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Let $k=\mathbb{Q}(\sqrt{d_1}, \sqrt{-d_0})$, be an imaginary biquadratic number field and $H^1(G, U_k)$ the first cohomology group with $G=\mathrm{G}al(k/\mathbb{Q})$ and $U_k$ is the unit group of $k$.

I want to know how I can calculate $H^1(G, U_k)$? and thank you.

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You will find the answer here mathoverflow.net/questions/103345/… –  Arijit Apr 28 '13 at 22:35
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I don't think it is too bad to do the calculation straight from the definition of $H^1$ as cocycles modulo coboundaries. First, you need to work out $U$ and the action of $G$ on it. Then you identify the cocycles to a subgroup $C$ of $U^3$ by mapping $a$ to $a_{\sigma},a_{\tau},a_{\sigma\tau}$, where $G = \{ 1, {\sigma},{\tau},\sigma\tau \}$ (note $a_1 =1$). Which subgroup? Namely, the subgroup cut out by the cocycle conditions. Since $G$ is small, that is just a handful of conditions. Now you work out the coboundaries which is the image $B$ of $U$ in $U^3$ by the natural map and, finally, compute $C/B$. Note that $U$ is a finitely generated group of rank $2$, so all these groups are manageable.

People working on the Brauer-Manin obstruction for surfaces do this kind of calculation to compute $H^1(G,NS(X))$, where $NS$ is the Nerón-Severi group. Martin Bright, who shows up on MO from time to time has code for this: http://www.boojum.org.uk/maths/quartic-surfaces/index.html

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Following Felipe's answer, I would like to suggest a slightly different strategy which makes deep use of your very peculiar setting (and probably boils down to Felipe's advise). Warning: At the very end, I have a computation which I cannot perform, so this may not qualify as a proper "answer"

Let me call $L$ your quadratic field $L=\mathbb{Q}(\sqrt{-d_0})$ and write $H=\mathrm{Gal}(k/L)$. The advantage is that now both $H$ and $G/H$ are cyclic. Look at the Inf-Res sequence $$ 0\to \hat{H}^1(G/H,U_L)\to\hat{H}^1(G,U_k)\to\hat{H}^1(H,U_L)^{G/H}\to \hat{H}^2(G/H,U_L)\;. $$ Since both $H$ and $G/H$ are cyclic, you can shift the Tate cohomology (I am assuming you are familiar with this, and with the fact that in degrees $\geq 1$ Tate cohomology coincides with the usual group cohomology, so the second term in the above sequence is the guy you are asking about: if you are not, check either the wikipedia page or Atiyah-Wall's paper in Cassels-Fröhlich quoted in the references in wikipedia) and you can rewrite the above as $$ 0\to \hat{H}^{-1}(G/H,U_L)\to\hat{H}^1(G,U_k)\to\hat{H}^{-1}(H,U_L)^{G/H}\to \hat{H}^{0}(G/H,U_L)\;. $$ Now observe that you have "almost no units" both in $\mathbb{Q}$ and in $L$. So, the above groups are easy to compute. I make the explicit computation assuming $L$ is neither $\mathbb{Q}(i)$ nor $\mathbb{Q}(\sqrt{-3})$ and that the fundamental unit $\varepsilon$ of $k$ has norm to $L$ equal to $1$. The two special imaginary fields do not present differences in the results, and if the norm of $\varepsilon$ is $-1$ some computations are easier, so I leave this to you.

By definition $\hat{H}^{-1}$ is the kernel of the norm modulo the submodule generated by $(\sigma-1)$, where $\sigma$ generates $H$. The unique units of $L$ are $\pm 1$, so $\hat{H}^{-1}(G/H,U_L)=U_L=\{\pm 1\}$. Similarly, $\hat{H}^0$ are invariants modulo norms, so the last term is $\hat{H}^{0}(G/H,U_L)=\mathbb{Z}^\times=\{\pm 1\}$. Finally, by Dirichlet, we can write $U_k=\{\pm 1\}\times \varepsilon^{\mathbb{Z}}$ and the assumption $\operatorname{Norm}^k_L(\varepsilon)=1$ tells us $\sigma(\varepsilon)=\varepsilon^{-1}$. It follows that the submodule generated by $(\sigma-1)$ in $U_k[\operatorname{Norm}^k_L=0]=\{\pm 1\}\times \varepsilon^{\mathbb{Z}}$ coincides with $1\times \varepsilon^{2\mathbb{Z}}$ and finally $\hat{H}^{-1}(H,U_L)=\{\pm 1/}\times\mathbb{Z}/2$: we need to take $G/H$ invariants of this. But the first summand is really $\mathbb{Z}^\times$, so has trivial Galois action, and the second too has trivial action of $G/H$ since this is the Galois group of $L/\mathbb{Q}$ and $\varepsilon$ comes from the real quadratic extension. Moreover, we see that the last arrow in the above sequences is surjective, because tracking what happens to your units shows that the copy of $\mathbb{Z}^\times$ which we have just found in $\hat{H}^{-1}(H,U_L)$ is sent isomorphically to $\hat{H}^0(G/H,U_L)$. Putting all together, the Inf-Res sequence is now $$ 0\to\{\pm 1\}\to H^1(G,U_k)\to \{\pm 1}\times\mathbb{Z}/2\twoheadrightarrow \{\pm 1} $$ showing that your group $H^1(G,U_k)$ is an extension of a cyclic group of order $2$ by another one. Now, I guess this group is cyclic or not according as whether the norm of $\varepsilon$ to the real quadratic field is the fundamental unit $u$ of $\mathbb{Q}(\sqrt{d_1})$: just look at the cocyle sending $\sigma\mapsto -1,\tau\mapsto u$ where $\tau$ generates $G/H$. It has order $2$ or $4$ (modulo coboundaries) following the value of $\tau(\varepsilon)\cdot\varepsilon$ but I cannot turn this into a complete proof that, if it has order $2$, then there is no other possible candidate for a cocylce of order $4$.

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