# Tag Info

### 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 ...
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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 ...
• 7,704
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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 $S = \mathbb{Z}[x]/(x^2+5)$,...
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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 ...
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### 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 ...
• 15.3k
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• 41
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 $n\leq m$. ...
• 41

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