In G. M. L. Powell's note 'Steenrod operations in motivic cohomology', he stated that if $\mathrm{char}(k)=0$, $$H^{*,*}(k,\mathbb{Z}/2)=K_*^M(k)/2[\tau]$$ where $\tau\in H^{0,1}$ is the unique nonzero element.
I wonder whether this result holds when $char(k)>0$?
This holds if the characteristic of $k$ is not 2, and it follows from the Milnor conjecture proved by Voevodsky.
Voevodsky ultimately proved the following (Theorem 6.17 in https://annals.math.princeton.edu/wp-content/uploads/annals-v174-n1-p11-s.pdf): If $m>0$ and $X$ is smooth over a field $k$ of characteristic prime to $m$, then the map $$ H^n(X, \mathbb Z/m(i)) \to H^n_{\mathrm{et}}(X,\mu_m^{\otimes i}) $$ is an isomorphism provided that $n\leq i$. When $X=\operatorname{Spec}(k)$ the right-hand side is $K^M_*(k)/m[\tau^{\pm 1}]$ if there is a primitive $m$th root of unity $\tau\in\mu_m(k)$ (this follows from the isomorphism for $n=i$), and the left-hand side is zero for $n>i$, so we know everything.
When the characteristic is 2, or more generally when the characteristic is $p$ and the coefficients are $\mathbb Z/p$, we also know everything by Geisser and Levine: in this case the motivic cohomology vanishes when $n\neq i$, so there is only Milnor K-theory.
