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Splitting of primes in abelian extensions

The abelian extension $K$ gives you a subgroup $\mathrm{Gal}(\mathbb Q(\zeta_m)/K)\subset \mathrm{Gal}(\mathbb Q(\zeta_m)/\mathbb Q)=(\mathbb Z/m\mathbb Z)^\times$. This is precisely the subset you ...
Kenta S's user avatar
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Weil height and Norm of elements in a number field

You cannot write $d_v=[k_v:\Bbb{Q}_p]$, because you want to include the archimedean places as well. You could write $\Bbb{Q}_v$ without too much of a problem, though. To my knowledge, there isn't a ...
SomeCallMeTim's user avatar
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If p is prime then J = $\left \langle p, a + \sqrt{D} \right \rangle_{\mathbb{Z}}$ is ideal prime.

Your description of $J$ agrees with the more concise description $J = {\mathbf Z}p + \mathbf Z(a+\sqrt{D})$: it's the $\mathbf Z$-linear combinations of $p$ and $a+\sqrt{D}$. You should first prove ...
KCd's user avatar
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Is $H^1(G_{\Bbb{Q}_p},E[2])=0$ for good prime of $E/\Bbb{Q}$?

No - I don't think this is ever true. For an easy example, take any prime $p$ that splits completely in $\mathbb Q(E[2]))$. Then $H^1(G_{\mathbb Q_p}, E[2]) \cong H^1(G_{\mathbb Q_p}, \mathbb Z/2\...
Mathmo123's user avatar
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2 votes
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A limit problem in $p$-adic fields

Because $E$ is ramified over $F$, $E \ne F$. If $|\gamma - \bar \gamma| > |\gamma - a|$ then Krasner's lemma implies that $E=F$. This means the limit is at most $1$. Taking the computation you ...
Merosity's user avatar
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Field extension $K/\Bbb{Q}$ of given $e,f,g$ .

If $K = \mathbf Q(\alpha)$ and $\mathcal O_K = \mathbf Z[\alpha]$ (a "monogenic" number field, which includes all quadratic and cyclotomic fields), with $\alpha$ having minimal polynomial $h(...
KCd's user avatar
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3 votes
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Degree of field extension $\Bbb Q(\sum\limits_{k=1}^{\text{ord}_n(2)}\zeta_n^{2^k}):\Bbb Q$

The answer to your question is yes when $n$ is squarefree. Here is a general result. When $L/K$ is Galois with Galois group $G$, $L = K(\alpha)$, and $H$ is a subgroup of $G$, then we can ask ...
KCd's user avatar
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$F'/K'$ and $F/K$ be algebraic function fields (of one variable), $F⊆F'$, $K⊆K'$. Does $F'/F$ be finite separable implies the same for $K'/K$.

I believe it's true: Let $x \in K'$ and $f(X):=\min(x, K) \in K[X]$. Suppose that $f(X)=p(X)q(X)$, where $p(X), q(X) \in F[X]$ are monic irreducible polynomials over $F$ and $\deg p(X), \deg q(X) \ge ...
RiverOfTears's user avatar
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Inverse problem of the field extension $K/\Bbb{Q}$ of given number of prime above $p$

It is a theorem that for every number field, infinitely many primes split completely in it. So let $K$ be an arbitrary number field with degree $g$ over $\mathbf Q$, e.g., $\mathbf Q(\sqrt[g]{2})$ and ...
KCd's user avatar
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Proving that $\cos(2\pi/n)$ is algebraic

An idea which generalizes to other contexts aswell but uses a little machinery from algebra. In your particular case, probably just using binomial expansion is better. We know that $\cos(2\pi/n)=(e^{2\...
Kadmos's user avatar
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Does $\#\text{Ker}(f^*)\le \#\text{Im}(f)$ hold? Duality of profinite group

What is the motivation for your question? You must intend your profinite groups to be abelian, since working with continuous homomorphisms to $\mathbf Q/\mathbf Z$ can only see $M$ as far as its ...
KCd's user avatar
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Inverse of squares of fractional ideals

The ideal $\mathfrak a^2$ is generally not the ideal generated by the squares $a^2$ where $a \in \mathfrak a$: it is the ideal generated by all $aa'$ where $a$ and $a'$ are in $\mathfrak a$. Let $J$ ...
KCd's user avatar
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What is the order of $\text{Ker}(\hat{f})$ when order of annihilator of $\text{Ker}f$ is given?

$\DeclareMathOperator{\Im}{Im}\DeclareMathOperator{\Ker}{Ker}$In general, $\Ker(\hat f)$ need not have the same order as $\Im(f)$. First note that for $A = \mathbb Z / n$ you have $\hat A \cong \...
red_trumpet's user avatar
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Decomposition of prime $2$ in ring $\mathbb{Z}[\zeta_n]$

Let $\varphi$ be the Euler totient function and write $n=2^km$ with $\text{gcd}(2,m)=1$. Then $2=\mathfrak{p}_1^{e}\cdots\mathfrak{p}_{r}^e$, where $e=\varphi(2^k)$, and $r=\varphi(m)/f$ where $f$ is ...
Snacc's user avatar
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1 vote

For $\mathbb{Q}(\sqrt{d})$ a quadratic field, $p$ an odd prime, does $\vartheta_K/(p\vartheta_K) \cong \mathbb{Z}[\sqrt{d}]/(p\mathbb{Z}[\sqrt{d}])$?

