5
$\begingroup$

I have a question regarding Example 6.1.2 (page 252) from the book "The K-Book" by Charles Weibel. Here is the statement:

Example 6.1.2:

Let $F(t)$ be a rational function field in one variable $t$ over $F$. Then $K_2(F)$ is a direct summand of $K_2F(t)$.

$\text{ }$

I find it quite difficult to understand the argument, so I hope I can get some help from here.


His goal is to construct an inverse map $\lambda: K_2 F(t)\to K_2(F)$ to the natural map $K_2(F)\to K_2 F(t)$.

Make the following definitions $$f(t)=\frac{a_0t^n+\cdots + a_n}{b_0t^m+\cdots+ b_m},$$ $\operatorname{lead}(f):=\frac{a_0}{b_0}$ and $\lambda(\{f,g\})=\{\operatorname{lead}(f),\operatorname{lead}(g)\}$.

Now he says that we may check the representation of Matsumoto's Theorem, to show that $\lambda$ is a homomorphism.

We begin to show bilinearity. Let $f,f',g\in F(t)^\times$ then $$\lambda(\{f,g\}\{f',g\})=\lambda(\{ff',g\})=\{\operatorname{lead}(ff'),g\}=\{\operatorname{lead}(f)\operatorname{lead}(f'),g\}=\{\operatorname{lead}(f),g\}\{\operatorname{lead}(f'),g\}=\lambda(\{f,g\})\lambda(\{f',g\}).$$

By symmetry, the same argument works if we consider something like $\lambda(\{f,g\}\{f,g'\})$ instead.

Let us now see that the Steinberg identity is, also, satisfied. Consider $f\in F(t)^\times$ and assume $m<n$, then we have that $\operatorname{lead}(1-f)=-\operatorname{lead}(f)$, since $$1-\frac{a_0t^n+\cdots a_{n-m}t^{m}+\cdots + a_n}{b_0t^m+\cdots + b_m}=\frac{b_0t^m+\cdots + b_m}{b_0t^m+\cdots + b_m}-\frac{a_0t^n+\cdots a_{n-m}t^{m}+\cdots + a_n}{b_0t^m+\cdots + b_m}=\frac{-(a_0t^n+\cdots a_{n-(m+1)}t^{m+1})+(b_m-a_{n-m})t^{m}+\cdots + (b_m-a_n)}{b_0t^m+\cdots + b_m},$$ and so $\operatorname{lead}(1-f)=-\operatorname{lead}(f)$. It was showed earlier in the book that $\{x,-x\}=1$ for every $x\in E^\times$, where $E$ is some field. Thus $\{\operatorname{lead}(f),\operatorname{lead}(1-f)\}=\{\operatorname{lead}(f),-\operatorname{lead}(f)\}=1$. One can also show by similar reasoning that $\{\operatorname{lead}(f),\operatorname{lead}(1-f)\}=\{\operatorname{lead}(f),1-\operatorname{lead}(f)\}=1$ if $n=m$ and that $\{\operatorname{lead}(f),\operatorname{lead}(1-f)\}=\{\operatorname{lead}(f),1\}$ if $n<m$.

Lastly, they say that since $K_2$ commutes with filtered colimits, it follows that $K_2(F)$ injects into $K_2F(T)$ for every purely transcendental extension $F(T)$ of $F$.


Questions:

  1. Why does checking the criteria for Matsuomoto's Theorem prove that $\lambda$ is a homomorphism? Don't we want to pick symbols $\{f,g\},\{f',g'\}$ and check that it is a group homomorphism: $$\lambda(\{f,g\}\{f',g'\})=\lambda(\{f,g\})\lambda(\{f',g'\})?$$ By checking Matsumoto's Theorem don't we just make sure that we map elements from $K_2F(t)$ to $K_2(F)$?
  2. Why does the following equality hold $\{\operatorname{lead}(f),1\}=1$?
  3. How comes that the last paragraph implies that $K_2(F)$ injects into $K_2F(T)$? I don't really understand why $K_2$ commutes with filtered colimits but if I take it for granted for the moment, why does it prove the fact?
  4. Lastly, why does showing that $K_2(F)\to K_2 F(T)$ is injective show that $K_2(F)$ is a direct summand of $K_2F(t)$?

The answers are maybe/probably easy, but I am quite confused right now, new to $K$-theory and I don't manage to figure it out.

Best wishes,

Joel

Edit:

Matsumoto's Theorem (for completeness):

If $F$ is a field, then $K_2(F)$ is the abelian group generated by the set of Steinberg symbols $\{x,y\}$ with $x,y\in F^\times$, subject to (only) the following relations:

$$\quad\quad\text{(Bilinearity)}\quad\{xx',y\}=\{x,y\}\{x',y\}\text{ and }\{x,yy'\}=\{x,y\}\{x,y'\}$$ $$\text{(Steinberg identity)}\quad\{x,1-x\}=1\text{ for all }x\not = 0,1.$$

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.