3

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\...


1

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 $\mathbb Z$ ...


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