Why isn't the crossed product algebra $(K(\sqrt{a}), Gal(K(\sqrt{a})/K, k)$ independent of a?

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I am trying to work through an example to understand the correspondence between $H_2(Gal(K(\sqrt{a})/K),K(\sqrt{a}))$ and $Br(K(\sqrt{a}),K)$. Here is my confusion:

Let $E = K(\sqrt{a})$ be a quadratic extension and let $G = Gal(K(\sqrt{a})/K) = \{\gamma, id\}$. ($\gamma$ represents conjugation of $\sqrt{a}$).

Then let: $k_{id,id} = 1$, $k_{id, \gamma} = 1$, $k_{\gamma, id} = 1$, $k_{\gamma, \gamma} = -1$. From this factor set, we can construct the crossed product algebra $(E, G, k) = span_{k(\sqrt{a})}(u_{id}, u_\gamma) \cong \big( \frac{a, -1}{K} \big) \cong \big( \frac{a,a}{K} \big)$. Now, note that if a is a sum of two squares, we have that $ \big( \frac{a,a}{K} \big) \cong \big( \frac{1,1}{K} \big) \cong M_2(K) \cong (E, G, 1)$. However, this would imply then that $k$ is 1. (Which isn't true because $k_{\gamma,\gamma} = -1$).

In other words: I don't understand how $(E, G, k)$ can depend on $a$ if $k$ doesn't depend on $a$ (and there is an isomorphism sending $k$ to $(E, G, k)$). Also, for reference, I am using Jacobson's Basic Algebra II for this description of crossed product algebras. Thanks so much!

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$k$ does depend on $a$ because the codomain of the cocycle is an extension field that depends on $a$. I was forgetting that changing the codomain will also change the coboundary that we mod out by -- making it possible for my example k to become trivial depending on a.