TITLE: Uniruled + Picard number 1 = Fano?
QUESTION [6 upvotes]: Hello, I'm a newbie to mathoverflow. I reading a paper about Fano varieties (over C) and there is an assumption that uniruled varieties with picard number 1 are Fano...why is this true? Sorry if this is obvious thing. I am just learning the definitions still.

REPLY [2 votes]: Just to give yet another argument using more or less the definition of uniruledness.
$X$ is uniruled if there exists a dominant morphism from a variety $Y\times \mathbb P^1$ where $\dim Y=\dim X-1$. Therefore the image of $\{y\}\times \mathbb P^1$ for a general $y\in Y$ moves in a family of dimension $\dim Y$ in $X$ and hence has a semi-positive normal bundle and in particular the determinant of its normal bundle has to be a non-negative line bundle. Then the adjunction formula shows that $-K_X$ is positive on this curve.
If the Picard number is $1$, then $\mathrm{Pic}X\otimes \mathbb Q\simeq \mathbb Q$. If $L$ is a fixed ample divisor, then for any divisor $D$ there exists $a,b\in \mathbb N$ such that $aD\sim bL$. Since there exists at least one curve on which $-K_X$ is positive, it follows that $-K_X\sim \alpha L$ for some $\alpha\in \mathbb Q_+$ and hence it is ample itself. It follows that $X$ is Fano.