is the number algebraic? Is the number $\alpha=1+\sqrt{2}+\sqrt{3}$ algebraic?
My first attempt was to try a polynomial for which $p(\alpha)=0$ for some $p(x)=a_{0}+a_{1}b_{1}+\cdots +b_{n-1}x^{n-1}$ i. e $x=1+\sqrt{2}+\sqrt{3}$ and then square it many times to get rid of the irrationals. This procedure was futile. 
Second attempt: I remember from the lecture we had that if $L=K(\alpha, \beta)$ with $\alpha,\beta$ algebraic over $K$ then $[L:K]<\infty$ moreover $[K(\gamma):K]<\infty$ for $\gamma=\alpha\pm \beta$ and $\gamma=\alpha\beta$ and $\gamma=\frac{\alpha}{\beta},\beta\neq 0$ as $K(\gamma)\subseteq L$ hence $\gamma$ is algebraic over $K$.
So applying the above to the problem then $\alpha=1+\sqrt{2}+\sqrt{3}$ is algebraic then over $K$ Right?  Or is there any other way to show it?
But I would like to know if it is possible to find a minimal polynomial having $\alpha$ as a zero? How to do that? 
 A: Using the theorem is the fastest way to see that the number is algebraic.
To obtain an explicit polynomial having $\alpha$ as root, compute successive powers of $\alpha$, which are all of the form $a+b\sqrt 2+c\sqrt 3+d\sqrt 6$. With five different powers (namely $\alpha^0,\ldots,\alpha^4$) for four unknowns $a,b,c,d$, you should be able to find a dependency, which is nothing else but a nonzero polynomial having $\alpha$ as root.
A: I can't do this in my head but I hope the pattern is obvious
$$\begin{align}
& (x - 1 - \sqrt{2} - \sqrt{3})(x - 1 - \sqrt{2} + \sqrt{3})
  (x - 1 + \sqrt{2} - \sqrt{3})(x - 1 + \sqrt{2} + \sqrt{3})\\
= & x^4-4x^3-4x^2+16x-8
\end{align}$$
In general, given any two algebraic numbers $\alpha, \beta$ with minimal polynomial $f(x), g(x) \in \mathbb{Q}[x]$. If we let $\alpha_1, \alpha_2, \ldots, \alpha_{\deg f}$ and  $\beta_1, \beta_2, \ldots, \beta_{\deg g}$ be the two complete sets of roots of $f(x)$ and $g(x)$ in some splitting field of $f(x)g(x)$. i.e 
$$f(x) = \prod_{i=1}^{\deg f}(x - \alpha_i)\quad\text{ and }\quad g(x) = \prod_{j=1}^{\deg g}(x - \beta_j)$$
$\alpha + \beta$ will then be a root of a polynomial with degree $\deg f \cdot \deg g$. The polynomial can be defined as
$$h(x) = \prod_{i=1}^{\deg f}\prod_{j=1}^{\deg g} ( x - \alpha_i - \beta_j )
= \prod_{i=1}^{\deg f} g(x - \alpha_i)
= \prod_{j=1}^{\deg g} f(x - \beta_j)
$$
The coefficients of $h(x)$ are symmetric polynomials with integer coefficients in $\alpha_i$ and $\beta_j$. This means they can be expressed in terms of elementary symmetric polynomials in either $\alpha_i$ or in $\beta_j$. i.e. in terms of coefficients $f(x)$ and $g(x)$ which belongs to $\mathbb{Q}$. As a result, the polynomial $h(x) \in \mathbb{Q}[x]$ and hence $\alpha + \beta$ is algebraic.
The $h(x)$ so constructed need not be minimal polynomial of $\alpha + \beta$. However, one of its irreducible factor will be the one you want.
A: $\alpha=1+\sqrt{2}+\sqrt{3} \in \mathbb Q(\sqrt{2},\sqrt{3})$, which is a finite-dimensional vector space over $\mathbb Q$. Hence, the subspace $\mathbb Q(\alpha)$ is also finite-dimensional over $\mathbb Q$ and so $\alpha$ is algebraic, because the powers of $\alpha$ are linearly dependent.
A: $$\alpha=1+\sqrt2+\sqrt3$$ $$\alpha-1=\sqrt2+\sqrt3$$ $$(\alpha-1)^2=5+2\sqrt6$$ $$(\alpha-1)^2-5=2\sqrt6$$ $$((\alpha-1)^2-5)^2=24$$
