user103697
• Member for 3 years, 9 months
• Last seen more than 3 years ago

184 views

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 $R$ ...

521 views

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 $$0\rightarrow \... View answer 1 answers 7 votes 600 views 3 votes First I have to say that my answer is really similar to the one already cited in the comments, I just want to focus a little more one the surjectivity part that was explicitly asked for in your ... View answer 1 answers 0 votes 51 views Accepted answer 3 votes Unfortunately, the assumption that there must be an i such that N_i\subset N\subseteq N_{i+1} is false. As a counterexample, consider the symmetric group G=S_4 which is solvable by the chain$$\...

243 views

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 $x^{n/m}$)...

136 views

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 $P(\mathrm{ker}(A'))=\mathrm{ker}(A)$, ...

371 views

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, $[\mathbb{Q}(\sqrt[5]{t}):\mathbb{Q}(t)]=5$, ...

318 views

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 ...

151 views

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 $\mathbb{Z}[i]... View answer 1 answers 1 votes 54 views Accepted answer 1 votes I hope I am not mistaking the comments above, if I got it right, the question is: Given some$\varphi\in\mathrm{Aut}_{\mathbb{Q}}(\mathbb{K})$, are there inifinitely many$\psi\in\mathrm{Aut}(\mathbb{...

79 views
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\...