Non isomorphic smooth cubic surfaces? Smooth cubic surfaces are known to have many wonderful properties in common. However, this means I cannot easily show any two to be non isomorphic. Presumably there are non isomorphic (over C) pairs of smooth cubic surfaces... Can someone exhibit a pair with proof?
 A: This is just an expansion on one of Ariyan's suggestions above. 
We can indeed show that cubic surfaces up to isomorphism form a 4-dimensional family just by counting moduli. The tricky point is precisely what the OP asked in comments: why must an isomorphism of cubic surface come from a projective equivalence? Let me answer that.
Every smooth cubic surface $S$ is anticanonically embeddeded in $\mathbf P^3$, meaning that $O_{\mathbf P^3}(1)_{\vert S} = -K_S$. 
Moroever, since the codimension is 1, this embedding is linearly normal, meaning that that the restriction map $H^0(\mathbf P^3, O(1)) \rightarrow H^0(S,-K_S)$ is an isomorphism. 
An isomorphism $S \cong S'$ of two cubic surfaces will induce an isomorphism $$H^0(S,-K_S) \cong H^0(S',-K_{S'}),$$ hence an automorphism of $H^0 (\mathbf P^3,O(1))$, and hence a projective equivalence of 
$$ \mathbf P^3 = \mathbf P \left( H^0 (\mathbf P^3,O(1))^\ast \right).$$
Keeping tabs on the various maps, we see that this projective equivalence induces exactly  the original isomorphism between $S$ and $S'$.
Remark: notice that this proof doesn't use the description of cubic surfaces as blowups of $\mathbf P^2$.
