Sum of Algebraic number over $\mathbb{Q}$ Let $j=e^{\frac{2i\pi}3} $. $j$ is algebraic over $\mathbb{Q} $ with $X^2+X+1$ as minimal polynomial.
$\sqrt{2}$ is algebraic over $\mathbb{Q}$ with $X^2-2$ as minimal polynomial. 
Find the minimal polynomial of $j+\sqrt{2}$. 
My solution 
I begin to show that $\mathbb{Q}[j+\sqrt{2}]=\mathbb{Q}[j,\sqrt{2}]$ but I can't.  One subset is evident. 
 A: If your task is to find a polynomial for $j+\sqrt2$, it’s not at all hard: Let $f(X)=X^2+X+1$, with $f(\,j)=0$. Consider $f(X-\sqrt2\,)=X^2+(1-\sqrt2\,)X+3-\sqrt2$ and its “conjugate” $f(X+\sqrt2\,)=X^2+(1+\sqrt2\,)X+3+\sqrt2$. Multiply these two $\Bbb Q(\sqrt2\,)$-polynomials, and you get $X^4+2X^3-X^2-2X+7$. You can see that this is a $\Bbb Q$-polynomial having $i+\sqrt2$ as a root. Irreducibility is another question.
A: There is a general construction to this using the Kronecker product:

If $A, B$ are matrices having eigenvalues $\alpha, \beta$ respectively, then
  $$ A \otimes I + I \otimes B $$
  has $\alpha + \beta$ as an eigenvalue.

Proof. Suppose $Av = \alpha v$ and $Bw = \beta w$ with $v, w \ne 0$. Then $v \otimes w \ne 0$ and
$$\qquad (A \otimes I + I \otimes B)(v \otimes w) = \alpha v \otimes w + v \otimes \beta w = (\alpha + \beta)(v \otimes w). \qquad\square $$
If we let $\alpha = e^{2\pi i/3}$ and $\beta = \sqrt 2$ then we can take
$$ A = \begin{pmatrix} 0 & -1 \\ 1 & -1 \end{pmatrix} \text{ and } B = \begin{pmatrix} 0 & 2 \\ 1 & 0 \end{pmatrix}. $$
We now have
$$ A \otimes I + I \otimes B =
\begin{pmatrix}
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1 \\
1 & 0 & -1 & 0 \\
0 & 1 & 0 & -1
\end{pmatrix} + 
\begin{pmatrix}
0 & 2 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 2 \\
0 & 0 & 1 & 0
\end{pmatrix}
$$
and after working through the calculations you get
$$ \operatorname{char}(A \otimes I + I \otimes B, t) = t^4 + 2t^3 - t^2 - 2t + 7. $$
A: As you observed $\mathbb{Q}[j+\sqrt{2}]\subset \mathbb{Q}[j,\sqrt{2}]$. So the extension $\mathbb{Q}[j+\sqrt{2}]$ can have either degree 2 or 4, but we can exclude the first case since $j+\sqrt{2}$ hasn't the form $a+b\sqrt{d}$, with $a,b\in \mathbb{Q}$, and we know all the quadratic extensions of $\mathbb{Q}$ have the form $\mathbb{Q}(\sqrt{d})$ for a squarefree integer $d$. Hence $\mathbb{Q}[j+\sqrt{2}]=\mathbb{Q}[j,\sqrt{2}]$, so we know the minimal polynomial $p(x)$ has degree 4 and $j+\sqrt{2}$ is a root. But we also know that the automorphisms of $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[j]$ send a root of $p(x)$ to another root of $p(x)$, so the roots of $p(x)$ are $j\pm \sqrt{2}$ and $j^2\pm \sqrt{2}$. Hence the minimal polynomial (it follows it actually is the minimal polynomial by degree considerations) is $p(x)=(x-(j+\sqrt{2}))(x-(j-\sqrt{2}))(x-(j^2+\sqrt{2}))(x-(j^2-\sqrt{2}))=x^4+2x^3-x^2-2x+7$
A: You can set up a system of linear equations as a last resort. Let $r=\sqrt{2}$. The powers of $x=j+r$ are then linear combinations of $jr,\ j,\ r, 1$ over $\mathbb Q$. In fact, using binomial expansion of $(j+r)^k$ and the identities $j^2+j+1=0,\ r^2=2$, we get
$$
\pmatrix{1\\ x\\ x^2\\ x^3\\ x^4\\ \vdots}
=\underbrace{\pmatrix{0&0&0&1\\ 0&1&1&0\\ 2&-1&0&1\\ -3&6&-1&1\\ 8&-11&4&-8\\ \vdots&\vdots&\vdots&\vdots}}_A\pmatrix{jr\\ j\\ r\\ 1}.
$$
Call the big infinite matrix in the above $A$. We can rewrite any monic polynomial $p(x)=x^k+c_{k-1}x^{k-1}+\cdots+c_1x+c_0$ as
$$
p(x)=(c_0,c_1,\ldots,c_{k-1},1)\ \pmatrix{1\\ x\\ \vdots\\ x^{k-1}\\ x^k}=(c_0,c_1,\ldots,c_{k-1},1)\ A_{k+1}\ \pmatrix{jr\\ j\\ r\\ 1},
$$
where $A_m$ denotes the submatrix taken from the first $m$ rows of $A$.
Since $jr,j,r,1$ are linearly independent over $\mathbb Q$, $p(x)=0$ if and only if $(c_0,c_1,\ldots,c_{k-1},1)A_{k+1}=0$, meaning that the $(k+1)$-th row of $A$ is a linear combination of the first $k$ rows. So, for $p$ to be the minimal polynomial, $k$ must be the first integer that exhibits such linear dependence.
It is not hard to verify that $A_4$ is invertible but $A_5$ is singular. Therefore, the degree of the minimal polynomial of $x=j+r$ is $k=4$. Solving $(c_0,c_1,c_2,c_3,1)\,A_5=0$, we get $(c_0,c_1,c_2,c_3,1)=(7,-2,-1,2,1)$. Hence the minimal polynomial is $x^4 + 2x^3 - x^2 - 2x + 7$.
