Dualizability and motivic cohomology

Question

Suppose $k$ is an algebraically closed field of characteristic $p$. Let $A=\mathbb{Z}/\ell\mathbb{Z}$, $\ell$ a prime coprime to $p$. Denote by $MA$ the motivic Eilenberg-Maclane spectrum over $k$. Is it true that in the category of $MA$-modules, $MA$ is dualizable? I want something weaker, actually. I want to justify that in the category of $MA$-modules and for smooth $k$-scheme $X$, we can 'dualize' $\text{Map}(MA, MA\wedge X_+)$ in the category of $MA$-modules to get $\text{Map}(MA\wedge X_+, MA)$. I want the process applied twice give back the original object. Am I missing something trivial?

Answer 1

For the equation you are asking about, you don't want $MA$ to be dualizable (which is lucky, because it's not), you want $MA\wedge X_+$ to be dualizable as an $MA$-module. This is true in the situation you are interested in but it is not trivial. This is proven in the following paper:

Marc Hoyois, Shane Kelly, Paul Arne Østvær - The motivic Steenrod algebra in positive characteristic.

Unfortunately a shift appears, which for smooth projective varieties is given by the dimension of the normal bundle, but I am not completely sure of what happens for non-proper varieties.

(I'm not sure if there is a simpler proof in your special case that appeared earlier in the literature. The idea is that $MX$ is always dualizable when $X$ is smooth and projective and then you can use alterations to "complete" any $X$ up to a factor prime to $l$).