user103697
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2 answers
8 votes
184 views
Showing $(3 + \sqrt{3})$ is not a prime ideal in $\mathbb{Z}[\sqrt{3}]$
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6 votes

In addition to what was already said in the comments, I think there is another possibly faster way worth mentioning, which is using the following criterion for some Ideal $I$ being prime in a ring $R$ ...

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1 answers
7 votes
521 views
Degree of a Projective Curve
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5 votes

Edit: Short answer: If we assume as known that $$\tag{1}\chi(\mathbb{P}_k^n,\mathscr{O}_{\mathbb{P}_k^n}(m))=\binom{n+m}{n},$$ then we can immediately use the short exact sequence $$0\rightarrow \...

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1 answers
7 votes
600 views
$Sp(2n)$ is embedded in $GL(2n)$ and has dimension $2n^2+n$
3 votes

First I have to say that my answer is really similar to the one already cited in the comments, I just want to focus a little more one the surjectivity part that was explicitly asked for in your ...

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1 answers
0 votes
51 views
What is wrong with my tower chain proof?
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3 votes

Unfortunately, the assumption that there must be an $i$ such that $N_i\subset N\subseteq N_{i+1}$ is false. As a counterexample, consider the symmetric group $G=S_4$ which is solvable by the chain $$\...

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1 answers
8 votes
243 views
$G$ is cyclic iff $f(H) = H$ for every automorphism $f$
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2 votes

One way to do this would be the following: "$\Rightarrow$": If $G=\langle x\rangle$ is cyclic of order $n$, it has only one subgroup of order $m$ for every $m\vert n$ (the one generated by $x^{n/m}$)...

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1 answers
1 votes
136 views
Solving matrix equations (or systems of linear equations) over $\mathbb Z$
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2 votes

To 1. You are right, $Q$ is a change of basis in $\mathbb{Z}^m$ and $P$ a change of basis in $\mathbb{Z}^n$, as depicted in the diagram. To 2. This means that $P(\mathrm{ker}(A'))=\mathrm{ker}(A)$, ...

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1 answers
3 votes
371 views
Splitting field of $X^5-t$ in $\mathbb{Q}(t)[X]$
2 votes

First just a short remark concerning b): You accidentally wrote $\mathbb{Q}$ in some places where it should have been $\mathbb{Q}(t)$ instead, for example, $[\mathbb{Q}(\sqrt[5]{t}):\mathbb{Q}(t)]=5$, ...

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2 answers
3 votes
318 views
Galois $\text{GF}(5)$, does $α^4=2$ make any sense?
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2 votes

I think there are some important things to remark concerning what you wrote above: To a). Your calculation is absolutely correct, but the fact that the question as you posted it in your comment ...

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1 answers
6 votes
151 views
Showing $\mathbb{Z}[i]/(1+2i) \oplus\mathbb{Z}[i]/(6-i)\cong\mathbb{Z}[i]/(8+11i)$
1 votes

In addition to everything said in the comments, I just wanted to remark one more thing that might be helpful concerning your last paragraph about the fact that $\mathbb{Z}[i]/(1+2i)$ and $\mathbb{Z}[i]...

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1 answers
1 votes
54 views
Restrictions of $Gal(\mathbb{C}/\mathbb{Q})$ to a finite extension $\mathbb{K}$ of $\mathbb{Q}$
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1 votes

I hope I am not mistaking the comments above, if I got it right, the question is: Given some $\varphi\in\mathrm{Aut}_{\mathbb{Q}}(\mathbb{K})$, are there inifinitely many $\psi\in\mathrm{Aut}(\mathbb{...

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1 answers
1 votes
79 views
Proof, that factor of irreducible polynomial is irreducible over field defined by root of irreducible
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1 votes

As $g(x)\in\mathbb{Q}(\rho_1)[x]$ is a polynomial of degree $2$ over a field, it is irreducible if and only if it has no roots in $\mathbb{Q}(\rho_1)$. So you are correct, proving that $\rho_2,\rho_3\...

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1 answers
4 votes
74 views
Local homology of a product having $\mathbb{R}^n$ as one of the factors
0 votes

First let me say that I'm really no expert on algebraic topology, only because there still wasn't an answer up to now I thought it might be helpful to give my suggestion for a proof anyways: Let us ...

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