Let $X$ be a nonsingular variety. (Perhaps some/all of this works over more general smooth schemes, but let's stick to the simple case.)


In, e.g., Fulton's *Intersection Theory* chapter 15, and Soule's *Lectures on Arakelov Geometry* chapter 2, there are brief expositions on the associated graded $\mathbb{Z}$-algebra $\text{Gr }K(X)$ of the Grothendieck group under its filtration by dimension of support, and its relationship to the Chow ring $A(X)$. These treatments are pretty much just glancing mentions, so I'm left somewhat confused about the situation. Specifically:


We have a surjective graded-group homomorphism $\phi:A(X)\to \text{Gr }K(X)$ given by $[V]\mapsto [\mathcal{O}\_V]$. Further, some geometry shows that the Chern character takes $[\mathcal{O}\_V]$ to $[V] +\alpha$ for $\alpha$ of lower-dimensional support, hence induces a graded-group homomorphism in the opposite direction $\text{ch}':\text{Gr }K(X)\to \text{Gr }A(X)\otimes \mathbb{Q}$, so that $\text{ch}'\circ \phi$ is the natural inclusion. So when we pass to rational coefficients everywhere, everything becomes an isomorphism, and hence $\text{ch}$ is also an isomorphism of graded groups since the associated graded module is a faithful functor, and hence it is also an isomorphism of rings. All of this is fine. (Unless I made a mistake!)


But now since $A(X)$ is already a graded algebra, it is naturally isomorphic to $\text{Gr }A(X)$. So we have that $K(X)\otimes \mathbb{Q}$ is also isomorphic to $\text{Gr }K(X)\otimes \mathbb{Q}$ as graded groups, which makes sense - after killing torsion, $K(X)$ already seems to be a graded group since direct sum of sheaves preserves dimension so long as we don't get killed off by one of the exactness relations. I am *pretty* sure this is all right.


So my clarifying questions are:


1) How exactly does taking the associated graded algebra of $K(X)$ by dimension of support change the multiplicative structure? And - does it? $K(X)\otimes \mathbb{Q}$ is already a graded $\mathbb{Q}$-algebra by the Chern character isomorphism, since the Chow ring is. But it seems that the induced gradation by this isomorphism is very weird since, e.g. $[\mathcal{O}\_V]$ corresponds not to $[V]$ but $[V]+\alpha$ as mentioned above, so it can't be the same gradation.


2) On a related note, $\phi,\text{ch'}$ can't be ring homomorphisms, right? Since then that would imply the impossible thing from #1. But this seems odd, because in the case of $\phi$ at least, $\phi([V]\cdot [W])=\phi([V])\cdot\phi([W])$ seems to be exactly the (true) Serre intersection multiplicity formula - or would be, if all the nonvanishing Tors had support on $V\cap W$, which I'm not sure is the case. What exactly is going on here? I'm particularly interested in whether there is a quick way to prove the Serre intersection multiplicity formula here quickly in full rigor. 



What does taking the graded algebra do to the Grothendieck group, and its relation to the Chow ring?
$\def\ZZ{\mathbb{Z}}\def\QQ{\mathbb{Q}}\def\cO{\mathcal{O}}\def\FF{\mathbb{F}}$Does this help?
$K(X) \otimes \mathbb{Q}$ is a graded ring in the sense that it is isomorphic (by the Chern character map) to $A(X) \otimes \mathbb{Q}$, which is graded. I would perhaps prefer to say that it is "gradeable", since the grading isn't very obvious in terms of $K$-theory. The most $K$-theoretic way I know to describe it is that the [Adams operators](https://en.wikipedia.org/wiki/Adams_operation) $\psi^k$ act by $k^j$ on the $j$-th graded piece. The corresponding descending filtration is the filtration by codimension.


