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Let U be a unirational variety over a field of characteristic 0. I have read that its canonical divisor cannot be ample, but I don't know why.

Any references will be most helpful.

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Smooth varieties

Let $X$ be a smooth uniruled variety. Then there is a free rational curve $f:\mathbb{P}^1\rightarrow X$, that is $H^1(\mathbb{P}^1,f^*T_X\otimes\mathcal{O}_{\mathbb{P}^1}(-1)) = 0$. Now $f^{*}T_X$ is a rank $n = dim(X)$ vector bundle on $\mathbb{P}^1$. Therefore we can write $$f^*T_X = \mathcal{O}_{\mathbb{P}^1}(a_1)\oplus...\oplus\mathcal{O}_{\mathbb{P}^1}(a_n).$$ Now, $f^{*}T_X$ contains $T_{\mathbb{P}^1}\cong\mathcal{O}_{\mathbb{P}^1}(2)$. We have $K_X\cdot f_*\mathbb{P}^1 = -\sum_{i=1}^{n}a_i\leq -2$. Therefore, $X$ uniruled implies that there is a free rational curve on $X$ which implies that $K_X$ is not nef. So, if $K_X$ is ample (in particular $K_X$ is nef) then $X$ is not uniruled.

Now, $X$ unirational $\Rightarrow$ $X$ rationally connected $\Rightarrow$ $X$ uniruled $\Rightarrow$ $K_X$ is not nef (in particular $K_X$ is not ample).

In other terms. If $X$ is uniruled then $X$ has nagative Kodaira dimension. If the canonical bundle of $X$ is ample (it is enough big) then $X$ has Kodaira dimension $dim(X)$.

Singular varieties

There are singular uniruled varieties with ample canonical bundle. In particular there are many examples of rational singular surfaces with ample canonical bundle. The following example is due to Kollar. You can find the details and many other interesting constructions here http://arxiv.org/abs/1007.1936

Let us consider the surface $$Y = Y(a_1, a_2, a_3, a_4) := (x_1^{a_1}x_2 + x_2^{a_2}x_3 + x_3^{a_3}x_4 + x_4^{a_4}x_1 = 0)$$ in the weighted projective space $\mathbb{P}(w_1, w_2, w_3, w_4)$, such that $$a_1w_1+w_2 = a_2w_2 + w_3 = a_3w_3 + w_4 = a_4w_4+w_1 = d$$ and $$\begin{array}{ll} w_1 = \frac{1}{w^*}(a_2 a_3 a_4-a_3 a_4+a_4-1), & w_2 = \frac{1}{w^*}(a_1 a_3 a_4-a_1 a_4+a_1-1),\\ w_3 = \frac{1}{w^*}(a_1 a_2 a_4-a_1 a_2 + a_2-1), & w_4 = \frac{1}{w^*}(a_1 a_2 a_3- a_2 a_3 + a_3-1), \end{array}$$ $$d = \frac{1}{w^*}(a_1 a_2 a_3 a_4-1)$$ where $w^* = \gcd(w_1, w_2, w_3, w_4).$ If $w^* = 1$, then $Y$ is a rational surface with $4$ cyclic singularities and $H_2(Y, \mathbb{Q}) \cong \mathbb{Q}^3$. The rational curves $$C_1 := (x_1 = x_3 = 0),\quad C_2 := (x_2 = x_4 = 0)$$ are extremal rays with respect the $K_Y + (1- \epsilon)(C_1 + C_2)$ minimal model program for $0 < \epsilon \ll 1$. Therefore $C_1$ and $C_2$ are contractible to quotient singularities and we get a rational surface of Picard number 1, $$\pi : Y=Y(a_1,a_2,a_3,a_4) \rightarrow X = X(a_1, a_2, a_3, a_4).$$ If $a_1, a_2, a_3, a_4 \geq 4$, then $K_{X}$ is ample.

For examples of big surfaces with big cotangent bundle you could take a look at this http://www.latp.univ-mrs.fr/~eroussea/big-surfaces.pdf

For a classification of complex compact Gorenstein surfaces with negative canonical bundle you can read this http://link.springer.com/article/10.1007%2FBF01421952#page-2. A classical example of a rational surface with negative canonical bundle is the quadric cone in $\mathbb{P}^3$. The canonical divisor $K_S$ is $-2H_{|S}$, where $H$ is a plane in $\mathbb{P}^3$ through the vertex.

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    $\begingroup$ It's obviously not true that $f^\ast K_X = K_{\mathbb P^1}$, e.g. if $X = \mathbb P^2$ and the embedding is a line. $\endgroup$ – user47305 May 12 '14 at 20:04
  • $\begingroup$ @Cobian: Could you please say why $f^*T_X$ is a vector bundle? Are you assuming $X$ is smooth? $\endgroup$ – user46578 May 29 '14 at 16:20
  • $\begingroup$ Yes, the proof works for smooth varieties. There are examples of singular rational varieties with ample canonical bundle. I added some more explanation in my answer. $\endgroup$ – F_L May 29 '14 at 20:43

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