Fix an algebraically closed field $k$. It is well known that the quadric hypersurface $Q=\mathbb{P}_k^1\times_k\mathbb{P}_k^1$ can be embedded in $\mathbb{P}^3$, and that any nonsingular projective curve $X$ can be embedded in $\mathbb{P}^3$. Can we sharpen this statement to say that any curve $X$ can be embedded in $Q$ or $\mathbb{P}^n$?
My reason for thinking this is that while plane curves only have a very narrow range of degrees, any nonsingular closed curve of degree $(a,b)$ in $Q$ will have genus $(a-1)(b-1)$, so in practice, if we want to find a curve that cannot be embedded in $\mathbb{P}^2$, we tend to look in $Q$. So do these give us all the examples?
My intuition is that this must be false, and any proof by counterexample would probably involve constructing some field $K$ such that $\operatorname{tr.deg}_k K=1$, and showing that such a curve could not exist in $Q$. However, this would seem to require more advanced theory than what is taught in the basic theory of algebraic curves, so I don't right away see how to do this.
Yes, it is false. As you said, any curve in $\mathbb{P}^1\times\mathbb{P}^1$ of type $(a,b)$ has genus $(a-1)(b-1)$, So, if a curve has genus $p$, a prime, of this type, it follows that $a$ or $b$ must be 2. But then, by a suitable projection, the curve must be a double cover of $\mathbb{P}^1$, that is hyperelliptic. If $p\geq 3$, there exists curves which are not hyperelliptic.