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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 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)?

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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 $\deg(f_*1_Y)=d$.

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