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Here is an extract from my Galois Theory notes proving that $\mathbb{Q}(\sqrt{5}, \sqrt{6}) = \mathbb{Q}(\sqrt{5}+ \sqrt{6}) $

My question is after rearranging equation (1) has my lecturer omitted an $\alpha$ that should be on the denominator? I think this may be a typo but solutions to the general case in one of our assignments have been released and it is also missing the $\alpha$ in the denominator.

If it is not a typo then how does this equation hold

thanks in advance

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Peter A
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3 Answers3

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Yes, but correcting the typo leaves the proof correct since $\,\sqrt 5 = \dfrac{1}2(\alpha - \alpha^{-1})\in\Bbb Q(\alpha)$

Simpler: $\ \alpha^{-1} = \dfrac{6-5}{\sqrt 6 + \sqrt 5} = \sqrt{6}-\sqrt 5\ \Rightarrow\ \begin{align}\alpha+\alpha^{-1}=2\sqrt 6\\ \alpha-\alpha^{-1} = 2\sqrt{5}\end{align}$ $\,\Rightarrow\, \sqrt 6,\sqrt5 \in \Bbb Q(\alpha)$

Remark $\ $ This exploits the key idea that lies at the heart of the proof of the Primitive Element Theorem. Namely, if a field $F$ has two $F$-linear independent combinations of $\rm\, \sqrt{a},\ \sqrt{b}\, $ then we can solve for $\rm\, \sqrt{a},\ \sqrt{b}\, $ in $F.$ See here for further discussion.

Community
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Bill Dubuque
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    Your simpler, as it often happens, is immensely unmotivated. Why would anyone, starting from the statement of the theorem even think of writing $1/\alpha$ in that way? – Mariano Suárez-Álvarez Apr 07 '15 at 16:42
  • @Mariano Your "often happens", as often happens, is immensely incorrect. That is the point of the Remark and the linked answer. – Bill Dubuque Apr 07 '15 at 16:44
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    Your remark and the relevant part of your linked answer do not in the least motivate why one would try to compute $1/\alpha$. What you wrote is a verification of the truth of the statement in question which no one who did not already know it was true and how to verify it would be able to come up with. It is simpler only in the sense in which verifying an answer is simpler that finding the answer. – Mariano Suárez-Álvarez Apr 07 '15 at 16:48
  • @Mariano I have no interest whatsoever in arguing with you on such subjective matters since, as you are well aware, we strongly disagree on many such topics. – Bill Dubuque Apr 07 '15 at 16:50
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Yes, it's a typo. In the general case of $\mathbb{Q}(\sqrt{a},\sqrt{b})$ where $a$ and $b$ are squarefree integers, we have, after $$ \alpha=\sqrt{a}+\sqrt{b}, $$ that $$ (\alpha-\sqrt{a})^2=b $$ so $$ \alpha^2-2\alpha\sqrt{a}+a=b $$ hence $$ \sqrt{a}=\frac{\alpha^2+a-b}{2\alpha}\in\mathbb{Q}(\alpha). $$ Thus also $$ \sqrt{b}=\alpha-\sqrt{a}\in\mathbb{Q}(\alpha) $$ (no need to repeat the proof).

In order to find the minimal polynomial of $\alpha$, we have to assume $a\ne b$ (otherwise the result would be trivial). Then $\sqrt{b}\notin\mathbb{Q}(\sqrt{a})$, because $$ (x+y\sqrt{a})^2=b $$ would mean $x^2+ay^2+2xy\sqrt{a}=b$, forcing $\sqrt{a}$ to be rational, unless $x=0$ or $y=0$; if $x=0$ we'd have $ay^2=b$ which is impossible as $a$ and $b$ are squarefree and unequal; if $y=0$, we'd have $x^2=b$, likewise impossible.

Then the degree of $\mathbb{Q}(\alpha)$ over $\mathbb{Q}$ is $4$. Then, squaring $$ 2\alpha\sqrt{a}=\alpha^2+a-b $$ we find an expression of degree $4$ in $\alpha$ that gives us a monic polynomial of degree $4$ having $\alpha$ as a root. This is the minimal polynomial.

egreg
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Here is another argument that $\Bbb Q(\sqrt5+\sqrt6\,)=\Bbb Q(\sqrt5,\sqrt6\,)$, but like Dubuque’s second argument, it is not an answer to your question. On the other hand, I think that Suárez-Álvarez will agree that it is well-enough motivated and reasonably natural.

$K=\Bbb Q(\sqrt5,\sqrt6\,)$ is certainly quartic over $\Bbb Q$, in fact biquadratic, which means that there’s another quadratic subfield of $K$ beyond $\Bbb Q(\sqrt5\,)$ and $\Bbb Q(\sqrt6\,)$, namely $\Bbb Q(\sqrt{30}\,)$. And these are the only quadratic subfields of $K$. So if $\Bbb Q(\sqrt5+\sqrt6\,)$ is unequal to $K$, it must be equal to $\Bbb Q(\sqrt{30}\,)$, a field in which $\{1,\sqrt{30}\,\}$ is a good $\Bbb Q$-basis. This would mean that $\sqrt5+\sqrt6=a + b\sqrt{30}$ for rational numbers $a$ and $b$. But $\{1,\sqrt5,\sqrt6,\sqrt{30}\,\}$ is a $\Bbb Q$-basis for $K$, and the preceding equation contradicts linear independence of these four numbers.

Lubin
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