Is -(E,E) greater or equal to 2 for a minimal resolution

I'm quite confused by the terminology minimal resolution and minimal model.

Let $f:X\longrightarrow Y$ be a minimal resolution of singularities, where $Y$ is a normal surface.

Let $E$ be an exceptional component.

Do we have that the intersection number $-(E,E)\geq 2$?

By Castelnuovo' criterion, this has to do with the minimality of $X$. I know one could blow down all the curves such that $-(E,E) = 1$ to obtain a minimal model, but if $Y$ is already regular and contains such curves, then so will $X$ right?

So minimal resolution and minimal model aren't always the same, right?

flag	asked Jul 24 2011 at 5:30 Enchanted 73●3

You're partially right. Smooth rational components of the exceptional divisor must have $E^2\le -2$, but other components could have $E^2=-1$. Furthermore, the notion of minimal resolution is relative to $Y$, so it is not necessarily a minimal model. – Donu Arapura Jul 24 2011 at 9:26

I see! Actually in the article I'm reading the surface has only rational singularities. that's why they get to use that $E^2 \leq -2$ because the components of the exceptional divisor are all rational. Thank you for clearing this up for me. – Enchanted Jul 24 2011 at 11:23