Skip to main content
Rollback to Revision 4
Source Link
Wojowu
  • 28.2k
  • 3
  • 103
  • 185

Question removed Non-trivial subfield of ${\bf Q}(\sqrt[3]{a+\sqrt{b}})$

**************Question removedLet $a$, $b$ be positive rational numbers such that $b$ is not the square of a rational number and $a^2-b$ is not a cube. Are these conditions sufficient to insure that the field ${\bf Q}(\sqrt[3]{a+\sqrt{b}})$ has a single non-trivial subfield and is there a way of showing that the only non-trivial subfield of ${\bf Q}(\sqrt[3]{a+\sqrt{b}})$ is ${\bf Q}(\sqrt{b})$ without computing the Galois group of the normal closure?

If there exists a subfield of degree 3, it must be ${\bf Q}(\sqrt[3]{a^2-b})$ by a norm argument, but I can't see how to conclude from this.

Question removed

**************Question removed

Non-trivial subfield of ${\bf Q}(\sqrt[3]{a+\sqrt{b}})$

Let $a$, $b$ be positive rational numbers such that $b$ is not the square of a rational number and $a^2-b$ is not a cube. Are these conditions sufficient to insure that the field ${\bf Q}(\sqrt[3]{a+\sqrt{b}})$ has a single non-trivial subfield and is there a way of showing that the only non-trivial subfield of ${\bf Q}(\sqrt[3]{a+\sqrt{b}})$ is ${\bf Q}(\sqrt{b})$ without computing the Galois group of the normal closure?

If there exists a subfield of degree 3, it must be ${\bf Q}(\sqrt[3]{a^2-b})$ by a norm argument, but I can't see how to conclude from this.

edited tags
Link
coudy
  • 18.7k
  • 5
  • 75
  • 135
Became Hot Network Question
deleted 217 characters in body; edited title
Source Link
coudy
  • 18.7k
  • 5
  • 75
  • 135

Non-trivial subfield of ${\bf Q}(\sqrt[3]{a+\sqrt{b}})$ Question removed

Let $a$, $b$ be positive rational numbers such that $b$ is not the square of a rational number and $a^2-b$ is not a cube. Are these conditions sufficient to insure that the field ${\bf Q}(\sqrt[3]{a+\sqrt{b}})$ has a single non-trivial subfield and is there a way of showing that the only non-trivial subfield of ${\bf Q}(\sqrt[3]{a+\sqrt{b}})$ is ${\bf Q}(\sqrt{b})$ without computing the Galois group of the normal closure?

If there exists a subfield of degree 3, it must be ${\bf Q}(\sqrt[3]{a^2-b})$ by a norm argument, but I can't see how to conclude from this.**************Question removed

Non-trivial subfield of ${\bf Q}(\sqrt[3]{a+\sqrt{b}})$

Let $a$, $b$ be positive rational numbers such that $b$ is not the square of a rational number and $a^2-b$ is not a cube. Are these conditions sufficient to insure that the field ${\bf Q}(\sqrt[3]{a+\sqrt{b}})$ has a single non-trivial subfield and is there a way of showing that the only non-trivial subfield of ${\bf Q}(\sqrt[3]{a+\sqrt{b}})$ is ${\bf Q}(\sqrt{b})$ without computing the Galois group of the normal closure?

If there exists a subfield of degree 3, it must be ${\bf Q}(\sqrt[3]{a^2-b})$ by a norm argument, but I can't see how to conclude from this.

Question removed

**************Question removed

Make the question more precise
Source Link
coudy
  • 18.7k
  • 5
  • 75
  • 135
Loading
added 26 characters in body
Source Link
coudy
  • 18.7k
  • 5
  • 75
  • 135
Loading
edited tags
Link
coudy
  • 18.7k
  • 5
  • 75
  • 135
Loading
Source Link
coudy
  • 18.7k
  • 5
  • 75
  • 135
Loading