Course talk:MATH600D

Hey folks in Math 600 D. I would love to give the lecture on "projective and flat modules, Noetherian rings". If you happened to be interested in giving lectures on this topic too, maybe we could split it and do it together :).

I've been posting the solutions to past homeworks in my website. Here is the link:

http://www.math.ubc.ca/~marioga/math_600D.html

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 $K_{0}(R,I)\xrightarrow {\pi _{2*}} K_{0}(R)\to K_{0}(R/I)~$

$\pi _{2}:D_{R}(I)\to R~$ is given by $\pi _{2}(x,y)=y~$ and $\pi _{2*}[-]=[R\otimes _{D_{R}(I)}^{\pi _{2}}-]~$

Question 1.

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

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

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

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

$0\to \mathbb {Z} \xrightarrow {\times 2} \mathbb {Z} ~$

Question 2.

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

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

since rank of $P~$ is one ( so constant ) over $R~$ then $\wedge _{R}^{k}P=0\;\forall \;k>1~$. [ 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. $R^{n}\simeq R^{m}\Leftrightarrow n=m~$, then the rank of a stably free $R~$-module $P~$ defined by ${\text{rank}}_{R}(P)=m-n~$ where $m,n\in \mathbb {N} \cup {0}~$ 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. $(\mathbb {Z} \oplus \mathbb {Z} /2\mathbb {Z} )\otimes _{\mathbb {Z} }2\mathbb {Z} \simeq \mathbb {Z} ~$ but $\mathbb {Z} \oplus \mathbb {Z} /2\mathbb {Z} ~$ is not a projective $\mathbb {Z} ~$-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.

Suslin, Andrei A.; Wodzicki, Mariusz (1992), "Excision in Algebraic K-Theory" (PDF), Annals of Mathematics, 136 (1): 51–122

Gabriel, Pierre (1962), "Des catégories abéliennes", Bulletin de la Société Mathématique de France (in French), 90: 323–448, MR 0381144, Zbl 0201.35602, retrieved 2011-10-08CS1 maint: unrecognized language (link) ^{author's bio, look at chapter III}

Charles, Weibel; Dongyuan, Yao (1992), Localization for the K-theory of noncommutative rings (PDF), Contemporary Math., 126, Providence, RI: Amer. Math. Soc., pp. 219–230, MR 1156514

C.T.C. Wall: Finiteness conditions for CW complexes I, Annals of Mathematics 81 (1965), 56-59.

C.T.C.Wall: Finitness conditions for CW complexes II, Proc. Royal Soc. A 295 (1966), 129-139.

J. Milnor, Whitehead torsion, bull. AMS. 72 (1966), 358-426.

You may find interesting things about surgery and L-theory on Andrew Ranicki’s webpage:

http://www.maths.ed.ac.uk/~aar/

Here is what you should understand by an algebraic bundle over a commutative ring R: An algebraic bundle over R is a finitely generated projective module of constant rank. Thus an algebraic line bundle is a finitely generated projective module of constant rank 1.

http://www.math.upenn.edu/~chai/624_08/mumford-oda_chap1-6.pdf

http://www.math.upenn.edu/~chai/624_08/math624_08.html

Vector Bundles & K-Theory

http://www.math.cornell.edu/~hatcher/VBKT/VBpage.html

questions 1, step 1, correction of a typo

$f_{n}(a)=\sum _{j=0}^{n}{2n \choose j}a^{2n-j}(1-a)^{j}=1-\sum _{j=n+1}^{2n}{2n \choose j}a^{2n-j}(1-a)^{j}~$

questions 3, correction of a typo

${\begin{pmatrix}e_{1}&0\\0&0\end{pmatrix}}=g{\begin{pmatrix}e&0\\0&0\end{pmatrix}}g^{-1}$