For example, let $L$ be a line bundle with $c\_1(L) = D$. Then $ch(L) = e^D$. So $ch(\log L) 
=D$ and $\log [L]$, defined as the class $([L]-1) - ([L]-1)^2/2 + ([L]-1)^3/3 - \cdots$ in $K(X)$, is pure of degree $1$. Indeed, $\psi^k [L] = [L^k]$, so $\psi^k \log [L] = \log [L^k] = k \log [L]$. Let's abbreviate $\log [L]$ by $z$. The structure sheaf of the vanishing locus of a section of $L$ has $K$-class $1-L^{-1} = 1-e^z = z-z^2/2+z^3/6-z^4/24+\cdots$. So this class is in the filtered part that has degree $\geq 1$, but is not of pure degree.


A natural question, to which I don't know the answer, is whether the integer $K$-ring has a natural grading $K(X) \cong \bigoplus K\_i(X)$, which turns into this grading when we tensor by $\mathbb{Q}$.




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I now have some concrete examples of algebraic varieties where $K(M, \mathbb{Z})$ is not isomorphic to $A(M, \mathbb{Z})$, and where $K(M, \mathbb{Z})$ does not support a grading with the correct Betti numbers.


The first example, due to Matt Larson, is to take $Q$ a quadratic $3$-fold in $\mathbb{P}^4$. Then $Q$ is a homogenous space for $SO(5)$, so it has a cellular stratification, as follows: Consider a chain $P \subset L \subset H \subset Q$, where $P$ is a point, $L$ is an isotropic line, $H$ is a singular hyperplane section containing $L$, and $Q$ is the whole space. Then $L\setminus P \cong \mathbb{A}^1$, $H \setminus L \cong \mathbb{A}^2$ and $Q \setminus H \cong \mathbb{A}^3$. The singular hyperplane section $H$ is rationally equivalent to a smooth hyperplane section $H'$ through $L$, in which case $H'$ is a quadratic surface and $L$ is one of the two rulings of $H'$; I'll prefer $H'$ for multipication computations, but $Q \setminus H'$ isn't $\mathbb{A}^3$, so I need $H$ to get the cellular stratification.


Thus, the Chow groups of $Q$ are $(\ZZ [Q], \ZZ [H], \ZZ [L], \ZZ [P])$. The non-obvious products are
$$[H]^2 = 2[L],\ [H][L] = [P].$$
To see the former relation, the normal bundle to $H'$ is $\mathcal{O}(1,1)$, so the self-intersection of $H'$ is the sum of a line from one ruling and a line from the other ruling, both of which are rationally equivalent to $L$ within $Q$.


$K^0(Q)$ also has $\ZZ$-basis the structure sheaves of $Q$, $H$, $L$ and $P$. This time, the products are
$$[\cO\_H]^2 = 2[\cO\_L]-[\cO\_P],\ [\cO\_H][\cO\_L] = [\cO\_P].$$
To see the former relation, note that the self-intersection of $\cO\_{H'}$ is the sum of one line from each ruling, minus the point where they intersect.


If these rings were isomorphic, their tensor products with $\mathbb{F}\_2$ would be isomorphic. A useful invariant of an $\mathbb{F}\_2$ algebra is the dimension of the vector space spanned by the squares of all elements. In Chow, we have $[H]^2 \equiv [L]^2 \equiv [P]^2 \equiv 0 \bmod 2$, so this vector space is $1$-dimensiona with basis $1$; in $K$-theory, we have $[\cO\_H]^2 \equiv [\cO\_L] \bmod 2$ and $[\cO\_L]^2 \equiv [\cO\_P]^2 \equiv 0 \bmod 2$, so this vector space is $2$-dimensional with basis $1$, $[\cO\_L]$. (Because squaring is a linear map modulo $2$, we only need to square the generators.)


