A variety $ Z $ over a field $ k $ of characteristic zero is ruled if there is a variety $ M $ and a dominant, birational map $ \phi: M \times \mathbb{P}^{1}_{k} \dashrightarrow Z $. A variety $ Z $ over a field $ k $ of characteristic zero is uniruled if there is a variety $ M $ and a dominant, rational map $ \phi: M \times \mathbb{P}^{1}_{k} \dashrightarrow Z $. Does anyone know of a projective variety $ Z $ over a field $ k $ of characteristic zero which is uniruled, but not ruled?

3$\begingroup$ I don't know much about this, but I would guess a cubic threefold. $\endgroup$ – roy smith Jul 6 '20 at 21:17
I think you just need to know that there exists a threefold $X$ which is unirational but not rational (e.g. the cubic threefold). If $X$ is birational to $S\times \mathbb{P}^1$, there is a dominant rational map $X \!\!\!\!> S$, thus $S$ is unirational, hence rational by Castelnuovo's theorem. Therefore $X$ is rational.
Some conic bundles that are not birationally trivial do the job. For explicit examples, see pages 143148 of
K. Matsuki: Introduction to the Mori program, Universitext. New York, NY: Springer (ISBN 0387984658/hbk). xxiii, 478 p. (2002). ZBL0988.14007,
The case of a cubic threefold $W_3 \subset \mathbb{P}^4$, cited by Roy Smith in his comment, belongs to this family of counterexamples. In fact, the blowup $X=\mathrm{Bl}_L(W_3)$ of $W_3$ along a line $L \subset W_3$ is a conic bundle over $\mathbb{P}^2$. By ClemensGriffths we know that $W_3$ is not rational, so $X$ is not rational, and this implies that its conic bundle $X \to \mathbb{P}^2$ is not birationally trivial.