In the documentation of the command "MininimalPolynomial" in Mathematica it says "Find the characteristic polynomial of Sqrt[2] over the extension E^(I Pi/4). The code: MinimalPolynomial[Sqrt[2], x, Extension -> E^(I Pi/4)] returns -1 - 2 x^2 + x^4.
I know that Sqrt(2) is in the field Q(E^(I Pi/4). In fact, Q(Sqrt(2)) is one of the 3 intermediate fields between Q and Q(E^(I Pi/4).
I do not know what "characteristic polynomial" means in this context. I am about half way through my first course in Field Theory. My textbook has not defined this term.
Given a linear transformation $T:V\to V$ where $V$ is a finite-dimensional vector space over a field $F,$ the characteristic polynomial of $T$ is defined as the determinant of $\lambda I-T$. This is independent of the basis chosen for $V$, and is a monic polynomial in $\lambda$ of degree equal to the dimension of $V$ over $F$.
For $L$ an finite-dimension field extension of $F$, then, and $x\in L$, we can see $L$ as a finite-dimensional vector space over $F$, and the characteristic polynomial of $x$ in $L$ over $F$ is the characteristic polynomial of multiplication of $T_x:L\to L$ defined as $T_x(a)=xa$.
If there is no $L$ given, just $F$ and an algebraic $x$ over $F$, the characteristic polynomial is taken as the characteristic polynomial of $x$ in $L=F(x)$.
The characteristic polynomial in this latter case - where $L=F(x)$ - is the same as the minimal polynomial for $x$.
In this case, let $\zeta=e^{i\pi/4}$. Then $\zeta-\zeta^3=\sqrt{2}$. So if you choose the basis $1,\zeta,\zeta^2,\zeta^3$ for the basis of $L=\mathbb Q[\zeta]$, the matrix for multiplication by $\sqrt{2}$ can be written as:
$$\begin{pmatrix}0&1&0&-1\\ 1&0&1&0\\ 0&1&0&1\\ -1&0&1&0\end{pmatrix}$$
It turns out, however, there is an easier way to compute the characteristic polynomial for linear transformations of this type - it is alway a perfect power of the minimal polynomial. In this case, that means the characteristic polynomial will be $(\lambda^2-2)^2$, since the minimal polynomial for $\sqrt{2}$ is $\lambda^2-2$ and the characteristic polynomial will be a fourth-degree monic polynomial.
You can see this more directly by seeing that $\mathbb Q(\zeta)=\mathbb Q(\sqrt{2},i)$, and use the basis: $1,\sqrt{2},i,\sqrt{2}i$. Then the matrix becomes:
$$A=\begin{pmatrix}0&1&0&0\\ 2&0&0&0\\ 0&0&0&1\\ 0&0&2&0\end{pmatrix}$$
It is much more obvious that $\det(\lambda I-A)=\lambda^4-4\lambda^2+4$.