Is there any rational map from the nonsingular Segre quadric surface in $\mathbb{CP}^3$ to a nonsingular surface of degree greater or equal to 4? Is there any rational map from the nonsingular Segre quadric surface in $\mathbb{CP}^3$ to a nonsingular surface in $\mathbb{CP}^3$ of degree greater or equal to 4?
Someone told me that the answer is negative and that is trivial but i can't find the way to prove it.
 A: Either the question has an error, or your informant was talking entirely through their hat: after all, given any two varieties $X$ and $Y$, one can get a regular map $X \rightarrow Y$ by choosing any point on $Y$, and mapping $X$ to that point!
However, if we insert the word dominant before rational map, then the question makes sense, and indeed the answer is negative. One way to prove it is (although I don't know if it counts as "trivial") is to put together the following facts. Here $K$ denotes the canonical bundle of a variety.


*

*If $X \dashrightarrow Y$ is a dominant rational map between nonsingular varieties over $\mathbf C$, then $\text{dim} \ H^0(Y,K_Y) \leq \text{dim} \ H^0(X,K_X)$. 

*The quadric surface $Q$ has $\text{dim} \ H^0(Q,K_Q)=0$ (by the adjunction formula).

*A nonsingular surface $Y$ of degree $\geq 4$ has $\text{dim} \ H^0(Y,K_Y) \geq 1$ (again by the adjunction formula).


Remark: Amazingly, this assertion is completely false over other fields! For example, Shioda showed that over a field of characteristic 3, the Fermat quartic surface $Y$ defined by the equation
$$x^4+y^4+z^4+w^4=0$$ 
is unirational: that is, there is a dominant rational map $\mathbf P^2 \dashrightarrow Y$. Since $\mathbf P^2$ and $\mathbf P^1 \times \mathbf P^1$ are birational, this shows that the OP's question has a positive answer in this context.
