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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.

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    $\begingroup$ You've put the Galois theory tag on this. What do you know about the degree of an extension created by adjoining finitely many square roots? $\endgroup$ – hardmath Sep 6 '12 at 20:10
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    $\begingroup$ What if $L$ is algebraically closed? $\endgroup$ – Hagen von Eitzen Sep 6 '12 at 20:16
  • $\begingroup$ If $L$ is a finite field this is easy. All the elements have square roots in $L$, but $L$ has a quadratic extension. As Hagen points out, the claim is false, if $L$ is algebraically closed. Actually separably closed seems to be enough. $\endgroup$ – Jyrki Lahtonen Sep 6 '12 at 20:27
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    $\begingroup$ Another way to see that this works for finite $L$ is to see that the mapping $x\mapsto x^2+x$ is two-to-one (it is a homomorphism of additive groups, and has kernel of two elements), hence not onto. Therefore there exists $a\in L$ such that $x^2+x+a$ has no zeros in $L$. $\endgroup$ – Jyrki Lahtonen Sep 6 '12 at 20:31
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    $\begingroup$ @Gaston: If $L$ is finite, the Frobenius automorphishm $F:x\mapsto x^2$ is injective (trivial kernel), hence onto. $\endgroup$ – Jyrki Lahtonen Sep 6 '12 at 20:32
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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 $x^2+x+a=0$.

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