I asked a similar question at MSE, as the question seemed quite basic to me, but did not get any hint in 24 hours, except for one upvote for the question itself. I still think I am stuck with some basic understanding of irreducible and/or non-singular varieties.
In $\mathbb{P}^3$ we have an irreducible variety $A$ given by two equations $tx=p$ and $ty=q$ where $p$ and $q$ are quadratic forms in $x$, $y$ and $z$ (but not $t$). We need to argue that the map $\phi(x,y,z,t)\to(x,y,z)$ restricted to $A$ is an isomorphism of $A$ into the plane variety $B$ given by $yp=xq$.
It is straightforward to see that $\phi$ restricted to $A$ maps into $B$. We can further define the inverse map by $(x,y,z)\to(x^2,xy,xz,p)$ for $x\neq0$, and $(x,y,z)\to(xy,y^2,yz,q)$ for $y\neq0$. This gives us the inverse rational map for all $x \neq 0 \neq y$. What I want is to further argue that $(0,0,1)$ is not in the image of $\phi$ using irreducibility (otherwise, there is a whole subspace mapped to this point as argued below, and, unless I am missing something, the restricted map is not an isomorphism).
The way I try to think about it is that if $(0,0,1)$ is in the image, then every point $(0,0,1,t)$ is in $A$. In particular, both $p$ and $q$ are missing the term $z^2$. Now, I do not have enough vision of irreducibility to conclude this argument. It seems that I miss some basic understanding which is unfortunately not provided by a book I read.
I have also checked when the intersection of two quadratic forms is irreducible. I found a result where it says that if the two quadratic forms are given by matrices $A$ and $B$, then their intersection is irreducible iff $\det(A-\lambda B)=0$ has different roots (unless, again, I misread the result). Assuming this is true, and assuming that $p$ and $q$ are both missing the term $z^2$, taking $A$ and $B$ for quadratic forms $p-tx$ and $q-ty$, indeed, we have the determinant $l(\lambda)^2$ where $l(\lambda)$ is a linear function in $\lambda$. This means that the determinant equals zero at a single point $\lambda$. This seems to support my idea (that $(0,0,1,t)$ cannot be solutions if $A$ is irreducible), but the problem is that this result is not a part of what I want to use (it is from a paper I found), so I hope there is some way to argue what I want without using any additional results, but on an intuitive level. For example, by showing explicitly a decomposition of $A$ in this case.
P.S. The reason I also mention non-singularity is because I do not understand exactly the relation between two notions. For example, in one place I read that an irreducible variety can have one singular point only, but at the same time I read that $V(x_0^2+\cdots+x_k^2)$ is irreducible in $\mathbb{P}^n$ for every $k \ge 1$. Then, there is another question: show that any irreducible quadric $Q\subset\mathbb{P}^{n+1}$ is rational by taking the linear projection from a non-singular point, which induces a birational map $Q-\to\mathbb{P}^n$. I saw many similar results, but they all assume that $Q$ is also non-singular, so again I am confused how exactly they use this additional assumption, or whether it is needed at all. Finally, in the question above if there was the additional assumption of non-singularity, it would probably help me.
