As part of working with dot products of vectors in an algebraic field, having previously solved the algebraic quotient, I need to find the algebraic square root as shown in (1)
Let $r_i$ be a root of polynomial P and $s_j$ be a root of polynomial Q i.e., P($r_i$)=0, Q($s_j$)=0.
I seek to find a third polynomial R and its root $t_k$, such that R($t_k$)=0, so that
(1) $t_k$ = $\sqrt{1 - r_i^2 - s_j^2}$
is satisfied. How can R be found, knowing $t_k$?
Lemma 1: If $P(x)$ is an integer polynomial with root $r$, then $P(x^2)$ is an integer polynomial with root $\sqrt{r}$.
Lemma 2: If $P(x)$ is an integer polynomial with root $r$, then $P(\sqrt{x})P(-\sqrt{x})$ is an integer polynomial with root $r^2$.
Lemma 3: If $P(x)$ is an integer polynomial with root $r$, then $P(x - t)$ is an integer polynomial with root $r + t$, for $t$ an integer.
Lemma 4: If $P, Q$ are integer polynomials with roots $r, s$, then an integer polynomial with root $r + s$ is given by the characteristic polynomial of $A \otimes I + I \otimes B$ where $A, B$ are the companion matrices of $P, Q$ and $\otimes$ denotes the Kronecker product. Multiplication by $r + s$ defines a $\mathbb{Q}$-linear transformation on $\mathbb{Q}[r, s]$, which has $\mathbb{Q}$-basis $r^i s^j$ where $i, j$ range from $0$ to one less than the degrees of $P$ and $q$, and the matrix above is the matrix of this linear transformation in that basis.
So apply Lemma 2 twice, then Lemma 3, then Lemma 4, then Lemma 1.
