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How to solve this problem?

Let $F$ be a field of odd characteristic and $K$ is a field extension over $F$ of degree $2$. Prove that there exists an element $a$ of $F$ such that $K\cong F(\sqrt a).$

Attempt:

Here $[K:F]=2$. Assume that there exists no $a\in F$ such that $K\cong F(\sqrt a).$ Then consider the tower $[K:F]=[K:F(\sqrt a)][F(\sqrt a):F]$.

Since $\sqrt a$ is a root of the polynomial $x^2-a$ over $ F$. Hence $[F(\sqrt a):F]=2\implies [K:F(\sqrt a)]=1\implies K=F(\sqrt a)$

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    What is the assumption and what is the contradiction? – 2017-01-15
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    Your attempt doesn't make sense because you never defined what $a$ is... – 2017-01-15
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    Did I ever say that @Mustafa – 2017-01-15

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The problem with your attempt is that you are assuming that $K$ contains a square root of $a$ for some $a\in F$ which is not a square in $F$. This assumption is essentially what you're trying to prove.

Instead, if $[K:F]=2$ then choose some $\alpha\in K$ such that $\alpha\not\in F$. Then the set $\{1,\alpha,\alpha^2\}$ is linearly dependent over $F$, and since $\alpha\not\in F$ (so $\alpha$ and $1$ are linearly independent over $F$) it follows that there is a quadratic polynomial $f(x)=x^2+bx+c\in F[x]$ such that $f(\alpha)=0$.

Finally, since $F$ doesn't have characteristic $2$, you can complete the square to find an element $a\in F$ such that $K=F(\sqrt{a})$.

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    I am not getting the last line ;yes i can compute $a$ to be $a=\dfrac{-b\pm\sqrt{b^2-4}}{2}$ and since $2\neq 0$ so $a$ is defined ;but how is $K=F(\sqrt a)$ – 2017-01-15
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    We have $\alpha^2+b\alpha+c=0$, so completing the square yields $(\alpha+\frac{b}{2})^2=\frac{b^2}{4}-c$. Let $a=\frac{b^2}{4}-c$, then this shows that $\sqrt{a}=\alpha+\frac{b}{2}\in K$, and $\sqrt{a}\not\in F$ since $\alpha\not\in F$. – 2017-01-15
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    Thank you very much ;One last question I solved Dummit Foote chapter 13 on field extensions but still could not do it? – 2017-01-15
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    I have an exam coming up ?Which book should i do ? – 2017-01-15
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    Dummit and Foote is a good book, but you could also take a look at Lang's Algebra book as well. It's a bit more difficult. – 2017-01-15