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4 votes

Calculate the ideal class group of $K=\mathbb{Q}(\sqrt[3]{11})$

By your previous analysis, $\alpha-1 \in \mathfrak{p}_2\cap\mathfrak{p}_5 = \mathfrak{p}_2\mathfrak{p}_5$, so there is an ideal $I$ such that $\mathfrak{p}_2\mathfrak{p}_5I=(\alpha-1)$. By computing ...
Aphelli's user avatar
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4 votes
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Rational matrix whose power is an integer matrix

Since the characteristic polynomial has integer coefficients, there are matrices $B \in M_n(\mathbb{Z})$ and $P \in GL_n(\mathbb{Q})$ with $A = PBP^{-1}$ (by the MO answer linked in the question). Let ...
arkeet's user avatar
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3 votes
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the order of $R=\Bbb{Z}[x]/(ax+b, x^2+5)$ is $5a^2+b^2$

As noted in the other answer, $R$ is in general not cyclic (as an abelian group under addition). Here is one proof of the result, which also reveals when $R$ is cyclic. Let $S = \mathbb{Z}[x]/(x^2+5)$,...
arkeet's user avatar
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2 votes
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Is it true that $A[2]\cong \varinjlim_i (A_i[2])$ if $A\cong \varinjlim_{i\in I} A_i$?

This is true if the colimit is filtered. Then $M[2] = \textrm{ker}(M \stackrel{2}{\to} M )$ is the kernel of a canonical map, and kernels commute with filtered colimits for abelian groups. In other ...
Andrea Marino's user avatar
2 votes

the order of $R=\Bbb{Z}[x]/(ax+b, x^2+5)$ is $5a^2+b^2$

You cannot prove the isomorphism $R\simeq \mathbb{Z}/(5a^2+b^2)\mathbb{Z}$, because its false. As you noted implicitely, $R\simeq \mathbb{Z}[\sqrt{-5}]/(a\sqrt{-5}+b)$. So, you want to prove that ...
GreginGre's user avatar
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2 votes
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Ramification for a Galois extension of number fields

You can find a proof in these lecture notes: https://websites.math.leidenuniv.nl/algebra/ant.pdf See theorem 4.14, and also theorem 4.10 if you need the relation between the discriminants of $f$ and $...
Servaes's user avatar
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1 vote

Proposition 2.6(b) in The Arithmetic of Elliptic Curves - fiber cardinality is almost always equal to separable degree for a map of curves

Yes, your final paragraph is almost correct. A purely inseparable morphism of schemes which is also projective with integral source and integral, normal, locally noetherian target has geometrically ...
KReiser's user avatar
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1 vote

Is it true that $A[2]\cong \varinjlim_i (A_i[2])$ if $A\cong \varinjlim_{i\in I} A_i$?

I think this should always be true. We have a natural map $\lim\limits_{\rightarrow}(A_i[2])\rightarrow(\lim\limits_{\rightarrow}A_i)[2]$ induced by the inclusions $A_i[2]\hookrightarrow A_i$. We can ...
LittleBear's user avatar
1 vote

The integral basis of $\mathbb{Q}(\zeta+\zeta^{-1})$

First, a priori you want to show that every element of $\mathcal{O}_K$ is of the form $\sum_{i=0}^{\frac{p-3}{2}} t_i (\zeta+\zeta^{-1})^i$ with $t_i\in\mathbb{Z}$, and not of the form $t_0+ \sum_{i=1}...
EML's user avatar
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1 vote

Let $F_\infty=\bigcup_{n\geq1}\operatorname{Q}(2^{1/2^n})$ , $K_\infty=\bigcup_{n\geq1}\operatorname{Q}(\zeta_{2^n})$, what is the intersection?

The answer to your first question is yes, $F_{\infty}$ has only one subfield of degree $2^n$. Indeed, writing $F_n = \mathbb{Q}(2^{\frac{1}{2^n}})$, you have $F_n \subset F_m$ whenever $n\leq m$. ...
EML's user avatar
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