I want know if my logic in showing that $\Bbb Q(\sqrt 2,\sqrt 3) = \Bbb Q (\sqrt 2 + \sqrt 3)$ is correct

Question

I want to know if my logic in showing that $\Bbb Q(\sqrt 2,\sqrt 3) = \Bbb Q (\sqrt 2 + \sqrt 3)$ is correct

Now firstly, it seems that $\Bbb Q(\sqrt 2,\sqrt 3)=(\Bbb Q (\sqrt 2))(\sqrt3){\overset{{\huge\color\red?}}{=}}\{a+b\sqrt2+c\sqrt 3:a,b,c\in \Bbb Q\}$

But from my last question it seems that we need to check this for multiplicative closure, $(a+b\sqrt2 + c\sqrt3)(d+e\sqrt 2 + f\sqrt 3)=ad+ae\sqrt 2+ af\sqrt 3 + 2eb+bf\sqrt 3+cd\sqrt 3+ ce\sqrt2\sqrt3+ 3cf$

So we actually find I believe that:

$\Bbb Q(\sqrt2,\sqrt3)=\{a+b\sqrt 2+c\sqrt 3+ d\sqrt2\sqrt3:a,b,c,d\in \Bbb Q\}$?

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I believe that $\Bbb Q(\sqrt 2 +\sqrt 3)\overset{\huge{\color\red ?}}=\{a+b(\sqrt 2+ \sqrt 3):a,b\in \Bbb Q\}$

So let's verify multiplicative closure $(a+b\sqrt 2+ b\sqrt 3)(c+d\sqrt 2+ d\sqrt 3)$

$$=ac+ad\sqrt 2+ad\sqrt 3+bc\sqrt2+2bd+bd\sqrt3+bc\sqrt3+bd\sqrt2\sqrt3+3bd$$ $$=(ac+3bd+2bd)+(ad+bc)\sqrt 2 + (ad+bd+bc)\sqrt 3 +(bd)\sqrt2\sqrt3$$

Since $\sqrt2$ and $\sqrt 3$ and $\sqrt2\sqrt3$ don't share common coefficients, they are linearly independent and hence $\Bbb Q(\sqrt2+\sqrt 3)=\Bbb Q(\sqrt2,\sqrt3)$(from above, i.e we have deduced that:) $\Bbb Q(\sqrt{2}+\sqrt{3})=\{a+b\sqrt2+c\sqrt3:a,b,c\in\Bbb Q\}$.

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Is that the correct way to show this? Also this means we have:

$$\begin{matrix}&&\left(\Bbb Q(\sqrt{2}+\sqrt{3})=\Bbb Q(\sqrt2,\sqrt3)\right)\\\\&{\huge\diagup}&&{\huge\diagdown}\\\Bbb Q(\sqrt2)&&&&\Bbb Q(\sqrt 3)\end{matrix}$$

Answer 1

The "usual" proof that $\Bbb Q(\sqrt{2},\sqrt{3}) \subseteq \Bbb Q(\sqrt{2}+ \sqrt{3})$ ), the other inclusion is obvious.

It suffices to show that $\sqrt{2} \in \Bbb Q(\sqrt{2}+ \sqrt{3})$, since then $\sqrt{3} = (\sqrt{2} + \sqrt{3}) - \sqrt{2}$ must be as well.

Now $(\sqrt{2} + \sqrt{3})^3 = 2\sqrt{2} + 6\sqrt{3} + 9\sqrt{2} + 3\sqrt{3}$

$= 11\sqrt{2} + 9\sqrt{3}$.

Hence $\dfrac{1}{2}[(\sqrt{2} + \sqrt{3})^3 - 9(\sqrt{2} + \sqrt{3})] = \sqrt{2}$, QED.

Answer 2

Your belief that $$\def\Q{\mathbb{Q}} \Q(\sqrt{2}+\sqrt{3})=\{a+b(\sqrt{2}+ \sqrt{3}):a,b\in \Q\} $$ is wrong, as well as $$ \Q(\sqrt{2}+\sqrt{3})=\{a+b\sqrt{2}+ c\sqrt{3}:a,b,c\in \Q\} $$ The second claim can be immediately dismissed, because this would mean that $[\Q(\sqrt{2}+\sqrt{3}):\Q]=3$, but clearly $\Q(\sqrt{2}+\sqrt{3})\subseteq\Q(\sqrt{2},\sqrt{3})]$ and this extension has degree a divisor of $4$, because $$ [\Q(\sqrt{2},\sqrt{3}):\Q]= [\Q(\sqrt{2},\sqrt{3}):\Q(\sqrt{2})]\, [\Q(\sqrt{2}):\Q] $$ and both extensions in the right-hand side are of degree either $1$ or $2$ (both $2$, actually, but it's not important).

