Trouble with Vakil's FOAG exercise 11.3.C I'm having trouble with the exercise in the title, even with part (a), which asks to prove that if $X$ is a closed subset of $\mathbb{P}^n_k$ of dimension at least 1 and $H$ is a non-empty hypersurface, then $X$ and $H$ meet. 
Following the hint, I've done the following: By assumption, $X$ is given by $V(I)$ for some homogenous prime ideal $I\subset k[x_0,\dots x_n]=:A$ and $H$ by $V(f)$ for some homogenous polynomial $f\in A$. We wish to prove that there is a homogenous prime ideal $\mathfrak{p}$ (that's not the irrelevant one) containing both $I$ and $f$.
Consider the affine cone $\mathrm{Spec}\ A/I$ of $X$ in $\mathrm{Spec}\ A$. Being homogenous, both $I$ and $f$ are contained in the ideal $(x_0,\dots x_n)$. By Krull's principal ideal theorem, every minimal prime ideal containing $f\mod I$ has codimension one. $X$ having dimension at least one implies that there are homogenous prime prime ideals $\mathfrak{q}\supsetneq \mathfrak{r}\supset I$. This implies that $(x_o,\dots x_n)$ has codimension at least two. Thus there are prime ideals containing $f\mod I$ that are contained in $(x_o,\dots x_n)$. However, I'd need to show that there are homogenous such prime ideals, right? I'd rather get hints than full solutions.
 A: Here's a proof since I think the current answers are possibly a little lacking (and the approach taken in the question itself seems slightly off):
As in the question, let $A = k[X_0,\dots,X_n]$, $X = V(I)$ and $H = V(f)$. For a closed subset $Z = V(J) \subset \mathbb{P}^n_k$, write $\overline{Z}=V(J)\subset \mathbb{A}^{n+1}_k$. First we proceed to show that $\overline{X}\cap\overline{H}(=\overline{X\cap H})$ has dimension at least $1$.
Notice that $\overline{X\cap H} = V(I+(f))$, and since $I$ and $f$ are homogeneous, we have $I+(f) \subset (X_0,\dots,X_n)$, so there is some prime $I+(f) \subset \mathfrak{p} \subset (X_0,\dots,X_n)$ minimal over $I+(f)$ (corresponding to an irreducible component of the affine cone over the intersection of $X$ and $H$). If $\mathfrak{p} \subsetneq (X_0,\dots,X_n)$, then $V(I+(f))$ has dimension at least $1$, as claimed. If not, $\mathfrak{p} = (X_0,\dots,X_n)$ and so by Krull's principal ideal theorem applied to $A/I$, any chain of primes between $I$ and $(X_0,\dots,X_n)$ has length at most $1$. But since $X$ has dimension at least $1$, there are homogeneous primes $I \subset \mathfrak{q}\subsetneq\mathfrak{q}'$ not containing $(X_0,\dots,X_n)$, giving a chain of primes from $I$ to $(X_0,\dots,X_n)$ of length $2$, a contradiction.
This dimension argument gives that there is a prime ideal inbetween $I+(f)$ and $(X_0,\dots X_n)$, and so the question is: is there a homogeneous prime ideal between them? There's a high tech way and a low tech way to proceed, but know that the high tech approach is really just hiding the low tech one behind some fancy language.
High tech: We have a map $\pi:\mathbb{A}^{n+1}_k\setminus\{0\}\rightarrow\mathbb{P}^n_k$, such that for any closed subset $Z$, $\pi^{-1}(Z) = \overline{Z}\setminus\{0\}$, so the cone over the intersection having dimension $1$ means that it contains a non-zero point, and so this maps to the intersection under $\pi$, meaning that the intersection is therefore non-empty.
Low tech: In the process of the $\rm{Proj}(S_{\bullet})$ construction, we get a bijection between the primes of $S_{(f)}$ and the homogeneous primes of $S_{f}$, where the map one way is given by taking the degree $0$ component. But in fact, given any prime ideal of $S_f$, taking the degree $0$ component gives a prime ideal of $S_{(f)}$ and so, by the bijection, there is some homogenous prime of $S_f$ with the same degree $0$ component. Thus, in our construction, the prime $\mathfrak{p}\subsetneq (X_0,\dots, X_n)$ lying over $I+(f)$ doesn't contain some $X_i$, and so corresponds to a prime in $A_{X_i}$ lying over $(I+(f))_{X_i}$. Then by the remarks above there is then a homogeneous prime of $A_{X_i}$ with the same degree $0$ component, which therefore also contains $(I+(f))_{X_i}$. This prime then corresponds to a homogeneous prime of $A$ (also not containing $X_i$) that contains $I+(f)$, as desired. Actually checking that all of the correspondences work out as I claim is somewhat of a hassle (it's not hard per say, but is a bit of a faff) but bear in mind that you would have to do all of this anyway if you wanted to prove that the projection map used in the high tech approach actually has the desired property, so I don't think it can really be avoided.
A: Hint: Do $\operatorname{Spec} A/I$ and $\operatorname{Spec} A/f$ meet inside $\operatorname{Spec} A$? What do you know about intersections in affine space?
A: The intersection  $Y=c(X)\cap V(f)$   of the cone        $c(X)\subset \mathbb A^{n+1}_k$  over $X$ with the hypersurface $V(f) \subset \mathbb A^{n+1}_k$ has each of its irreducible components $Y_i$  of dimension at least $\operatorname {dim}c(X)-1= \operatorname {dim}(X)\gt 0$ by Krull's principal ideal theorem.
One of those irreducible components $Y_0$ contains the origin and thus, since it has positive dimension, also another point $p\in Y_0\subset c(X)\cap V(f)$.
The image of this point is the required point of intersection  $[p]\in X\cap V_+(f)$
A: This is how I did it, following Vakil's hint (hopefully it's a sufficiently different phrasing from the other answers):
In $k[x_0,...,x_n]$, consider a prime ideal $\frak p$ which is minimal over the ideal $I+(f)$.   Since $I+(f)$ is contained in the irrelevant ideal $\frak{m}$ $:=(x_0,...,x_n)$, we know $\frak p$ exists, and since $I+(f)$ is homogeneous, $\frak p$ is homogeneous. We have only to show that $\frak{p}$ $ \neq \frak m$.
But if $\frak m$ is minimal over $I+(f)$, then $\frak m$ is minimal over $f$ in $k[x_0,...,x_n]/I$, so that by Krull's principal ideal theorem we know that the codimension of $\frak m$ in $k[x_0,...,x_n]/I$ is not greater than $1$.  But  $V(I)$ has dimension at least $1$ in $\mathbb{P}^n$, meaning there are homogeneous prime ideals $I\subset \frak q_0 \subsetneq q_1 \subsetneq m$, which is absurd.