Let's see further that the $K$-theory ring can't be graded as $K\_0(Q) = \ZZ \oplus \ZZ \gamma\_1 \oplus \ZZ \gamma\_2 \oplus \ZZ \gamma\_3$, with $\gamma\_i$ in degree $i$. Degree considerations show that $\gamma\_2$ and $\gamma\_3$ square to $0$, so they must lie in $\ZZ [\cO\_L] + \ZZ [\cO\_P]$, and thus must span it. Since $\gamma\_1$ is nilpotent, it must be of the form $a [\cO\_H] + b[\cO\_L] + c[\cO\_P]$; since $\ZZ \gamma\_1 \oplus \ZZ \gamma\_2 \oplus \ZZ \gamma\_3$ is the maximal ideal, we must have $a = \pm 1$ and, WLOG, we can take $a=1$. Then $\gamma\_1^2$ should be $m \gamma\_2$ for some integer $m$. But $\gamma\_1^2 = ([\cO\_H] + b[\cO\_L] + c[\cO\_P])^2 = 2 [\cO\_L] + (2b-1) [\cO\_P]$ and $GCD(2, 2b-1) = 1$ so we must have $m=1$ and $\gamma\_2 = 2 [\cO\_L] + (2b-1) [\cO\_P]$. Similarly, $\gamma\_3$ must be $\gamma\_1^3/2 = [\cO\_P]$. But $([\cO\_H] + b[\cO\_L] + c[\cO\_P], 2 [\cO\_L] + (2b-1) [\cO\_P], [\cO\_P])$ only span an index $2$ sublattice in $K^0$, a contradiction.




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I have asked various people for toric counterexamples, and now I have one. I'll do this more quickly. Let $C$ be the toric variety whose fan is combinatorially the normal fan to the cube, but with rays in directions
$$(1,0,0), (0,1,0), (0,0,1), (-1,0,0), (1,-1,0), (0,2,-1).$$
This is a so-called "Bott tower".


This space is a tower of $\mathbb{P}^1$ bundles, so I can compute $K$-theory and cohomology/Chow using the projective bundle theorem. ([Cohomology](https://math.stackexchange.com/questions/2644888/cohomology-of-projective-bundle-only-depends-on-base-and-fiber)) ([$K$-theory](https://mathoverflow.net/questions/235612/closed-formulas-for-topological-k-theory?noredirect=1&lq=1)).


For cohomology/Chow, I get
$$H^{\ast}(C) = A^{\ast}(C) = \ZZ[X,Y,Z]/\langle X^2, Y(Y+X), Z(Z+2Y) \rangle$$
In other words,
$$X^2=0, Y^2 = -XY,\ Z^2 = -2YZ.$$
Note that the square free monomials in $X$, $Y$, $Z$ are a basis.


In $K$-theory, I get
$$K^{0}(C) = \ZZ[u,v,w]/\langle (u-1)^2,\ (v-1)(vu-1),\ (w-1)(wv^2-1) \rangle.$$


Put $u = 1+x$, $v = 1+y$ and $w = 1+z$. Since $u$, $v$, $w$ are classes of line bundles, $x$, $y$ and $z$ are in codimension $1$. We get the relations
$$x^2 = 0,\ y(y+x+xy)=0,\ z(z+2y+y^2+2yz+y^2z)=0.$$
So
$$x^2=0$$
$$y^2 = -xy-xy^2 = -xy+x(xy+xy^2) = -xy.$$
$$z^2 = -2yz - y^2 z - 2y z^2 = -2yz + xyz - 2y(-2yz +\cdots) = -2yz + xyz +4 y^2 z = -2yz - 3xyz.$$
Here the "$+\cdots$" are terms that I know will die at the next step for dimension reasons. To summarize
$$x^2=0,\ y^2 = -xy,\ z^2 = -2yz-3xyz.$$


We can see that the Chow relations are the associated graded of the $K$-theory relations.


However, we can also see that the rings are not isomorphic. Tensor
with $\FF\_2 = \ZZ/2 \ZZ$ to get
$$\FF\_2[X,Y,Z]/(X^2=Z^2=0,\ Y^2 = YZ)$$
and
$$\FF\_2[x,y,z]/(x^2=0, y^2=yz, z^2 = xyz)$$
respectively. The two rings are not isomorphic, because there squares form a $2$-dimensional $\FF\_2$ vector space in the first case (basis $\{ 1, YZ \}$) and $3$-dimensional vector space in the second (basis $\{ 1,yz, xyz \}$).