The first belief is wrong, too: this would mean that $\sqrt{2}+\sqrt{3}$ has a degree two minimal polynomial and this contradicts the fact you have to prove. Indeed $$ [\Q(\sqrt{2},\sqrt{3}):\Q(\sqrt{2})]=2 $$ because no element of $\Q(\sqrt{2})$ is the square root of $3$: $$ (a+b\sqrt{2})^2=3 $$ means $$ a^2+2b^2+2ab\sqrt{2}=3 $$ hence either $a=0$ or $b=0$, leaving either $2b^2=3$ or $a^2=3$; neither equation has a solution in $\Q$.

Now this is a proof that $\Q(\sqrt{2}+\sqrt{3})=\Q(\sqrt{2},\sqrt{3})$. Indeed, we know that $[\Q(\sqrt{2},\sqrt{3}):\Q]=4$ and that $$ [\Q(\sqrt{2}+\sqrt{3}):\Q]>2 $$ (well, assuming we know that $\sqrt{2}+\sqrt{3}$ is not rational, which is almost obvious). Therefore $[\Q(\sqrt{2}+\sqrt{3}):\Q]$ is a divisor of $4$, but greater than $2$, which leaves just one possibility.

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Why doesn't $\sqrt{2}+\sqrt{3}$ have degree $\le2$ over $\Q$? Since $(\sqrt{2}+\sqrt{3})^2=5+2\sqrt{6}$, we see that $$ \Q\subset\Q(\sqrt{6})\subseteq\Q(\sqrt{2}+\sqrt{3}) $$ and, since $\sqrt{6}$ is clearly irrational (a standard proof based on Eisenstein's criterion applies), we have that the degree of $\sqrt{2}+\sqrt{3}$ (which we already know is a divisor of $4$) can be either $2$ or $4$. If it is $2$, we find that $\sqrt{2}+\sqrt{3}=a+b\sqrt{6}$, for some $a,b\in\Q$. Squaring gives $$ 5+2\sqrt{6}=a^2+6b^2+2ab\sqrt{6} $$ that implies $ab=1$, so $$ a^2+\frac{6}{a^2}-5=0 $$ that is, $$ a^4-5a^2+6=0 $$ or $$ (a^2-2)(a^2-3)=0 $$ a contradiction.

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Of course, the above is much more complicated than the direct observation that $$ (\sqrt{2}+\sqrt{3})+(\sqrt{2}+\sqrt{3})^{-1}= \sqrt{2}+\sqrt{3}+\frac{1}{\sqrt{3}+\sqrt{2}}= \sqrt{2}+\sqrt{3}+\sqrt{3}-\sqrt{2}=2\sqrt{3} $$ so $\sqrt{3}\in\Q(\sqrt{2}+\sqrt{3})$ and therefore also $\sqrt{2}\in\Q(\sqrt{2}+\sqrt{3})$, ending the proof. However, the main object was to show the issues in your argument.

Answer 3

Let $x=\sqrt 2+\sqrt 3$. Evaluate $\frac{x+x^{-1}}2$.

Answer 4

Just to add to completeness of the other answers let me find minimal polynomial of $\alpha = \sqrt 2 + \sqrt 3$ over $\mathbb Q$ directly.

\begin{align} \alpha = \sqrt 2 + \sqrt 3 &\implies \alpha - \sqrt 2 = \sqrt 3\\ &\implies \alpha^2 -2\alpha\sqrt 2 + 2 = 3\\ &\implies \alpha^2 - 1 = 2\alpha\sqrt 2\\ &\implies \alpha^4 - 2\alpha^2 + 1 = 8\alpha^2\\ &\implies \alpha^4 - 10\alpha^2 + 1 = 0 \end{align}

Thus, $\alpha$ is a root of $p(x) = x^4 - 10x^2 + 1$, so if we show that it is irreducible over $\mathbb Q$, it will follow that it is minimal polynomial. By Gauss's lemma, since $p$ is primitive, it is enough to prove that it is irreducible over $\mathbb Z$. Assume the contrary, that $p = fg$ for non-invertible $f, g\in\mathbb Z[x]$. Neither $f$ nor $g$ can be of degree $0$ since $p$ is primitive, nor of degree $1$ since $p$ has no rational roots by Rational root theorem. So we have $\deg f = \deg g = 2$ and thus can write $$ p(x) = (x^2 + ax + b)(x^2 + cx + d),\quad a,b,c,d\in\mathbb Z $$ By expanding we get $$x^4 - 10x^2 + 1 = x^4 + (a+c)x^3 + (b+d+ac)x^2 + (ad + bc)x + bd$$ or equivalently

\begin{align} a + c &= 0\\ b + d + ac &= -10\\ ad + bc &= 0\\ bd &= 1 \end{align}

Now, we note that $b = d = \pm 1$ and that neither cases yield integer solutions of above system and thus arrive at contradiction that $p$ is reducible.