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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 $b$.

JJ Beck
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2 Answers2

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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?

user68061
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  • Suppose the minimum polynomial of $a$ is $x^2+ex+f=0$. So $a^2+ea+f=0$. Let $e=2g$. (Here we use the $char\neq 2$.) So $(a+g)^2+(f-g^2)=0$, so $a\in F(\sqrt{g^2-f})$. Since $F(\sqrt{g^2-f})$ is a field containing $F$ and $a$, we have $F(a)\subseteq F(\sqrt{g^2-f})$. Now why must they be equal? – JJ Beck Jan 17 '14 at 16:05
  • Very good, now there are two ways to do that:

    1. direct: you can just write the formula for $\sqrt{g^2-f}$ in terms of $a$ and get the inverse inclusion

    2. What is the degree of this inclusion? If it is $m$ then from the tower of extensions we have $F \subset F(a) \subset F(\sqrt{b})$ so $[F(a):F] \cdot m=[F(\sqrt{b}):F]$, $2m=m$, $m=0$

     – user68061 Jan 17 '14 at 16:15
  • I haven't read up to inclusion degree yet (will do soon), so first way suffices for now. Thanks! – JJ Beck Jan 17 '14 at 16:19
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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 $\,a,b\in F.$

Bill Dubuque
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