In definition of Galois cohomology, is continuity necessary? Let $K$ be a number field.
Galois cohomology $H^1(Gal(\overline{K}/K), E)$ is composed of representative of continuous maps $Gal(\overline{K}/K) \to E$ which satisfies cocycle relation.
If we discard the condition 'continuous', what will happen ?
Continuous is essential condition of definition ?
Here, $Gal( \overline{K}/K)$ with Krull topology, $E$ with discrete topology.
P.S Thank you for comments, if we drop continuity, for example, what kind of property we lose ? (For example, Hilbert theorem 90)
 A: This is a good question, so thank you for asking.
I'd rather beg: what properties would you have if you drop the continuity? Consider the case where $E$ is some abelian group with the trivial action of $G=\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$. Then you could consider two different $H^1$. The one practically always considered is the continuous $H^1$ which gives you
$$H^1(G,E) = \mathrm{Hom}_{\mathrm{cont}-\mathrm{group}}(G,E).$$
I guess the question is asking about the abstract cohomology, that I will write $h^1$ for lack of a better symbol. It will be given by
$$h^1(G,E) = \mathrm{Hom}_{\mathrm{group}}(G,E).$$
In the $H^1(G,E)$ any homomorphism $\varphi: G \rightarrow E$ factors through the topological abelianization which is
$$G^{\mathrm{ab}} = G/(\text{closure of commutator subgroup}) = \mathrm{Gal}(\mathbf{Q}^{\mathrm{ab}}/\mathbf{Q})$$
where $\mathbf{Q}^{\mathrm{ab}}$ is the maximal abelian extension of $\mathbf{Q}$, known by a theorem named after Kronecker-Weber to be the field generated by the cyclotomic extension $\mathbf{Q}(\zeta_m)$ for all $m$-th roots of unity $\zeta_m$ and all $m$. We know this group very well, it is topologically pro-cyclic, isomorphic to the profinite completion $\widehat{\mathbf{Z}}$ of the integers.
On the other hand, the $h^1(G,E)$ is going to calculate the abstract abelian quotient. I haven't a clue what to say about it. This abstract abelian group $\Gamma$ is just $G$ modulo its commutator subgroup, but that subgroup is not closed (according to this mathoverflow question) and so the quotient
$$\Gamma \twoheadrightarrow G^{\mathrm{ab}}$$
is really some non-trivial quotient.
I wish I had a concrete phenomenon to tell you about, but the concrete phenomena all happen in the continuous setting. For instance, when $E = \mathbf{C}^\times$ any continuous map $G^{\mathrm{ab}} \rightarrow E$ will be finite image and I somehow doubt that for $\Gamma$. (Anyone have a proof or counter-example? Edit: the continuous ones of course end up being identifying in a super concrete way with Dirichlet characters.) I doubt the cohomology groups in the abstract setting have finite-dimensional properties when characteristic zero vector spaces are plugged in. And so on and so forth. This concrete behavior is crucial to getting the machine of Galois cohomology off the ground, letting you reasonably ask questions about things like a Theorem 90 in this setting. So again, I'm begging: what properties would you actually have?
