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I understand that every degree 2 extension of fields of characteristic not equal to 2 can be obtained by adjoining the square root of a single element. It seems that for extensions in characteristic 2, this statement can be true and it can be false. May I please ask for two specific examples for either case, please? (P.S. I am a beginner in field theory so it would help if some explanations on the two examples can be given.)

Here are a few confusions:

  1. It seems that I cannot adjoin a square which already exists in the smaller field. I can only adjoin a square root of an element (of the smaller field) which does not exist in the smaller field (By the following proposition from Artin: enter image description here

Does that mean I cannot adjoin any square root of the elements in $\Bbb Z_2$ because it contains all its square roots?

  1. I still don't know how to formally prove that the field $\Bbb Z_2[x]/<x^2+x+1>$ cannot be obtained by adjoining a square root.

  2. In which case an extension field can be obtained by adjoining a square root? Detailed Explanation needed. Thanks so much.

Y.X.
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  • I'm sure this has been handled on our site earlier. Anyway, show that $p(x)=x^2+x+1$ is irreducible over $\Bbb{F}_2$, but the extension field (of four elemens) you get by adjoining a zero of $p(x)$ cannot be gotten by adjoining a square root. OTOH, it is possible to extend the field $K=\Bbb{F}_2(x)$, $x$ transcendental, by adjoining $\sqrt x$. – Jyrki Lahtonen Mar 19 '17 at 09:28
  • A reasonably good match. – Jyrki Lahtonen Mar 19 '17 at 09:34
  • And another. I think this is a duplicate, but I am not sure which candidate is a better pick. – Jyrki Lahtonen Mar 19 '17 at 09:38
  • @JyrkiLahtonen I had a look at that question before but the answers are just beyond my understanding. I am wondering why that extension field (of four elements) cannot be obtained by adjoining a square root? – Y.X. Mar 19 '17 at 09:38
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    Basically because $\Bbb{F}_2$ already contains the square roots of all its elements. You won't get anything new by adjoining one! The same applies to all the finite fields of characteristic two. All the square roots are already included. – Jyrki Lahtonen Mar 19 '17 at 09:39
  • @JyrkiLahtonen OK. I know that 1 is its own square root and 0 is its own square root in $\Bbb Z_2$. If I cannot get anything new by adjoining a square root, then that statement should be "false" instead of "sometimes false". In which case, adjoining a square root (of 0 or 1 in $\Bbb Z_2$) will give me an extension field? – Y.X. Mar 19 '17 at 09:43
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    It is sometimes false, sometimes true. You cannot extend a finite field of characteristic two by adjoining a square root, but some infinite fields of characteristic two can be so extended. That's why I also mentioned $\Bbb{F}_2(x)$. – Jyrki Lahtonen Mar 19 '17 at 09:48
  • Or $\Bbb{Z}_2$ has no extensions by square roots, but $\Bbb{Z}_2(x)$ has. The truth value of the claim depends on which field of characteristic two you want to extend. – Jyrki Lahtonen Mar 19 '17 at 09:53
  • @JyrkiLahtonen I have edited my question. So that should clarify my question. – Y.X. Mar 19 '17 at 10:45
  • @JyrkiLahtonen In your first comment:"but the extension field (of four elemens) you get by adjoining a zero of p(x) cannot be gotten by adjoining a square root." Can you prove it? –  Oct 17 '21 at 05:53
  • @Avenger Yes. So can you. Look at the field of two elements $\Bbb{F}_2={0,1}$. Which element is lacking a square root that you could adjoin? – Jyrki Lahtonen Oct 17 '21 at 06:16
  • @JyrkiLahtonen I think square root of both the elements exists! –  Oct 17 '21 at 06:20
  • @Avenger Correct. – Jyrki Lahtonen Oct 17 '21 at 06:58
  • @JyrkiLahtonen You didn't answered what I was asking though? –  Oct 17 '21 at 07:18
  • @Avenger Yes, I did. You asked for a proof that $\Bbb{F}_4$ cannot be constructed by adjoining a square root to $\Bbb{F}_2$. – Jyrki Lahtonen Oct 17 '21 at 09:40

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