Let ${\mathbb K}$ be a normal extension of $\mathbb Q$. Let $\mathbb L$ be the subfield of ${\mathbb C}$ generated by the real and imaginary parts of elements of $\mathbb K$. Thus $\mathbb L$ is a subfield of $\mathbb R$. Must $\mathbb L$ be a normal extension of $\mathbb Q$ also ?

  • $\begingroup$ If $i\sqrt7\in \Bbb{K}$, is it your intention that $\sqrt7\in\Bbb{L}$? Or in other words, do you chop the $i$ factor off the imaginary parts or not? $\endgroup$ Sep 5 '14 at 20:04
  • 2
    $\begingroup$ @JyrkiLahtonen I'm surprised you should ask that, because in every definition I’ve seen, the imaginary part is a real number, the $i$ is always "chopped off". I've never seen a book or article that uses ${\mathfrak Im}(z)$ to denote a purely imaginary number. $\endgroup$ Sep 6 '14 at 5:08
  • $\begingroup$ Ok. Sure. I was just confused for a moment. May be the late hour :-) $\endgroup$ Sep 6 '14 at 5:22
  • 1
    $\begingroup$ What if $\Bbb{K}=\Bbb{Q}(\root3\of2,\sqrt{-3})$? Then $\root3\of2\in\Bbb{L}$, but its conjugates are not real, so they are $\notin\Bbb{L}$. Did I misunderstand something else? OTOH if you wanted $\Bbb{L}$ to be gotten from $\Bbb{K}$ by adjoining the real and imaginary parts, then $\Bbb{L}=\Bbb{K}(i)$, and that is normal as a compositum of two Galois extensions. $\endgroup$ Sep 6 '14 at 5:42
  • $\begingroup$ Hmm. The argument about compositum of Galois extensions may assume that the extensions are finite. Not sure about the infinite case. $\endgroup$ Sep 6 '14 at 5:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.