If the linear system $|2P_0|$ had base points, an elliptic curve would be rational?

Hartshorne writes the beginning chapter IV.4 Elliptic curves

Let $X$ be an elliptic curve over the algebraically closed field $k$. Let $P_0 \in X$ be a point and consider the linear system $|2P_0|$ on $X$. The divisor $2P_0$ is nonspecial, so by Riemann-Roch, this linear system has dimension $1$. It has no base points, because otherwise the curve would be rational.

I don't understand the last sentence here. How does the existence of base points imply that $X$ is rational?

I can use Corollary 3.2 to see that $|2P_0|$ has no base points, because $\deg(2 P_0) = 2 \geq 2 \cdot g(X)$. But what argument might Hartshorne have in mind regarding the rationality of $X$?

Clearly any base point of $|2P_0|$ would have to be $P_0$. Since the linear system $|2P_0|$ is $1$-dimensional, there exists aother effective divisor $D \neq 2P_0$, linearly equivalent to $2 P_0$. But as $P_0$ is a base point we have $D = P_0 + Q$ for some other point $Q \in X$. And hence $Q \sim P_0$, which is a criterion for rationality (example II 6.10.1).