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Traditional complex numbers may be considered an extension of real numbers, which make square root extraction always possible and quadratic polynomials always reducible.

What extensions with similar properties are known for finite fields?

Although we can take a field element $g$ which has no square root and build an extension around it ($i\times i=g$), we make quadratic polynomials factorizable, but extracting a square root of an extended element is still not always possible. So this does not work as good as traditional complex numbers.

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Roman Maltsev
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    It's called the quadratic closure. The same phenomenon (adding a single new square root does not make all quadratic polynomials split) happens with the rationals. – Arturo Magidin Jun 02 '22 at 04:50
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    I think the OP's first sentence is true but misleading. The complex numbers are an extension of $\Bbb R$ that makes every polynomial always reducible and allows for any root extraction. Consequently, the analog for finite fields (or any field) is the algebraic closure of that field. – Greg Martin Jun 02 '22 at 05:40
  • What Arturo Magidin said. You should think of $\Bbb{F}_p$ as an analogue of $\Bbb{Q}$ rather than $\Bbb{R}$. The latter is very exceptional in the sense the fields with the property that an algebraic closure is a finite extension share quite a few properties with $\Bbb{R}$. – Jyrki Lahtonen Jun 02 '22 at 15:23
  • Given that Moishe Kohan lends their support, I will use my dupehammer. The upshot in the duplicate thread is that unless $q$ is even we can go from $\Bbb{F}q$ to $\Bbb{F}{q^2}$ by adjoining a square root. If you want to include roots other than square roots that makes it necessary to go all the way to the algebraic closure, which is an infinite field. – Jyrki Lahtonen Jun 02 '22 at 15:26
  • If some aspects of the question are not clarified by the chosen duplicate, please comment. I will be @-pinged, and will help. – Jyrki Lahtonen Jun 02 '22 at 16:56

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