Alternative way to show $\sqrt{2}+ \sqrt{3}$ is algebraic over $\mathbb{Q}$ of degree 4. I want to show that $\sqrt{2}+ \sqrt{3}$ is algebraic over $\mathbb{Q}$ of degree 4. I am aware of the technique of setting $x = \sqrt{2} + \sqrt{3}$ and working backwards to obtain the fourth degree polynomial. I was wondering if there is a different way of doing this, maybe working with field extensions? Particularly, I was thinking that maybe showing that $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] =4$ and that $\mathbb{Q}(\sqrt{2},\sqrt{3}) =  \mathbb{Q}(\sqrt{2} + \sqrt{3})$ would help me with this. Thanks for any suggestion!
 A: There's a method I like quite a bit.  A spanning set (but not necessairly a basis, it may be too large) for $\mathbb Q(\sqrt{2},\sqrt{3})$ is undoubtedly $\lbrace 1,\sqrt{2},\sqrt{3},\sqrt{2}\sqrt{3}\rbrace$.  This is just "every multiplicative combination" of the bases for $\mathbb Q(\sqrt{2})$ and $\mathbb Q(\sqrt{3})$.
What we now want to do is find a matrix that has $\sqrt{2}+\sqrt{3}$ as an eigenvalue.
The way we do this is by first writing:
$$(\sqrt{2}+\sqrt{3})\begin{pmatrix} 1 \\ \sqrt{2} \\ \sqrt{3} \\ \sqrt{6}\end{pmatrix}$$
Now, multiply out each row. As an example, the top row is $1\times (\sqrt 2+\sqrt 3) = \sqrt 2+\sqrt 3$.  The second row is $\sqrt 2(\sqrt 2+\sqrt 3) = 2+\sqrt 6$.  The third row is $\sqrt 3 (\sqrt 2+\sqrt 3) = \sqrt 6+ 3$, and the fourth row is $\sqrt 6(\sqrt 2+\sqrt 3) = 2\sqrt 3+3\sqrt 2$.
We can write this as:
$$(\sqrt 2+\sqrt 3) = \begin{pmatrix} 1 \\ \sqrt 2 \\ \sqrt 3 \\ \sqrt 6\end{pmatrix} = \begin{pmatrix} \sqrt 2+\sqrt 3 \\ 2+\sqrt 6 \\ 3+\sqrt 6 \\ 2\sqrt 3+3\sqrt 2\end{pmatrix}$$ 
We can "factor" this last vector as the following:
$$\begin{pmatrix} 0 & 1 & 1 & 0 \\ 2 & 0 & 0 & 1 \\3 & 0 & 0 & 1 \\0 & 3 & 2 & 0 \end{pmatrix}\begin{pmatrix} 1 \\ \sqrt 2\\ \sqrt 3\\ \sqrt 6\end{pmatrix}$$
This matrix has $\sqrt 2+\sqrt 3$ as an eigenvalue, so its characteristic polynomial has this as a root.
This will be a degree $4$ polynomial.  It's not necessairly the minimal polynomial though, so you still have to check this.
A: HINT.-A distinct way could be the following.
You have with operations of field three irrational linearly independant over $\Bbb Q$. You can verify that 
if $a,b,c,d$ are rational then$$a\sqrt2+b\sqrt3=0\iff a=b=0\\a\sqrt2+b\sqrt3+c\sqrt2\cdot\sqrt3=0\iff a=b=c=0\\a\sqrt2+b\sqrt3+c\sqrt2\cdot\sqrt3+d=0\iff a=b=c=d=0$$ This shows that the four elements $\{\sqrt2,\sqrt3,\sqrt6,1\}$ is a basis of $\Bbb Q(\sqrt2,\sqrt3)$ over $\Bbb Q$ so you get the number four you require.
A: Yes, that approach would work. In fact, we have:
A field extension $K\subseteq L$ is algebraic iff $[L:K]$ is finite.
where we say $K\subseteq L$ is algebraic if $\alpha$ algebraic over $K$ for any $\alpha \in L$.
Let's prove the backward direction, which is relevant to your question. Suppose $[L:K] = n$, and let $\alpha\in L$. Then we know that $1,\alpha,\ldots,\alpha^n$ must be linearly dependent. Therefore, we have $$ \alpha^n + a_{n-1}\alpha^{n-1} + \cdots + a_1\alpha + a_0 = 0$$ and so we have found a polynomial $f(x) = x^n + \cdots + a_0$ with $f(\alpha) = 0$, so $\alpha$ algebraic. Since $\alpha$ was arbitrary, the extension $K\subseteq L$ is algebraic.
Thus, since $[\mathbb{Q}(\sqrt2 + \sqrt3) : \mathbb{Q}]$ finite, we know that $\sqrt2 + \sqrt3$ algebraic.
In fact, we know more. We also know that $[K(\alpha):K] = deg\,p_{\alpha}$, where $p_{\alpha}$ is the minimal polynomial of $\alpha$. You can see this from the isomorphism of fields $$K(\alpha) \rightarrow \frac{K[x]}{(p_{\alpha})}$$So in your case, with $K=\mathbb{Q}$ and $\alpha = \sqrt2 + \sqrt3$, we find that the minimal polynomial of $\sqrt2 + \sqrt3$ has degree 4 since $[\mathbb{Q}(\sqrt2 + \sqrt3) : \mathbb{Q}] = 4$.
