Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
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… – yoyostein Sep 3 '12 at 11:56
@anon Yes you are right, thanks! – yoyostein Sep 3 '12 at 11:57
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


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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.