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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!

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In (a) you mean $K\subset L$ right? Are you trying to type $L^*\setminus K^*$ (the nonzero elements of $L$ that are not of $K$)? [The code could be \setminus or \backslash if you wish.] Note that $\{1,\beta\}$ is LI over $K$ for every $\beta\in L^*\setminus K^*$, so there is no need for the phrase "such that." As a counterexample to your reasoning, consider $L=\Bbb Q$ and $K=\Bbb Q(\varphi)$ with $\beta:=\varphi=\frac{1+\sqrt{5}}{2}$ the golden ratio satisfying $\varphi^2-1=\varphi$, where $\{1,\beta^2\}$ is also a basis and $[L:K]=2$. –  anon Sep 3 '12 at 3:39
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Ah! Now I see where your other question came from. Please keep them linked (in the future), so that there won't be any orphaned questions here :-) –  Jyrki Lahtonen Sep 3 '12 at 5:44
    
A philosophical remark. In part A) you should be (or become) aware that the condition "$\beta^2\in K$" is much more restrictive than the condition "$\{1,\beta\}$ is a basis." This is something you might have picked up in an earlier course in linear algebra. Vector space bases abound! In a 2-dimensional vector space you can expand any which basis of a 1-d subspace (here $K$ with a basis $\{1\}$) by appending any vector not in that subspace to it! But here few of those vectors have the extra property that their square is in $K$, so that is the property you need to work hard to get. –  Jyrki Lahtonen Sep 3 '12 at 5:56
    
@JyrkiLahtonen Thanks for the suggestion! The link is math.stackexchange.com/questions/190053/… –  yoyostein Sep 3 '12 at 11:56
    
@anon Yes you are right, thanks! –  yoyostein Sep 3 '12 at 11:57
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1 Answer

up vote 6 down vote accepted

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 $K=\{0,1\}$.

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