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Let $K:F$ be a field extensions with $|K:F|=2$

I want to show that $K=F(\alpha)$ for $\alpha \in K$ and $\alpha ^2 \in F$ given that char $F\neq2$

Then that when F=$\mathbb{Q}$ you can choose $\alpha$ with $\alpha ^2 \in \mathbb{Z}$

I have set $K=a+b\alpha$ with $a,b \in F$ but I don't know where to go from here

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    $\begingroup$ When you say, "I have set $K = \left\{a + b\alpha \mid a,b\in F\right\}$", it sounds like you're constructing $K$ from $\alpha$. If so, you've got it backwards. You're given $K$, and you need to show that such an $\alpha$ exists. $\endgroup$ – Matthew Leingang Oct 28 '17 at 23:24
  • $\begingroup$ Doesn't the degree of extension imply that such an $\alpha$ exists, I thought I just needed to show that $\alpha ^2 \in F$ $\endgroup$ – Bill Hayes Oct 28 '17 at 23:44
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    $\begingroup$ The degree of the extension implies that for any $\beta$ in $K$ that's not in $F$, its degree over $K$ is $2$. But it's not true that $\beta^2 \in F$ for all $\beta$. Consider $F= \mathbb{Q}$ and $\beta = \sqrt{5}+1$. $\endgroup$ – Matthew Leingang Oct 29 '17 at 9:49
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Let

$\beta \in K \setminus F; \tag 1$

then, since $[K:F] = 2$, there must exist a linear dependence over $F$ 'twixt $1$, $\beta$, and $\beta^2$; that is, there are $a, b, c \in F$, not all zero, such that

$a \beta^2 + b\beta + c = 0; \tag 2$

if now $a = 0$,

$b\beta + c = 0; \tag 3$

then $b = 0$ forces $c = 0$, so $a = b = c = 0$, contrary to our assumption; if $b \ne 0$, then (3) implies $\beta \in F$, another contradiction; thus we may rule out the case $a = 0$.

With $a \ne 0$, (2) yields

$\beta^2 + d \beta + e = 0, \tag 4$

where $d = a^{-1}b$ and $e = a^{-1}c$; then we may write

$\beta^2 + d \beta = -e, \tag 5$

whence, with $\text{char}(F) \ne 2$,

$(\beta + \dfrac{d}{2})^2 = \beta^2 + d \beta+ \dfrac{d^2}{4} = \dfrac{d^2}{4} - e \in F; \tag 6$

we note that $4 = 2^2 \ne 0$ in $F$; otherwise $2^2 = 0 \Longrightarrow 2 = 0$, and $\text{char}(F) = 2$; now setting

$\alpha = \beta + \dfrac{d}{2}, \tag 7$

we see from (6) that $\alpha^2 \in F$; also,

$\alpha \in K \setminus F, \tag 8$

thus $[F(\alpha):F] = 2$, so

$K = F(\alpha), \tag 9$

and we have produced the requisite $\alpha$.

If $F = \Bbb Q$, then

$\alpha^2 = \dfrac{p}{q} \tag{10}$

with $p, q \in \Bbb Z$; note $q \ne 0$; then

$q \alpha^2 = p, \tag{11}$

so

$(q \alpha)^2 = q^2 \alpha^2 = pq \in \Bbb Z; \tag{12}$

finally,

$K = F(\alpha) = F(q\alpha). \tag{13}$

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First remember that a finite field extension is algebraic. Then there exists $\alpha\in K$ with $\min(\alpha,F)\in F[x]$ a degree 2 polynomial.

Now we write the minimal polynomial: $\min(\alpha,F) = x^2+bx+c$ with $b,c\in F$. We know that the roots of this polynomial are $\frac{-b+\sqrt{b^2-4c}}{2},\frac{-b-\sqrt{b^2-4c}}{2}$.

The primitive element theorem guarantees the existence of a primitive element since we know the roots for $\alpha$'s minimal polynomial are different, which means it's separable, and $K|F$ is finite.

With either of these roots, since we know $b,c\in F$, we can see that the part which is not in $F$ is $\sqrt{b^2-4c}$. We can take this element as our primitive element for the extension.

Therefore, $K=F(\sqrt{b^2-4c})$, the splitting field for the polynomial $x^2-(b^2-4c)$.

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  • $\begingroup$ Why do you need to invoke the primitive element theorem and separability ? Here pick any element $\alpha \in K \setminus F$ its minimal polynomial is of degree $2$ and $[F(\sqrt{b^2-4ac}):F] = 2$ thus $K = F$ $\endgroup$ – reuns Oct 29 '17 at 6:47
  • $\begingroup$ I felt it was necessary to mention it as a justification of why such element exists. For the rest of the proof I am constructing such element without using it. $\endgroup$ – Ignacio Rojas Oct 29 '17 at 17:14
  • $\begingroup$ What happens if the $charF=2$ $\endgroup$ – Bill Hayes Oct 30 '17 at 22:40
  • $\begingroup$ I didn't think about that case when writing the answer. I found this question where the answer by Math Gems helped me understand what was going on. I hope you can find it useful as well! $\endgroup$ – Ignacio Rojas Nov 2 '17 at 4:05

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