@AriyanJavanpeykar Hi Ariyan, thanks for following up. Indeed, I checked the nonsingularity yesterday and the result was as expected. The surface is nonsingular along $(s: t: 0: 0: 0)$ unless $s=0$ or $t=0$ or $s=t$ so your argument was perfect. Thank you.

@AriyanJavanpeykar Hi Ariyan, thank you so much for your nice argument and for your patient proof. Let's take it as an answer!

@AriyanJavanpeykar Hi Ariyan, thank you for your comments. But I'm not sure I understand correctly the idea. How the existence of a line of smooth points would imply the geometric irreducibility? Thank you in advance for your further indications.

@Libli Hi Libli, this surface is the surface of lines passing through a nodal point in a particular nodal cubic fourfold. I'm looking at a particular family of nodal cubic fourfolds and I want to check that for the general fibre in this particular family, the surface of lines passing through the nodal point is irreducible.

@CraniumClamp Thank you for your comments. Indeed, this variety comes from the degeneration of a family of (2,3)-surfaces. I'm looking at a particular family of nodal cubic fourfolds and the surface I'm considering is the surface of lines on the cubic fourfold passing through the nodal point. I want to check that for the general fibre in this particular family, the surface of lines passing through the nodal point is irreducible. I think it suffices to check for one fibre. The surface in OP is such a surface which is relatively easy to find the equations.

@AriyanJavanpeykar Hi Ariyan, thank you for your comments. I do like your idea to check the geometrical irreducibility by looking at the rational point. But I don't think the surface $\Sigma$ is normal since it is singular at the line $(0: 0: 0: s: t)$.

@JasonStarr Thank you for pointing it out ! I've corrected it. I feel sorry for the confusions.