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I think the center $Z(\mathrm{GL}_{2}(\mathbb{R}))$ in your bijection should be replaced by its identity component $Z(\mathrm{GL}_{2}(\mathbb{R}))^{\circ}$. Otherwise you will get the bijection: $$\pm\mathrm{Id}_{2}\mathrm{GL}_{2}(\mathbb{Q})\backslash\mathrm{GL}_{2}(\mathbb{A}_{\mathbb{Q}})^{1}\simeq\mathrm{GL}_{2}(\mathbb{Q})Z(\mathrm{GL}_{2}(\mathbb{R}))\... 2 Clearly the ring class field of conductor p\infty of \mathbb Q is contained in the ray class field with conductor p of K. But the ray class field of conductor p\infty of \mathbb Q is the field of p-th roots of unity, which contains a unique quadratic subextension k = {\mathbb Q}(\sqrt{p^*}). Thus the ring class field of conductor p\infty ... 2 Observe that not only 23 but also 7=\alpha-3-(\alpha-10)=7 is a element of \mathcal{O}. Note also that 23 and 7 are coprime. Thus there are integers u,v such that 1=23u+7v implying that 1\in \mathcal{O}. 2 P(X)=X^n-a is a monic polynomial which is separable mod the maximal ideal. The residue field of K^{ur} is algebraically closed and thus P has a root in the residue field. Taking a lift of this root, we find some x \in O_{L} such that P(x) is not invertible and P’(x) is (invertible), where L/K is a finite unramified extension. Then Hensel’s ... 1 We will use the fact that R/I is a domain if and only if I is a prime ideal of the ring R. Let f:\mathbb Z\to \mathbb Z[\sqrt d]/(p,\sqrt d) be a mapping given by f(x)=x+(p,\sqrt d). It is routine to check that this is an onto ring homomorphism. Moreover, \ker(f)=(p,d)\subset \mathbb Z. By the First Isomorphism Theorem,$$\mathbb Z/(p,d)\cong\...
One important ingredient is the following theorem of Swan (Theorem 8.1. in Induced representations and projective modules): If $P$ is any projective f.g. $RG$-module, then $P \otimes_R K$ is free. With that understood, we can do that following: We actually have a splitting $K_0 (\mathcal P_R ) \cong \mathbb{Z} \oplus \widetilde{K_0 RG}$, where $\mathbb Z$ ...