Let $I = \langle x^2 +y^2 +z^2 −1,3x^2 +y^2 −z^2 −1\rangle \triangleleft \mathbb{C}[x,y,z]$ and let $W = \mathcal{V}(I) \subset \mathbb{A}^3$ be the variety defined by $I$. The generators themselves are irreducible, but $$x^2 −z^2 = \frac{(3x^2 +y^2 −z^2 −1)−(x^2 +y^2 +z^2 −1)}{2} \in I$$ too, and $x^2 − z^2 = (x + z)(x − z)$. Therefore by the argument of Proposition 1.9, $W$ can be decomposed into the union of $W_1 = V(⟨x^2 +y^2 +z^2 −1,3x^2 + y^2 −z^2 −1,x+z⟩)$ and $W_2 = V(⟨x^2 +y^2 +z^2 −1,3x^2 +y^2 −z^2 −1,x−z⟩)$.
*CONTEXT\NOTATION: *$\langle x^2 +y^2 +z^2 −1,3x^2 +y^2 −z^2 −1 \rangle$ is the ideal generated by $x^2 +y^2 +z^2 −1$ and $3x^2 +y^2 −z^2 −1$. $\mathcal{V}(I)$ is the set of points that every function in $I$ vanishes over, where $I$ is an ideal over $\mathbb{C}[x,y,z]$. $\mathbb{A}^3$ contains all triples of complex numbers. In Proposition 1.9, it is said that, given an affine algebraic variety $W$, then if $f_1f_2\in\mathcal{I}(W)$ but $f_1,f_2\notin\mathcal{I}(W)$, then $W= \mathcal{V}\left(\langle\mathcal{I}(W),f_1\rangle\right)\bigcup\mathcal{V}\left(\langle\mathcal{I}(W),f_2\rangle\right) $. $\mathcal{I}(W)$ is the set of polynomials in $[x,y,z]$ with complex coefficients that vanish over every point in $W$.
What I don't understand is why the above example (highlighted) uses $I$ instead of $\mathcal{I}(\mathcal{V}(I))$ when reducing $W$. From Hilbert's Nullstellensatz, we know that $\mathcal{I}(\mathcal{V}(I))=\sqrt{I}$. So if $I$ is a radical ideal (i.e. $\sqrt{I}=I$), then it would make sense to use $I$ instead of $\mathcal{I}(\mathcal{V}(I))$.
But in this case, is $I$ a radical ideal?
Yes, $I$ is radical. Observe that $$\Bbb C[x,y,z]/I\cong \Bbb C[x,y]/(2x^2+y^2-1)$$ and this is a reduced ring, as the polynomial $2x^2+y^2-1$ is irreducible.