you may look at the K-book chapter 1, pages: 4 (ex 1.6), 15, 16

For discussion of rank look at the K-book chapter 1, pages: 2, 9

Exercise. If $R$ has the invariant basis property (IBP), i.e. $R^n\simeq R^m\iff n=m$, then the rank of a stably free $R$-module $P$ defined by ${\text{rank}}_R(P)=m-n$ where $m,n\in \mathbb{\mathbb{N}}\cup 0$ are such that $P\oplus R^n\simeq R^m$ is well defined.

Question 3.

$(a)\nRightarrow (b)$ ===> So one needs to add more assumptions.

example. $(\mathbb{\mathbb{Z}}\oplus \mathbb{\mathbb{Z}}/2\mathbb{\mathbb{Z}}){\otimes }_{\mathbb{\mathbb{Z}}}2\mathbb{\mathbb{Z}}\simeq \mathbb{\mathbb{Z}}$ but $\mathbb{\mathbb{Z}}\oplus \mathbb{\mathbb{Z}}/2\mathbb{\mathbb{Z}}$ is not a projective $\mathbb{\mathbb{Z}}$-module.

This question is the exercise 3.1 of the K-book chapter 1 on page 23, also look at pages 15-23 for the review of line bundles and the picard group.