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$\mathbb{Q}(\sqrt2 + \sqrt3) \subset \mathbb{Q}(\sqrt2,\sqrt3)$ is obvious. Now for the converse. Since $p(x) = x^4 - 10x^2 + 1$ has $\sqrt2 + \sqrt3$ as a root, is monic and irreducible in $\mathbb{Q}$, $[\mathbb{Q}(\sqrt2 + \sqrt3) : \mathbb{Q}] = \text{deg}(p(x)) = 4$. However, $[\mathbb{Q}(\sqrt2,\sqrt3) : \mathbb{Q}] = 4$ and since $\mathbb{Q}(\sqrt2 + \sqrt3)$ is a subspace of $\mathbb{Q}(\sqrt2, \sqrt3)$ from the first implication, we must have equality as they have equal dimensions.

Is this correct? Thank you for your time.

Kenta S
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Ben Xiao
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    How do you prove irreducibility of $p(x)$? – Ingix Jul 04 '18 at 11:19
  • @Ingix You know all the roots, none is rational, and multiplying any two of the linear factors you get a quadratic with non-rational coefficients. –  Jul 04 '18 at 11:23
  • See also https://math.stackexchange.com/questions/1204279/show-that-x4-10x21-is-irreducible-over-mathbbq – lhf Jul 04 '18 at 11:48

2 Answers2

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You ought to give a reason for $p$ being irreducible. Also, your $p(x)$ has a typo in it; it should be $x^4-10x^{\color{red}2} + 1$).

You can prove it more directly. Let's set $t = \sqrt2+\sqrt3$ because it's easier to write. Then $t^3 = 11\sqrt2+9\sqrt3$. This gives us $$ \sqrt2 = \frac12(t^3-9t)\in\Bbb Q(t)\\ \sqrt3 = \frac12(11t-t^3)\in \Bbb Q(t) $$ which proves that $\Bbb Q(\sqrt2, \sqrt3)\subseteq \Bbb Q(t)$

Arthur
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  • Sorry, I did mistype it. I can prove irreducibility by noting that there are no rational roots for if $p/q$ were a rational root with $(p,q) = 1$ then we'd have $p | q$, a contradiction. Then, if it were reducible, it would be a product of two binomials. I showed that the resulting equation derived from the coefficients have no solution. – Ben Xiao Jul 04 '18 at 11:37
  • @BenXiao If you did that (and included the actual calculations, of course), then that would be more than enough in my book. Whether it's enough for you is something you would have to ask your professor, because ultimately they are the ones deciding your grade. – Arthur Jul 04 '18 at 11:45
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I'll offer a slight variant on Arthur's argument. Let's prove $\sqrt{a},\,\sqrt{b}\in\mathbb{Q}(\sqrt{a}+\sqrt {b})$ for any $a,\,b\in\mathbb{N}$ that aren't perfect squares with $a<b$, e.g. $a=2,\,b=3$. Define $t:=\sqrt{a}+\sqrt{b}$ so $$\sqrt{b}-\sqrt{a}=\frac{b-a}{t},\,\sqrt{a}=\frac{1}{2}(t-\frac{b-a}{t}),\,\sqrt{b}=\frac{1}{2}(t+\frac{b-a}{t}).$$

J.G.
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  • You do not need the second paragraph. $\mathbb Q (t) $ is a field, so it contains $t^{-1} $. – Andrés E. Caicedo Jul 04 '18 at 12:01
  • Also, your field contains $t^2$, so it contains $\sqrt {ab}$, so it contains $t\sqrt {ab}$. This is a linear combination of $\sqrt a $ and $\sqrt b $ that is not a rational multiple of $t $, and solving the resulting $2\times2$ system shows both roots are in the field. The point is that one does not actually need to work to exhibit both roots as explicit expressions in the field, as you did. It is enough to know that such expressions exist. – Andrés E. Caicedo Jul 04 '18 at 12:06
  • @AndrésE.Caicedo Your first comment has been heeded (that paragraph was more in case a reader is unconvinced $\mathbb{Q}(\sqrt{a}+\sqrt{b})$ is a field); your second comment could just be another proof. – J.G. Jul 04 '18 at 12:09