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4 questions
0
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1
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137
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$k(F_i)_{i=1}^{n}=k(G_j)_{j=1}^{m}$ iff there exist $a_i,b_j \in k$ such that $\langle F_i-a_i \rangle_{i=1}^{n} = \langle G_j-b_j \rangle_{j=1}^{m}$
Let $k$ be an algebraically closed field of characteristic zero, for example $k=\mathbb{C}$ and let $F_1,\ldots,F_n,G_1,\ldots,G_m \in \mathbb{C}[x,y]$, $n,m \in \mathbb{N}-\{0\}$.
Claim:
$\mathbb{C}(...
0
votes
1
answer
162
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$\mathbb{C}(u(x,y),v(x,y),f(x)+g(y))=\mathbb{C}(x,y)$ implies $\mathbb{C}(u(x,y),v(x,y))=\mathbb{C}(x,y)$?
The following question is a direct continuation of this question:
Let $u,v \in \mathbb{C}[x,y]$.
Assume that for every $f \in \mathbb{C}[x]$ and every $g \in \mathbb{C}[y]$ (excluding the cases where $...
2
votes
1
answer
138
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On maximal ideals of $k [X_i : i \in I ] $ where $k$ is a field , $I$ is an infinite set with $|k| > |I|$
Let $k$ be a field and $I$ be an infinite set such that $|k| > |I|$ . Let $R := k [X_i : i \in I ] $ and $m$ be a maximal ideal of $R$ ; then is it true that $m \cap k[X_i] \ne 0 , \forall i \in I$ ...
-1
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1
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403
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Prime dividing norm of algebraic integer [closed]
Let $k=\mathbf Q[i]$ be the field of Gaussian numbers.
I've proved the following easy lemma:
"If $x \in \mathcal{O}_k$ (the ring of integers of $k$) and $p$ is an odd rational prime dividing the norm ...