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 $t\_k$?



Algebraic square root question
**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](http://en.wikipedia.org/wiki/Companion_matrix) of $P, Q$ and $\otimes$ denotes the [Kronecker product](http://en.wikipedia.org/wiki/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.


