Show that for a field $L$ of characteristic $2$ there exist quadratic equations which cannot be solved by adjoining square roots of elements in the field $L$.
In $\mathbb{Z_2}$ adjoining all square roots we obtain again $\mathbb{Z_2}$, but $t^2+t+1$ does not have roots in this field. I don't know how to do the general case.
Best Answer
If $L$ is a finite field of characteristic two, then consider the mapping $$ p:L\rightarrow L, x\mapsto x+x^2. $$ Because $F:x\mapsto x^2$ respects sums: $$F(x+y)=(x+y)^2=x^2+2xy+y^2=x^2+y^2=F(x)+F(y),$$ the mapping $p$ is a homomorphism of additive groups. We see that $x\in \mathrm{Ker}\ p$, if and only if $x=0$ or $x=1$. So $|\mathrm{Ker}\ p|=2$. Therefore (one of the basic isomorphism theorems) $|\mathrm{Im}\ p|=|L|/2$. In particular, the mapping $p$ is not onto.
Let $a\in L$ be such that it is not in the image of $p$. Then the quadratic equation $$ x^2+x+a=0 $$ has no zeros in $L$.
Because the mapping $F$ is onto (its kernel is trivial), all the elements of $L$ have a square root in $L$. Thus adjoinin square roots of elements of $L$ won't allow us to find roots of the above equation.
In general the claim may not hold. By elementary Artin-Schreier theory we can actually show that quadratic polynomials of the prescribed type exist exactly, when the above mapping $p$ is not onto (irrespective of whether $L$ is finite or not). This is because, unless the quadratic $r(x)$ is of the form $x^2+a$ (when joining square roots, if needed, will help), its splitting field is separable, hence cyclic Galois of degree two. Thus the cited theorem of Artin-Schreier theory says that the splitting field of $r(x)$ can be gotten by joining a root of a polynomial of the form $x^2+x+a=0$.