Uniruled + Picard number 1 = Fano? 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.
 A: Here is another way of seeing this: If $X$ us uniruled, then as in Polizzi's answer we find a rational curve $f:\mathbb{P}^1\to X$ through a general point on $X$. Since $f^*K_X$ has negative degree on $\mathbb{P}^1$, any global section in $H^0(X,mK_X)$ must vanish on the image of $f$, hence on a dense subset of $X$, and hence on $X$. In particular, $K_X$ cannot be ample, and so $X$ is Fano.
In fact, the above argument shows that if $X$ is uniruled, then all the plurigenera $p_m=h^0(X,mK_X)$ vanish for all $m\ge 0$. In terms of Kodaira dimension, this means that $\kappa(X)=-\infty$.
An open problem in birational geometry is whether the converse also holds (see Debarre's book). For curves, this is obvious, since $p_1$ is just the genus of $X$. In dimension 2, Castelnuovo’s criterion asserts that a surface $X$ is rational if and only if $q(X) = p_{12}(X) = 0$, where $q(X)$ is the irregularity $h^0(X, \Omega_X^1)$. In dimension 3, this true by a deep result by Miyaoka and the MMP. For higher dimensions, only a few partial results are known.
A: 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.
A: Since $X$ is a projective variety with $\textrm{Num}(X) \cong \mathbb{Z}$, the canonical divisor $K_X$ is either ample, or anti-ample or numerically trivial. 
On the other hand we know that $X$ is uniruled, so it contains a free rational curve, namely  the image of a map  $$f \colon \mathbb{P}^1 \longrightarrow X$$
such that $f^*T_X$ is generated by global sections. In this case,
$$f^*T_X=\mathcal{O}_{P^1}(a_1) \oplus \cdots \oplus \mathcal{O}_{P^1}(a_n),$$
with $a_1 \geq a_2 \geq \cdots \geq a_n \geq 0.$ 
Moreover, since $T_{P^1}$ embeds into $f^*T_X$, it follows $a_1 \geq 2$. Therefore
$$K_X \cdot f_{*} \mathbb{P}^1= - \sum_{i=1}^n a_i \leq -2,$$ 
hence $K_X$ is not nef. This implies that $K_X$ is anti-ample, i.e. $X$ is Fano.
