Irreducibility of a quadric I am struggling with a problem in Shafarevich's Basic Algebraic Geometry. First, some context: Fix $k$ an algebraically closed field. Lines in $\mathbb{P}^3$ correspond to planes through the origin in $4$-dimensional space. Thus lines in $\mathbb{P}^3$ have an interpretation as points of the $(2,4)$-Grassmannian, which has an embedding in $\mathbb{P}^5$ given by Plücker coordinates. Call this embedding $\Pi$. In section 1.6 of Shafarevich's book, it is detailed how points corresponding to lines in a surface in $\mathbb{P}^3$ are given by a projective subvariety of $\Pi$.
The problem I am struggling with asks you to show that points in the Plücker surface corresponding to lines in an irreducible quadric $Q$ over $\mathbb{P}^3$ are given by two disjoint conics. 
One way to approach this seems to find a "nice" form for $Q$ where the solution becomes obvious, for example by using the fact that one can pick coordinates to "diagonalize" $Q$ and then solve the problem there. However, I am having the following issues:
(i) I am not sure what the rank of the quadratic form corresponding to an irreducible quadric should be. According to Georges Elencwajg's answer, Quadrics are birational to projective space it seems that the rank should be 4, but $x_0^2+x_1^2+x_2^2$ which would correspond with a quadratic form of rank 3 seems pretty irreducible to me. I understand that the rank cannot be less than 3 though (2 is a problem because then we have a change of coordinates to $x_0^2+x_1^2$ which is obviously reducible, and 1, well...).
(ii) Diagonalizing seems to require a "Gram-Schmidt" process using the bilinear form associated with the quadratic form, which only is available for fields with characteristic different from 2, and since Shafarevich does not specify this in his statement of the exercise, this approach to the problem does not work in all cases. Thus another approach is necessary, and I don't have one.
Any help would be appreciated.
 A: Let me try to answer your first question at least (and leave the second to someone who knows more about that). If $Q$ were a quadric cut out by a quadratic form of rank 3, then yes it would be irreducible, but it would not be smooth. It would be a projective cone over a plane conic and of course the vertex of this cone is the unique singularity.
E.g. $Q := V(x_0^2 + x_1^2 + x_2^2)$ can be viewed as the cone over the conic $Q\cap V(x_3)$ cut out by the same equation in the plane $V(x_3)$ thought of as $\mathbb{P}^2$ sitting inside $\mathbb{P}^3$ with $Q$ having vertex at $[0:0:0:1]$.
In this case, any line contained in the cone $Q$ contains the vertex. (Suppose not, then let $v$ be the vertex point and $p \in Q$ a non-vertex point with $L\subset Q$ a line containing $p$ but not $v$. If $H = \overline{vL}$ is the unique plane containing $v$ and $L$ then it intersects $Q$ in two (not necessarily distinct) lines which meet at $v$ and also contains $L$ - that's a degree (at least) 3 intersection, which contradicts Bezout's theorem). This means that the parameter space $\Phi$ (in the Grassmannian $G(2,4)$) of lines lying in $Q$ can't be disconnected because if you take the family of lines over $\Phi$ (i.e. $\Psi := \{(p,[L]) \in Q \times \Phi : p \in L\}\subset \mathbb{P}^3\times G(2,4)
$), it maps  surjectively to $Q$ (via $(p,[L]) \mapsto p$) so that the preimage of a non-vertex point $p \in Q$ contains points $(p,[L]) \in \Psi$ where $L$ is any line in $Q$ containing $p$. Since we saw above that $L$ is unique for $p$ a non-vertex point, this is an isomorphism away from the preimage of the vertex and it's not hard to check that this couldn't happen if $\Phi$ (and hence $\Psi$) were disconnected (check, for example, that the map would not contract a connected component to the vertex point).
Indeed, $\Psi$ can be realized as the strict transform of $Q$ under the blowup of $\mathbb{P}^3$ at the vertex of $Q$ and is isomorphic to $\mathbb{P}_C(\mathcal{O}\oplus \mathcal{O}(1)) = \mathbb{F}_2$ for $C$ the conic in $\mathbb{P}^2$ that $Q$ is a cone over.
