For $X$ a topological space, from the short exact sequence
$$ 0 \rightarrow \mathbb{Z}/2 \rightarrow \mathbb{Z}/4 \rightarrow \mathbb{Z}/2 \rightarrow 0 $$
we get a Bockstein homomorphism
$$H^i(X, \mathbb{Z}/2) \rightarrow H^{i+1}(X, \mathbb{Z}/2)$$
This is also known as the Steenrod square $Sq^1$.
Now suppose instead that $X$ is a variety over a (not algebraically closed) field. We still get a sequence
$$ 0 \rightarrow \mathbb{Z}/2 \rightarrow \mathbb{Z}/4 \rightarrow \mathbb{Z}/2 \rightarrow 0 $$
inducing a Bockstein homomorphism in etale cohomology
$$H^i_{et}(X, \mathbb{Z}/2) \rightarrow H^{i+1}_{et}(X, \mathbb{Z}/2).$$
However, there is also a short exact sequence
$$ 0 \rightarrow \mathbb{Z}/2 \rightarrow \mathbb{Z}/4(1) \rightarrow \mathbb{Z}/2 \rightarrow 0 $$
where $\mathbb{Z}/4(1)$ denotes the Tate twist, probably less confusingly written as $\mu_4$. This also induces a [presumably different] Bockstein map in etale cohomology
$$H^i_{et}(X, \mathbb{Z}/2) \rightarrow H^{i+1}_{et}(X, \mathbb{Z}/2).$$
Question: Which of these is the "right" Bockstein homomorphism?
This question is a little open-ended. My main criterion for "right" is that the map be "$Sq^1$ on etale cohomology", meaning it fits into an action of the Steenrod algebra.
There are other possible criteria. For instance, a literature search revealed that people have defined notions of "Bockstein homomorphism" and "Steenrod operations" on Chow rings, motivic cohomology, … so "right" could also mean "compatible with these other things". (Hopefully the answer is the same.)
Relevant literature:
- Jardine's paper https://link.springer.com/chapter/10.1007%2F978-94-009-2399-7_5 gives a version with $\mathbb{Z}/4$
- Guillou and Weibeil (http://arxiv.org/pdf/1301.0872.pdf) describe a $Sq^1 : H^k(\mu_2^i) \rightarrow H^{k+1}(\mu_2^{2i})$, so the weight has doubled. I don't know if it fits into a Bockstein story.
Unfortunately, I can't really parse what's happening in these papers.
Some motivation/example: For $X \subset Y$ a closed embedding of smooth varieties of codimension $r$, I have a cycle class $[X] \in H^{2r}_X(Y; \mathbb{Z}/2)$. I would like to show that $Sq^1 [X]=0$ (it may not be true). Since the cycle class lifts to $H^{2r}_X(Y;\mathbb{Z}_{2}(r))$, this depends on which sequence is related to $Sq^1$.
Best Answer
You maybe want to have a look at
There are two sequences of cohomology operations in motivic and étale cohomology which can rightfully be called Steenrod operations. One comes from a "geometric" model of the classifying space of the symmetric groups, the other one from a "simplicial" model. The first one is related to Voevodsky's motivic Steenrod algebra, the second one to the more classical Steenrod algebra (as in Denis Nardin's answer).
The paper mentioned above provides a comparison between these sequences of cohomology operations (see Theorem 1.1 part iii for the étale cohomology) in terms of cup products with powers of a motivic Bott element in $H^0(\operatorname{Spec} k,\mathbb{Z}/\ell\mathbb{Z}(1))$.
Addendum: Actually, the Bockstein operations are the same in both sequences of cohomology operations, and they are the ones defined by the $\mathbb{Z}/4\mathbb{Z}$ extension. (So probably the first one should be the "right".) This follows from the paper of Brosnan and Joshua as well as the paper by Guillou and Weibel in the question. Note that $\mu_2^{\otimes i}\cong\mu_2^{\otimes 2 i}\cong\mathbb{Z}/2\mathbb{Z}$ and the $Sq^1$ in the paper of Guillou-Weibel actually fits the Bockstein story.
Another note: If the motivation/example is defined over a field with $4$-th root of unity then $\mathbb{Z}/4\cong\mu_4$ (compatible with the extensions) and so both operations you defined agree.