Krdytkyu

I want to show the following:
Let $E/K$ be a field extension and $1+1 neq 0$. Let $alpha , beta in K^* $. Show

$K(sqrt{alpha}$)= $K(sqrt{beta}$)$Leftrightarrow$ $exists$ a $gamma in K^*$ with $alpha = gamma ^2 beta$

Now I had some ideas but they didn't solve the problem for me. A lot would be easier if I could just assume that the $K$-Basis of $K(alpha$) ist {1,$alpha$}. But I don't know if that is the case. I would appreciate any help to solve this problem.

Edit: I am really sorry that I forgot to put on the square.. sorry guys. I also would like to add, that we had no minimal polynomial yet.

Best
KingDingeling

$E$ is irrelevant now, so I will ignore it. It may serve a technical purpose equivalent to assuming that all of the below occurs inside a fixed algebraic closure, so that we may actually write that the field extensions are equal instead of just abstractly isomorphic. We will make this assumption. Let $K,alpha,beta$ be as in the revised statement.

$K(sqrt{alpha})$ and $K(sqrt{beta})$ have degrees at most $2$ over $K,$ as the minimal polynomial $f$ of $sqrt{alpha}$ divides $x^2 - alpha$ (and similarly for $sqrt{beta}.$ The degree is $2$ if $sqrt{alpha}notin K,$ and $1$ if $sqrt{alpha}in K.$

First, suppose that $K(sqrt{alpha}) = K(sqrt{beta}) = K.$ Then $frac{sqrt{alpha}}{sqrt{beta}}in K,$ and $alpha = left(frac{sqrt{alpha}}{sqrt{beta}}right)^2beta,$ as required.

Now, suppose that $K(sqrt{alpha}) = K(sqrt{beta})$ is a degree $2$ extension of $K.$ In this case, $sqrt{alpha}notin K,$ so that ${1,sqrt{alpha}}$ is a $K$-basis of $K(sqrt{alpha})$ (similarly for $beta$). This implies that we may write
$$
sqrt{beta} = a + bsqrt{alpha}
$$
for some $a,bin K.$ Squaring both sides, we find
$$
beta = a^2 + 2absqrt{alpha} + alpha.
$$
This implies that $2absqrt{alpha} = beta - alpha - a^2in K,$ and since we already observed that ${1,sqrt{alpha}}$ is a basis for $K(sqrt{alpha})/K,$ this means that $a$ or $b$ is zero (this is where we use the fact that the characteristic is not $2$).

If $b = 0,$ we have $sqrt{beta} = ain K,$ which contradicts the assumption that $K(sqrt{alpha}) = K(sqrt{beta})$ is a degree $2$ extension of $K.$ Thus, we must have
$$
sqrt{beta} = bsqrt{alpha},
$$
and squaring both sides gives the desired result.

There appear to be some typos in your post, so if you update your question I think that this answer will no longer be valid.

Let $K=mathbb{Q}$ and $E=mathbb{Q}(pi)$. Take $alpha=pi$ and $beta=frac{pi+1}{pi-1}$. Note that $2(beta-1)^{-1}+1=alpha$, so that $K(alpha)=K(beta)$.

However, there cannot be a $gammain K^times$ such that $alpha=gamma^2beta$ or we would have
$$
pi=gamma^2frac{pi+1}{pi-1} quad Rightarrow quad pi^2-(gamma^2+1)pi-gamma^2=0
$$
which is a contradiction, since $pi$ is transcendental over $mathbb{Q}$.

I have assumed some of the corrections left as a comment on your post, except that $[E:K]=2$. So if you meant that in addition to the fact that $K$ is not characteristic $2$ this answer obviously fails.

EDIT: In fact, I don't even think that the criterion of $[E:K]=2$ is enough to salvage this question. Consider $K=mathbb{Q}(pi^2)$ and $E=mathbb{Q}(pi)$. This is a degree $2$ extension, and if all other choices are made as above then you still end up with a contradiction (though the final contradiction takes a bit more effort to show that it is a nonzero polynomial).