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This question already has an answer here:

How can I show this field extension equality $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$?

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marked as duplicate by Chris Eagle, Hagen von Eitzen, Ayman Hourieh, Amzoti, Martin May 19 '13 at 13:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ This is definitely a duplicate. $\endgroup$ – Chris Eagle May 19 '13 at 12:41
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Note that $(\sqrt 2+\sqrt 3)^2=5+2\sqrt 6$ and $(\sqrt 2+\sqrt 3)\sqrt 6=3\sqrt 2+2\sqrt 3$, which you can combine linearly with $\sqrt 2+\sqrt 3$ to obtain $\sqrt 2$ and $\sqrt 3$.

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The easier direction is $\mathbb{Q}(\sqrt{2} + \sqrt{3} ) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3}).$ Can you tell us why this is true?

Now we have $\mathbb{Q} \subseteq \mathbb{Q}(\sqrt{2} + \sqrt{3} ) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3}).$


Since $\sqrt{2}$ is a root of the degree 2 polynomial $X^2-2$ which is irreducible over $\mathbb{Q}$ (because if it were reducible, it would have a linear factor corresponding to a rational root which the rational root theorem rules outs) we have $[\mathbb{Q}(\sqrt{2}): \mathbb{Q}]=2.$

Now $\sqrt{3}$ is a root of the degree 2 polynomial $X^2-3.$ Let's check this is irreducible over $\mathbb{Q}(\sqrt{2}).$ It is reducible if and only if it has a root in $\mathbb{Q}(\sqrt{2}).$ Again by the rational root theorem, it has no roots in $\mathbb{Q}.$ So if there is a root $a+b\sqrt{2} \in \mathbb{Q}(\sqrt{2})$ it must have $b\neq 0$ and satisfy $(a+b\sqrt{2})^2-3 = a^2+2ab\sqrt{2}+2b-3=0.$ If $a=0$ then $2b^2=3$ so $b=\pm \sqrt{3/2}$ which is rational if and only if $2b = \pm \sqrt{6}$ is rational. It is not hard to show that $\sqrt{n}$ is rational if and only if $n$ is a perfect square which $6$ is not, so $a+b\sqrt{2}$ is a root of $X^2-3$ then $a$ and $b$ must both be non-zero. But then we can rearrange to get $\sqrt{2} = \dfrac{3-2b+a^2}{2ab}$ and the right hand side is rational while the left is not. So $X^2-3$ is irreducible over $\mathbb{Q}(\sqrt{2})$ so $[\mathbb{Q}(\sqrt{2})(\sqrt{3}):\mathbb{Q}(\sqrt{2})]=2.$

Thus $$[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}] = 2\cdot 2=4.$$


Now we find the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}.$ The natural guess is $$(X-\sqrt{2}-\sqrt{3})(X-\sqrt{2}+\sqrt{3})(X+\sqrt{2}-\sqrt{3})(X+\sqrt{2}+\sqrt{3})=X^4+2X^2+25.$$ Let's check it is irreducible to confirm. Gauss's lemma allows us to simply check it is irreducible over $\mathbb{Z}.$ The rational root theorem gives $\pm 1, \pm 5, \pm 25$ as the only possibilities for integer roots, none of which work. So any factorization of $X^2+2X^2+25$ has no linear factors, and thus is of the form $$X^4+2X^2+25 = (X^2+aX+b)(X^2+cX+d)$$ where $a,b,c,d\in \mathbb{Z}.$ Equating coefficients of powers of $X$ gives $a=-c, ac+b+d=2, ad+bc=0$ and $bd=25.$ If $a=-c=0$ then $b+d=2$ and $bd=25$ which implies $4 = (b+d)^2 = b^2+2bd + d^2 = b^2+d^2+50 \implies b^2+d^2<0.$ Thus $a=-c\neq 0$ and from $ad+bc=0$ we get $b=d$ and from $bd=25$ we conclude $b=d=\pm 5.$ This makes $ac+b+d=2$ becomes $-a^2 = 2\pm 10$ which is impossible.

Thus $$ [\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}]=4.$$


Putting everything together, $$ 4 = [\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2}+\sqrt{3})][\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2}+\sqrt{3})]\cdot 4 $$

so $$[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2}+\sqrt{3})]=1$$ which implies $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3}).$

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A way to do this is computing the degrees:

$$\def\Q{\mathbb{Q}} [\Q(\sqrt{2}):\Q]=2 $$ because $x^2-2$ is irreducible over $\Q$. Moreover $$ [\Q(\sqrt{2},\sqrt{3}):\Q(\sqrt{2}]=2 $$ because $x^2-3$ is irreducible over $\Q(\sqrt{2})$ as it is easy to see.

It follows that $[\Q(\sqrt{2},\sqrt{3}):\Q]=4$.

Of course $\Q\subset\Q(\sqrt{2}+\sqrt{3})\subseteq\Q(\sqrt{2},\sqrt{3})$ and so

$$ [\Q(\sqrt{2},\sqrt{3}):\Q(\sqrt{2}+\sqrt{3})] $$

can be either $1$ or $2$. If it is $2$, then $\sqrt{2}+\sqrt{3}$ would have degree $2$ over $\Q$, hence it would satisfy a monic polynomial

$$ x^2 + ax + b $$ in $\Q[x]$. If a root is $\sqrt{2}+\sqrt{3}$, the other one is

$$ \frac{b}{\sqrt{2}+\sqrt{3}}=b(\sqrt{3}-\sqrt{2}). $$

This implies that $\sqrt{3}-\sqrt{2}\in\Q(\sqrt{2}+\sqrt{3})$, so both $\sqrt{2}$ and $\sqrt{3}$ belong to $\Q(\sqrt{2}+\sqrt{3})$, a contradiction.

Therefore

$$ [\Q(\sqrt{2},\sqrt{3}):\Q(\sqrt{2}+\sqrt{3})]=1 $$

and the proof is complete.

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One can use Galois theory to show that $\sqrt{2}+\sqrt{3}$ is a primitive element of $$\mathbb{Q}(\sqrt{2},\sqrt{3})$$ by finding explicitly what $$G:=Gal(\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q})$$ is and showing that this element is fixed under all elements of $G$

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