Field theory, extensions HiAll, I am stuck with this problem:
(a) Let $K$ be a field such that characteristic of $K$ is not 2. Prove that any extension $L$ of $K$ with $K\subset L$, and $[L:K]=2$ has the form $L=F(\beta)$ for some $\beta\in L^* \setminus K^*$  with $\beta^2\in K^*$. 
(b) Is this true when the ground field $K$ has characteristic 2?
Here is my attempt for part (a):
Choose an element $\beta\in L^*\setminus K^*$ such that $\{1,\beta\}$ is linearly independent over $K$.
Since $[L:K]=2$, so $\{1,\beta\}$ is a basis for $L$ over $K$.
Since $\beta$ cannot be written as $c_1 +c_2 \beta^2$, with $c_i\in K$ (not sure if this is true??), hence $\{1,\beta^2\}$ is not a basis and is hence linearly dependent.
Hence $\beta^2$ is a scalar multiple of $1$ and is hence in $K*$.
I am pretty sure it is not entirely right, as I didn't even make use of the fact that the characteristic of $K$ is not 2.
For (b), I think the answer is no, not true, but am not sure how to come up with an counterexample.
Sincere thanks for any help!
 A: Pick an arbitrary nonzero $x\in L\setminus K$. Show that $\{1,x\}$ is linearly independent over $K$ and hence a basis for the field $L$. However $x^2\in L$ implies $x^2=rx+s$ for some $r,s\in K$ (because remember that $\{1,x\}$ is a basis?); if $r\ne 0$ then we get $x^2\not\in K$ (if it were $\in K$ then $x=\frac{x^2-s}{r}\in K$, contrary to the choice of $x$ outside of $K$). We need to find a $\beta\in L\setminus K$ that squares to an element of $K$ using our element $x$. If we complete the square (using $2=1+1\in K$ because $\mathrm{char}\,K\ne2$) we get
$$s=x^2-rx=(x-r/2)^2-(r/2)^2.$$
Show that $\beta=x-r/2$ works just right.
Now suppose (for simplicity) we're working with $K=\Bbb F_2$, the prime field of characteristic two, and we look at an index two extension, which will necessarily be $L=\Bbb F_{2^2}\cong \Bbb F_2[x]/(x^2+x+1)$ (it is relatively straightforward to find a quadratic irreducible over $\Bbb F_2$). As a set, this is $\{0,1,x,x+1\}$, so show that both elements of $L\setminus K=\{x,x+1\}$ do not square to either element of $K=\{0,1\}$.
