A class in the motivic cohomology group $H^{0,1}(\operatorname{Spec}k;\mathbb{Z}/p)$ In the following paper
N. Yagita, Examples for the Mod p Motivic Cohomology of Classifying Spaces,
on the first page, below display (1.1), it says "It is known that there is an element $\tau\in H^{0,1}(\operatorname{Spec}k;\mathbb{Z}/p)$...".
So, where and how exactly is this class $\tau$ defined?
Since there is no motivic stable cohomology operation of bidegree $(0,1)$, is it true that this class $\tau$ vanishes after stabilization?
 A: The definition of the class is actually given in the cited sentence. For the relevant motivic cohomology group we have
$${\rm H}^{0,1}({\rm Spec} k,\mathbb{Z}/p\mathbb{Z})\cong {\rm H}^0_{\rm ét}({\rm Spec } k,\mu_p)\cong \mu_p(k)$$
The complex realization (denoted $t_{\mathbb{C}}^{\ast,\ast}$ in Yagita's paper) in this case boils down to an identification $\mu_p(k)\cong\mathbb{Z}/p\mathbb{Z}$, hence the class $\tau$ is basically a choice of primitive $p$-th root of unity in $k$.
The notation is pretty standard in the literature, it already appears in Voevodsky's "Motivic cohomology with $\mathbb{Z}/2$-coefficients" (but of course, in the $p=2$ situation, there is only one choice of primitive 2nd root of unity).
Not sure what is meant with the question on stabilization. Stability in the context of cohomology operations refers to compatibility with the suspension isomorphisms. Of course, as a class in the motivic cohomology of the base field, the class $\tau$ doesn't die under the suspension isomorphisms. I think there's some misconception concerning the motivic cohomology operations in the question. The motivic Steenrod algebra is an algebra over the motivic cohomology of the base field, hence "multiplication by $\tau$" actually is a bistable motivic cohomology operation of bidegree $(0,1)$.
