Identifying squares in field extension

Question

I'm trying to solve problem 4.11 of http://homepages.warwick.ac.uk/~masda/MA3D5/Galois.pdf.

The problem is, given $a,b\in k$ a field, and assuming that $\sqrt{a}\in k(\sqrt{b})$ prove that, either $\sqrt{a}\in k$ or $\sqrt{ab}\in k$. This is what I have achieved:

I may assume that $k\not=k(\sqrt{b})$ (if this doesn't happen then the conclussion follows trivially as $\sqrt{a}\in k(\sqrt{b})=k$). Now, write $$\sqrt{a}=c+d\sqrt{b}$$ with $c,d\in k$. Then $$a=c^2+2cd\sqrt{b}+d^2b$$ Reorganizing we obtain $$\sqrt{b}=\frac{a-c^2-d^2b}{2cd}$$ Since $a,b,c,d\in k$ then the right part of the previous equality is in $k$, but $\sqrt{b}\not\in k$(since we assumed $k\not=k(\sqrt{b})$). So $c=0$ or $d=0$. If the latter happens then $\sqrt{a}=c\in k$, if the former happens $\sqrt{a}=d\sqrt{b}\Rightarrow \frac{\sqrt{a}}{\sqrt{b}}=\sqrt{ab^{-1}}=d\in k$. I don't see why $\sqrt{ab}\in k$ in this case. Can you give me a hint? Also, is what I've done so far correct? Thanks in advance

@rogerl: Can you write this as an answer, so we can get this question out of the unanswered queue? — aleph_two, Dec 17 '18 at 04:40
Note that your proof doesn't work in characteristic $2$ and indeed the result is false there : take $k=\mathbb{F}_2(X)$, $a=X+1, b= X$. Then if $s^2= b$ we have $(s+1)^2= s^2+1= X+1=a$ so $\sqrt{a}\in k(\sqrt{b})$; however $\sqrt{a},\sqrt{ab}\notin k$ — Maxime Ramzi, Dec 17 '18 at 17:25

score 1 · Answer 1 · answered Dec 17 '18 at 15:02

1

If $\sqrt{a} = d\sqrt{b}$, then multiplying both sides by $\sqrt{b}$ gives $\sqrt{ab} = d\sqrt{b^2} = db$. Thus $\sqrt{ab}\in k$.

answered Dec 17 '18 at 15:02

rogerl

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Identifying squares in field extension

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