Let a prime $p\equiv 1$ mod 4. I am trying to study the cokernel of the following map: $f:\operatorname{H}^1(Spec (\mathbb{Z}_p);\mathbb{Z}/4)\to \operatorname{H}^1(Spec (\mathbb{Q}_p);\mathbb{Z}/4)$, which I know being isomorphic to the cyclic group $\mathbb{Z}/4$.
In particular, I am trying to understand whether a certain class is actually generating the whole cokernel.
Consider $\overline{\alpha}\in\operatorname{Coker}f$ with $\alpha \in \operatorname{H}^1(Spec (\mathbb{Q}_p);\mathbb{Z}/4)$ corresponding to $K_p:=K\mathbb{Q}_p$, with $K$ the unique $\mathbb{Z}/4$-Galois subextension of $\mathbb{Q}\subset \mathbb{Q}(\zeta_p)$.
Proving that $K_p$ is ramified at $p$ should be enough to have that $\overline{\alpha}$ in non-zero, but how can I verify whether it has actually order 4?
Why do you need the number field $K$ to enter the game ? I guess that your $H^1$'s are étale cohomology groups, but they can also be interpreted as the more "visible" Galois cohomology groups $H^1(G(\mathbf Q_p^{unr}/\mathbf Q_p), \mathbf Z/4)$ and $H^1(G(\mathbf Q_p^{sep}/\mathbf Q_p), \mathbf Z/4)$ with trivial action on $\mathbf Z/4$ ; here $G(.)$ denotes the Galois group, $\mathbf Q_p^{sep}$ is the separable closure, and $\mathbf Q_p^{unr}$ is the maximal unramified extension of $\mathbf Q_p$. Since the Galois action is trivial, these $H^1$'s are actually $Hom$ groups, and your map $f$ is just the (injective) natural map $Hom (G(\mathbf Q_p^{unr}/\mathbf Q_p), \mathbf Z/4) \to Hom (G(\mathbf Q_p^{sep}/\mathbf Q_p),\mathbf Z/4)$, whose cokernel is $Hom (G(\mathbf Q_p^{sep}/\mathbf Q_p^{unr}), \mathbf Z/4)$. Since the extension $\mathbf Q_p (\zeta_p) /\mathbf Q_p$ is totally ramified of degree $p-1$ , and $p\equiv 1\pmod 4$, this cokernel is obviously $\cong$ the dual of the Galois group over $\mathbf Q_p$ of the unique $\mathbf Z/4$-subextension of $\mathbf Q_p (\zeta_p) /\mathbf Q_p$ .