# What is the grading of x(x−1)R[x]? Loopspace for Karoubi-Villamayor K-theory.

I am reading the chapter on Karoubi-Villamayor K-theory in Weibel's K-book. In particular he defines $\Omega R=(x^2-x)R[x]$ for a ring. This will eventually lead to a model for the loopspace

Corollary 11.13.1 For $n\geq 2$ we have, $$KV_n(R)\cong KV_{n-1}(\Omega R).$$

The proof uses the long exact sequence of the GL-fibration $$(x^2-x)R[x]\to xR[x]\to R$$

and uses the following fact to show that $KV_n(xR[x])$ vanishes:

Exercise 11.5 Let $R=R_0\oplus R_1\oplus\cdots$ be a graded ring. Then for every homotopy invariant functor $F$ on rings, possibly without unit, (i.e. the map $R\to R[x]$ induces homotopy equivalences; true for $KV$) then we have $F(R)\simeq F(R_0)$.

Now the question: Can't we just apply the exercise to $\Omega R$ as well? Or is the grading different? I am really confused here.

Edit: If anybody is put off by the word Karoubi-Villamayor, just assume $R$ regular, as we then have $KV=K$.

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I see no grading on $\Omega R$. – Wilberd van der Kallen May 17 '12 at 13:37
What exactly stops us from taking the grading coming from $R[x]$, i.e. $(x^2−x)P$ has degree $gradP+2$? – Simon Markett May 17 '12 at 13:55
Simon: With this "grading", $(x^2-x)x$ has degree 3, which means its square should have degree 6. But in fact the square is $(x^2-x)(x^4-x^3)$, which is not even homogeneous. – Steven Landsburg May 17 '12 at 22:21
Duh, that was stupid of me. I was convinced that the "grading" works fine wrt multiplication, forgetting that it has to work wrt addition as well (i.e. that each degree has to consist of homogenous polynomials). Thanks for this lesson :/ – Simon Markett May 18 '12 at 10:20