*EDIT: The vague question Q1 below is partially answered, while the concrete question Q2 seems to be still open.*

Let $V$ be a geometrically integral variety over a field $K$. I consider the following properties:

(1) There exists a dominant rational map $\mathbb{P}_K^n\dashrightarrow V$ for $n={\rm dim}(V)$.

(2) There exists a dominant rational map $\mathbb{P}_K^n\dashrightarrow V$ for some $n$. ($V$ is unirational)

(3) There exists a dominant rational map $\mathbb{P}_K^n\dashrightarrow V$ with geometrically integral generic fiber, for some $n$.

(4) There exists a dominant rational map $\mathbb{P}_K^n\dashrightarrow V$ with a right inverse $V\dashrightarrow\mathbb{P}_K^n$, for some $n$. ($V$ is retract rational)

(5) There exists a birational map $\mathbb{P}_K^n\dashrightarrow V\times\mathbb{P}_K^m$ for some $m,n$. ($V$ is stably rational)

(6) There exists a birational map $\mathbb{P}_K^n\dashrightarrow V$ for some $n$. ($V$ is rational)

We have that $(6)\Rightarrow(5)\Rightarrow(4)\Rightarrow(3)\Rightarrow(2)\Leftrightarrow(1)$. For curves all of these properties are equivalent, but they diverge in higher dimension. From browsing the literature I gather that it is known that $(2)\not\Rightarrow(4)$ and $(5)\not\Rightarrow(6)$, and it is expected that $(4)\not\Rightarrow(5)$.

However, I am interested in property (3), which I could not find anywhere in the literature.

**Q1:** Does property (3) occur in the literature? Does it have a name? Is it equivalent to either (2) or (4)?

*EDIT: As the answer of Daniel Loughran shows, the Châtelet surfaces described below are examples for $(3)\not\Rightarrow(5)$ over $K=\mathbb{R}$.*

*Phrased more down to earth, here is a very concrete question I am interested in:*

**Q2:** Is every intermediate field $F$ of $\mathbb{R}(X,Y,Z)/\mathbb{R}$ which is algebraically closed in $\mathbb{R}(X,Y,Z)$ purely transcendental over $\mathbb{R}$?

Of course this is clear for $F$ of transcendence degree $0$, $1$ or $3$ over $\mathbb{R}$, so it is really just a question about surfaces. The equivalent question over $\mathbb{C}$ has a positive answer, as every unirational complex surface is known to be rational. The closest to a counterexample I found in the literature is the surface over $\mathbb{R}$ given by $x^2+y^2=f(z)$ with $f$ of degree $3$ with three real roots, which I think satisfies (2) but not (5), but I don't know if it satisfies (3).

*EDIT: As explained in the answer of Daniel Loughran, such a surface $V$ satisfies (3). But it seems not clear whether the $n$ in $\mathbb{P}_K^n\dashrightarrow V$ with geometrically integral generic fiber can be chosen to be 3, which would be needed to answer Q2 negatively.*