I recently saw in class that the degree $2$ extensions of a field of characteristic $\neq 2$ are given by square roots of non-squares in the base field.
I wonder what happens in the case of characteristic $2$ fields. For finite fields of characteristic $2$, it's clear what's going on, but what about others like $(\mathbb{Z}/2\mathbb{Z})(X)$?
There are two kinds of quadratic extensions in characteristic $2$.
The first are the same as in other characteristics: namely, if $\alpha \in F \setminus F^2$, then $F(\sqrt{\alpha})$ is a quadratic extension. It need not be the case that every element is a square in characteristic $2$. This occurs iff the field $F$ is perfect, so e.g. if the field is finite or algebraically closed. For an example of a nonperfect field, start with any field $k$ of characteristic $2$ -- e.g. $k = \mathbb{F}_2$ -- and take $F = k(t)$, the rational function field. Then the element $t$ is not a square, so $F(\sqrt{t})/F$ is a quadratic extension. These quadratic extensions are characterized by being purely inseparable.
(More generally, a degree $p$ extension in characteristic $p$ is purely inseparable iff it is of the form $F(\alpha^{\frac{1}{p}})$ for some $\alpha \in F \setminus F^p$.)
The second are the quadratic extensions which are separable and hence cyclic Galois of order $2$. Such extensions are all obtained as $K = F[t]/(t^2-t + \alpha)$ for some $\alpha \in F$. In order to get a field extension one needs the quadratic polynomial to be irreducible, which occurs iff $\alpha$ is not of the form $x^2-x$ for any $x \in F$. For instance, the (unique, up to isomorphism) quadratic extension of $\mathbb{F}_2$ is given by the polynomial $t^2-t + 1$.
(More generally, Artin-Schreier theory obtains all cyclic Galois degree $p$ extensions in characteristic $p$ as roots of Artin-Schreier polynomials $t^p - t + \alpha$. This is a beautiful theory which looks strange at first but in many ways works out more simply than cyclic $p$ extensions in characteristic different from $p$.)
It is quite easy to classify quadratic extensions $\rm\,E\,$ of a field $\rm\,F\,$ of characteristic $2.\:$ Namely, suppose $\rm\:E = F(\alpha)\,$ where $\,\alpha\,$ has minimal polynomial $\rm\:f(x) = x^2 + bx - c\in F[x].\:$ The quadratic formula no longer applies, i.e. we cannot change variables $\rm\:x = \bar x - b/2\,$ to reduce to the case $\rm\,b = 0,\,$ since we cannot divide by $\rm\:2 = 0.\:$ However, we can reduce every case $\rm\:b\ne 0\:$ to the case $\rm\:b = 1\!:\:$ divide $\rm\:f(x)\:$ by $\rm\:b^2\:$ and let $\rm\: y = x/b,\:$ i.e. $\rm\:f(x)/b^2 = (x/b)^2 + x/b + c/b^2 = y^2 + y + c',\,\ c' = c/b^2.$ Hence every quadratic extension of $\rm\,F\,$ is isomorphic to one of the following two types.
Type $\rm1\!:\,\ b = 0\ \Rightarrow\ E = F[x]/(x^2-c)\, \cong\, F[\bar x] \,\cong\, F[\sqrt{c]}$
Type $\rm2\!:\,\ b\ne 0\ \Rightarrow\ E = F[ y]/( y^2+ y - c') \cong F[\bar y] $
They're never isomorphic (over $\rm\,F)$ since elements in type $1$ have trace $= 0,\,$ but in fields of type $2$ the element $\rm\,\bar y\,$ has trace $\rm= \bar y+\bar y' = -1.\:$ Or, avoiding trace: elements of the first have square $\rm\in F\,$ by $\rm\: (d + e \sqrt{c})^2 = d^2 + e^2c \in F\:$ by $\rm\:2 = 0,\:$ but in the second $\rm\:\bar y^2\! = c' - \bar y \not\in F\,$ (else $\rm\,\bar y\in F).\:$
