An algebraic integer is an algebraic number that is a root of some monic polynomial $f \in \mathbb{Z}[x]$.
When I check that $\frac{\sqrt{-23}+\sqrt[23]{-2}}{2}$ is an algebraic integer or not by WolframAlpha, the result is false. But I can not prove it. I try to find the minimal polynomial of that element. But it does not seem to work. I mean $23$ is big and difficult to do it by hand. Is there any quicker way to check it?
Can someone help me? Thanks.
On
Let's call your number $\alpha$. Because $-23\equiv1\pmod4$ it is well known that $\beta=(1+\sqrt{-23})/2$ is an algebraic integer. Because algebraic integers form a ring, $\alpha$ is one if and only if $$ \gamma=\alpha-\beta=\frac{-1+\root{23}\of{-2}}2 $$ is an algebraic integer.
Eisenstein's criterion gives that the polynomial $x^{23}+2$ is irreducible, so $$[\Bbb{Q}(\gamma):\Bbb{Q}]=[\Bbb{Q}(\root{23}\of{-2}):\Bbb{Q}]=23.$$ Therefore the minimal polynomial $m(x)$ of $\gamma$ has degree $23$. Clearly, $(2\gamma+1)^{23}+2=0$, so (scaling to make the polynomial monic) $$ m(x)=\frac1{2^{23}}\left((2x+1)^{23}+2\right). $$ The constant term $m(0)=3/2^{23}$ is manifestly not a rational integer, so $\gamma$ is not an algberaic integer. Therefore neither is $\alpha$.
$\sqrt{-23}$ is a root of the polynomial $x^2 + 23$, so $\sqrt{-23}/2$ is a root of $(2x)^2 + 23 = 4 x^2 + 23$ and thus of $x^2 + 23/4$. Thus it is an eigenvalue of the companion matrix $$ A = \pmatrix{0 & -23/4\cr 1 & 0\cr}$$ of $x^2 + 23/4$. $\sqrt[23]{-2}$ is a root of $x^{23}+2$, and $\sqrt[23]{-2}/2$ is a root of $x^{23} + 2/2^{23} = x^{23} + 2^{-22}$ and an eigenvalue of the $23 \times 23$ companion matrix with $1$ on the first subdiagonal, $2^{-22}$ in position $(1,23)$, and $0$ elsewhere. Thus their sum is an eigenvalue of $A \otimes I + I \otimes B$, a $46 \times 46$ matrix. The characteristic polynomial of this is an irreducible (according to Maple) monic polynomial of degree $46$ which has non-integer coefficients. Thus your number is not an algebraic integer.