Let us consider the case $d=1+4k$, where $k$ is integer. We have for an odd $p$ the relation $2\cdot s=1$ modulo $p$, where $s=(p+1)/2$ is an integer. Below, we consider $s$ also as an element of $\...
dan_fulea's user avatar
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1 vote
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The idempotented algebra

Note that elements of $H_1 \otimes_k H_2$ can be written as finite sums $\sum_{1 \leq i \leq n} a_i \otimes b_i$ by pushing the coefficients inside the first component. From the definition of ...
Maximilien Mackie's user avatar
2 votes
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Generalised traceform

I guess that $R$ is supposed to be commutative, since otherwise this is false : $R=M_n(k)$ has a non degenerate trace form but is not a product of field extensions of $k$. In the sequel, I will assume ...
GreginGre's user avatar
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1 vote

How to find the minimal dimension of a primitive matrix for a given Perron number?

There is indeed a necessary condition, whether it is possible to transform a negative matrix $B$, (whose spectral radius is a perron root) into a similar primitive integral matrix! It is known as the ...
user714's user avatar
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2 votes
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Serre's elliptic curve example for integral $j$

By Néron-Ogg-Shafarevich (see the paper Good reduction of abelian varieties by Serre-Tate), $\Phi_2$ acts faithfully on every $E[\ell]$ with $\ell \geq 3$, and $\Phi_2 \cong SL(2,3)$ as is written. If ...
Aphelli's user avatar
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In Dedekind domain, fractional ideals are invertible. How can we precisely compute the inverse of a given ideal in Dedekind domains

$I=\langle 2,\sqrt{-5}\rangle$ is the whole ring, $I=\Bbb{Z}[\sqrt{-5}]$, as it clearly contains $-(2^2+\sqrt{-5}^2) = 1$. In particular, it's already principal, so its inverse is itself. Perhaps you ...
SomeCallMeTim's user avatar
1 vote

What is the intersection of $\mathbb Z$ with the ideal generated by $1-\zeta_n$ in $\Bbb Z[\zeta_n]$?

I enjoyed thinking about this problem and think my solution is sufficiently different to warrant posting. For every divisor $d|n$ with $t$ relatively prime to $d$ we have $1-\zeta_d^t \in (1-\zeta_n)$,...
Merosity's user avatar
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3 votes
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If $K/\mathbb{Q}_p$ totally ramified, is $\mathcal{O}_K / \pi^n \mathcal{O}_K \cong \mathbb{Z} / p^n \mathbb{Z}$?

True as you say for $n=1$, but for any $n \ge 2$, these are not even isomorphic as abelian groups unless the extension is trivial. Let's assume it's not, i.e. $\pi^e \mathcal{O} = p\mathcal{O}$ for ...
Torsten Schoeneberg's user avatar
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Finiteness of Ideal Class Group in Ring of Integers of Number Field

Chapter 5, Corollary 2 of Marcus states that "There are only finitely many ideal classes in R." In fact, each ideal class will have infinitely many members. For example, all principal ideals ...
Eric Zhu's user avatar
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Action of $G/H$ to $H^1(H,M)$, group cohomology

Let $\Gamma$ be a group acting on a set $S$ and let $\Gamma'$ be a normal subgroup of $\Gamma$ (I'm using weird letters and using just set-actions and not module-actions to emphasize that this is pure ...
hunter's user avatar
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2 votes
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Group action on cohomology is trivial

You were right to be suspicious of not having used the congruence mod $H$ anywhere, and in fact your proof is not valid. The coycle $X$ is a function on $H$ satisfying $X(h_1h_2) = h_1X(h_2) + X(h_1)$ ...
hunter's user avatar
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1 vote
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$N( (p) \mathcal{O}_F) = p^n$, where $n:=[ F : \mathbb{Q}]$, $F$ is a number field ( Norm of ideal )?

You can apply the result you cited ($N((\alpha)) = |N(\alpha)|$) immediately to finish the proof, since $N((p)\mathcal{O}_F) = N(p) = p^{[F : \mathbb{Q}]}$. Alternatively, you can use the fundamental ...
stillconfused's user avatar
2 votes
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Visualization of Polynomials and Algebraic Numbers

The beautiful theory of dessins d'enfants partially answers your request. Partially because the insights are not as straightforward as in linear algebra and they work only on a certain class of ...
Randy Marsh's user avatar
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A local sort-of Dedekind criterion

You can still say a number of things: The ramification index $e_i$ of $\mathfrak{P}/\mathfrak{p}$ is the ramification index of the local field extension $L_{\mathfrak{P}}/K_{\mathfrak{p}}$. The ...
CJ Dowd's user avatar
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1 vote

On principal ideals and GCD

Write $A=P_1^{a_1}P_2^{a_2}\cdots P_k^{a_k}$ and $AB=P_1^{b_1}P_2^{b_2}\cdots P_k^{b_k}Q_1^{c_1}Q_2^{c_2}\cdots Q_l^{c_l}$ for distinct prime ideals $P_1,P_2,\dots,P_k,Q_1,Q_2,\dots,Q_l$ and positive ...
TravorLZH's user avatar
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3 votes
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Additive characters over a Number Field

As pointed out in the comment, if you peel the cover, this is a standard result in representation theory. Since $K$ is a number field, $\mathfrak o / \mathfrak b$ is a finite abelian group. So this ...
Just a user's user avatar
3 votes

Additive characters over a Number Field

This is probably best understood abstractly, by Pontryagin duality for finite abelain groups. Let $A$ be a such a finite abelian group. (You can apply the discussion to $A = \mathfrak{o}/\mathfrak{b}$....
m.s's user avatar
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