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I know that if $K$ is an extension of $\mathbb{Q}$ of degree $2$, then $K = \mathbb{Q}(\sqrt{d})$ for some squarefree integer $d$. I understand that this is not the case for degree $2$ extensions of $\mathbb{R}$.

What can we say for degree $2$ extensions of fields of characteristic $p\in$ prime?

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  • $\begingroup$ It is the case for degree $2$ extensions of $\mathbb{R}$, those are $\mathbb{R}[\sqrt{-1}]$, and $-1$ is a squarefree integer. $\endgroup$ – Daniel Fischer Jul 26 '13 at 16:23
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    $\begingroup$ If $p\ne 2$, then the result is the same. The case $p=2$ is always odd :-) $\endgroup$ – lhf Jul 26 '13 at 16:24
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    $\begingroup$ @lhf: +1. All prime numbers are odd, and 2 is the oddest of them all :-) $\endgroup$ – Georges Elencwajg Jul 27 '13 at 7:59
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Given a field $k$ not of characteristic $2$, and given $[K:k]=2$, let $K=k(\beta)$ with $\beta$ a zero of $x^2+ax+b$. Replacing $x$ by $x-{a\over 2}$ converts this to the form $x^2-b'$. Thus, the extension is obtained by adjoining $\sqrt{b'}$.

In characteristic $2$, extensions obtained by adjoining square roots are inseparable, but they are still there.

In characteristic $2$, the separable quadratic extensions are obtained by adjoining zeros of Artin-Schreier polynomials $x^2-x+a$.

Similarly, in characteristic $p$, separable degree-$p$ extensions are obtained by adjoining zeros of Artin-Schreier $x^p-x+a$.

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