# Quadratic extension is generated by square root in field

Let $F$ be a field whose characteristic is $\neq 2$. Suppose the minimum polynomial of $a$ over $F$ has degree $2$. Prove that $F(a)$ is of the form $F(\sqrt{b})$ for some $b\in F$.

Well, $F(a)$ consists of all $x+ya$ for $x,y\in F$. I don’t understand how we can find such $b$.

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Hint: what is the formula for the solution of the second degree equation?
Try to first find $b$ such that $a \in F(\sqrt{b})$. Explain why $F(a)=F(\sqrt{b})$.

Where did the $char \neq 2$ become important?

Hint $\ 2\ne 0\,\Rightarrow\ 1/2\in F\,$ which enables one to complete the square and use the quadratic formula, yielding a root of the form $\, x = a + \sqrt{b}\,$ for $\,a,b\in F.$