I am trying to prove 5.15b) here:
http://www.math.u-bordeaux1.fr/~qliu/Book/Errata3/pages76-77.pdf
I am not sure if my argument is complete, and would therefore appreciate if someone could look into it.
Update: If you know the solution, but don't have the time to look into my solution, I would be very grateful for a solution / hint and I can compare your solution with my solution myself.
So, I start by assuming wlog that Y is irreducible of dimension r. Further, we can assume that Y is of the form $Proj B/I$ for some I since $Proj B/I$ has the same topological space as Y. Now, consider the map $B \rightarrow B/I$ and consider the images, which we will call $g_i$ , of $f_i$ for $i= 1, \ldots, r$. Now, $V_+(f_1) \cap Y = V_+(g_1) \cap Y = V_+(g_1)$. If $V_+(g_1) = \emptyset$, then by a), the dimension of Y is $\leq 0$, which is not true if $r > 0$. So, let us consider $V_+(g_1)$. This will then be a closed subvariety of Y of pure dimension $r-1$ if $g_1$ is not nilpotent, if nilpotent nothing will change dimensionwise. So consider $Proj B/(I,f_1)$. We can repeat the same argument to this scheme and continuing along these lines, one sees that the intersection can't be the empty set.
