Let $A^\ast$ be an algebraic oriented cohomology theory (i.e. it is equipped with certain push-forwards for projective morphisms of smooth varieties over the base field $k$; see section 2 of <http://www.math.uiuc.edu/K-theory/0535/orient.pdf> for more detail); let $f:Y\to X$ be a finite morphism of smooth varieties whose degree is $d$. I would like to prove the following conjecture: if $f^\ast h=0$ for $h\in A^\ast (X)$, then $d^lh=0$ for some $l>0$ (one cannot take $l=1$ here when $A^\*$ is the K-theory). To this end it suffices to verify that $f\_\ast 1\_Y-d$ is nilpotent in $A^0(X)$ (since $f\_\ast f^\ast h=f\_\ast 1\_Y\cdot h$ by the property (v) in the reference cited). It seems sufficient to prove the latter for $A^\ast$ being the algebraic cobordism (as defined by Levine and Morel), since this is the universal algebraic oriented cohomology theory.


I would like to say that $f\_\ast 1\_Y-d$ is supported in codimension 1. Does $A^0(X)$ possess a multiplicative coniveau filtration? If $f$ is generically etale, then I can use the fact that $f'\_{\*}1\_{Y'}=d$ for $f':Y'\to X'$ being the (etale) morphism of generic points; Levine proves this in his cobordism book. 


Is there a better way to prove my conjecture (that would work even if $f$ is not generically etale)?



For a finite flat (etale?) morphism $f:Y\to X$, is $f_*1_Y-\deg f . 1_X$ nilpotent in $A^0(X)$, where $A^*$ is the algebraic cobordism? 
I think for such a statement to be true, you would need the oriented cohomology theory $A^\*$ to be generically constant in the sense of Levine-Morel and it should also satisfy the localization property. In this case, we can apply the Generalized degree formula of Levine-Morel (Theorem 4.4.7 in their book). It implies that $f\_\*1\_Y-\deg(f\_\*1\_Y)$ is nilpotent in $A^0(X)$ where $\deg$ is the homomorphism $A\_\*(X)\to A\_{\*-\dim\_kX}(X)$. Also, they show that for $f$ seperable of relative dimension 0, this $\deg$ coincides with the usual notion of degree of a morphism of schemes, so that in your case $\deg(f\_\*1\_Y)=d$.


