TITLE: Quadratic extensions in characteristic $2$
QUESTION [17 upvotes]: 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 $(\mathbb{Z}/2\mathbb{Z})(X)$?

REPLY [10 votes]: 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 $\rm\,\bar y\in F).\:$