This is all about how to make consistent choices of square roots. If you want to attach a meaning to half-integer weight modular forms of level $\Gamma$, then you need to pick, in some consistent ...

First of all: $\rho$ is usually used for a Galois representation; in the context, I'm pretty sure it's supposed to be the Galois representation given by the Tate module of an elliptic curve over $\... View answer Accepted answer 13 votes There's no law against reading more than one book at a time! Although algebraic number theory and algebraic geometry both use commutative algebra heavily, the algebra needed for geometry is rather ... View answer Accepted answer 13 votes I'm going to suppose that we know some polynomial$F \in \mathbf{Q}[x]$such that$F(z)=0$. (This is somehow the only reasonable "certificate" for a number being algebraic -- I suppose it's possible ... View answer 11 votes As for "by linearity": if you want to find$x$that is$p$-adically close to$x_1$and$q$-adically close to$x_2$, then you can do this as follows: suppose you can find a$y$that is$p$-adically ... View answer 11 votes An algebraic variety (over a non-algebraically-closed field) is "more than just a set of points". If X is an algebraic variety over$K$, you can make sense of the set$X(L)$of$L$-points for any ... View answer Accepted answer 9 votes Weston's definition is more general than Greenberg's. If$V$is 1-dimensional and corresponds to a ramified finite-order character, then$V$is ordinary in Weston's sense, but not in Greenberg's. If ... View answer Accepted answer 8 votes There is a question of conventions here. In lifting a Dirichlet character$\chi$mod$N$to a character$\chi_{\mathbf{A}}$of$\mathbf{A}^\times / \mathbf{Q}^\times$, you can either (1) arrange that ... View answer Accepted answer 8 votes I'm not sure this can quite be correct. The problem is that$Q^{p^m}$is going to tend to 1, so$Q^{p^m} - 1$tends to 0, not$1/Q$. I think you may have misread the paper and what was meant was$1/Q =...

For $X(N)$, one can give a $\mathbb{Q}$-model, but there's a sense in which doing so is "cheating". The moduli interpretation of $X(N)$ only makes sense over $\mathbb{Q}(\zeta_N)$; if $R$ is an ...

I'd say that one of the key things to get your head around is the theory of Riemann surfaces. D + S use this very heavily in the chapter on dimension formulae, and in various other places too. So you ...

In each of these cases, you have a group $G$ (with various additional structures), and a space of functions on that group, which is therefore naturally a $G \times G$ module under left and right ...

Let me answer your question as best I can; but I'll start by correcting a misconception in your question. It actually never happens that "all units are powers of just one unit" except when the unit ...

Here are two standard examples of non-split but quasi-split groups: Non-split tori (a dumb example), e.g. U(1) over $\mathbf{R}$. If $T$ is any torus, then $T$ is a Borel subgroup of itself, and this ...

There won't be a characterization depending only on the degree, because there can be more than one totally ramified extension of any given degree (while there is a unique unramified degree $d$ ...

As Dylan points out, parts (1) and (2) are clear. Moreover, $\mathbb{Z}[\zeta + \zeta^{-1}]$ contains $\zeta^j + \zeta^{-j}$ for all $j \ge 1$ (by induction using the binomial theorem); these include ...

In fact the situation is even worse than Michalis' answer suggests: a globally irreducible polynomial can be (in fact usually is) locally reducible almost everywhere. For instance, consider the ...

You could try looking in Jantzen's book "Representations of algebraic groups". This assumes the ground field is algebraically closed. For the full story, see Tits' paper "Représentations linéaires ...

Suppose $f$ and $g$ both have weight $k$ and level $\Gamma$. As John M notes, it is clear that there must exist some constant $N$ depending on $\Gamma$ and $k$ such that if $a_i = b_i$ for $0 \le i \... View answer Accepted answer 6 votes This is really a local question: you may as well assume$X$is defined over a local field$L$, it doesn't matter whether it comes from a number field. The point is that if$S = \operatorname{Spec} O_L$... View answer Accepted answer 6 votes There are three cases that can arise. The point is that if$\Delta(E) < 0$, then$E(\mathbf{R})$is the circle group$\mathbf{R} / \mathbf{Z}$, so any infinite subgroup is dense; while if$\Delta(E)...

There do exist relations between these elements. If $T$ and $R$ are your generators, then $R T^{-1} = \left(\begin{array}{rr} 1 & -1 \\ 2 & -1 \end{array}\right)$ and hence $(RT^{-1})^2$ is ...

As posed, the answer is clearly "no": it's not enough to know all but finitely many prime coefficients. To see why, let $f$ be your favourite non-zero modular form (of some weight $k$ and level $N$); ...

Here are a few reasons why one shouldn't expect $Cl_S(k)$ to be related to $Gal(k_S / k)$, or even to $Gal(k_S^{\mathrm{ab}} / k)$ where $k_S^{\mathrm{ab}}$ is the maximal abelian extension unramified ...

This is defined if (and only if) $d$ is congruent to 1 mod $p$. You can see this by thinking about the limit (in the $p$-adic topology) of $d^{n_i}$, where $n_i$ is some sequence of integers ...

There is a lot of literature on this, motivated by aplications to automorphic forms. These groups go by the name of "Bianchi groups" and that keyword should help you to find more literature on them. ...

This is clearly false: if the condition that every prime in $\mathcal{H}$ is split in $L$ is satisfied, then it is also satisfied if we replace $\mathcal{H}$ with any smaller subgroup $\mathcal{H}'$ ...

No, it does not, because FLT does not hold for general cyclotomic fields. Here's a totally stupid counterexample: take a Pythagorean triple, e.g. $3^2 + 4^2 = 5^2$. Then the number field generated by $... View answer Accepted answer 6 votes Yes, these three relations are all true. Since$\eta^{24}$is the weight 12 level 1 cusp form$\Delta\$, you can write them as relations between level 1 modular forms, and these are easy to check ...