Hyperplane not containing a given set of points over a Noetherian scheme This is related to the answer in this question:
Showing that a power of an ample sheaf is equivalent to an effective Cartier divisor
Let $X$ be a quasiprojective scheme over a Noetherian ring A and suppose we have a very ample sheaf $\mathcal{L} \cong i^\ast O(1)$ for $i$ an immersion into projective space. Then, given a finite set of points $F$ (say the associated points of $X$) I want to show that there is a hyperplane $H \in \mathcal{O}(1)$ such that it does not meet any of these points, i.e that $Supp H \cap F = \emptyset.$ 
I was told that this is really tautologous, and I believe it is, but I am afraid I don't see it. I have seen arguments of the form previously, but never felt completely comfortable with them and thus I would be interested to see a careful proof (or as careful as you have the energy to give) of doing this. 
In the comments, it seems as if the statement I am making here is not true. Basically, I am interested in this just to get a detailed answer for the previous question, so feel free to reinterpret the question as long as the previous question gets a detailed answer.
 A: Embed $X$ into a projective scheme $\mathrm{Proj}(B)$ where $B$ is a homogeneous $A$-algebra. Let $\mathfrak p_1, \dots, \mathfrak p_n$ be the homogeneous prime ideals corresponding to the fintely many points you want to avoid. 
As $B_+$ (the irrelevant ideal) is not contained in any $\mathfrak p_i$, the homogeneous prime avoidance lemma says that there exists a homogeneous element $f\in B$, of some degree $d$, such that $f\notin \cup_i \mathfrak p_i$. Then the hypersurface of degree $d$ defined by $f$ avoids the given points. 
As said in the comment, in general it is not possible to find such an $f$ of degree $1$. Indeed, if $A$ is a finite field, and $X$ is a projective space over $A$, there is no hyperplane avoinding the (finite) set of the rational points of $X$. If $A$ contains an infinite field $k$, then we can view $B_1$ (homogeneous elements of degree $1$) and the $\mathfrak p_i$ as vector spaces over $k$, and of course there will then exists an $f\in B_1$ not in any $\mathfrak p_i$, thus a hyperplane avoinding the given points. 
A: As noted in cant_log's comment and answer, the statement is not true for all $A$,
although a modified version is.  (One has to replace $\mathcal O(1)$ by $\mathcal O(d)$ for an appropriate choice of $d$.)
Here is a proof when $k$ is an infinite field.
Let $X$ be a closed subset of $\mathbb P^n$, and let $X_i$ ($i = 1,\ldots,n$) denote the finite collection of irreducible and embedded components of $X$.
Let $(\mathbb P^n)^*$ denote the dual projective space, parameterizing hyperplanes
in $\mathbb P^n$.  Let $Z_i$ denote the collection of hyperplanes which contain $X_i$.  Since $X_i$ is non-empty, this is a proper closed subset of $(\mathbb P^n)^*$, hence its complement $U_i$ is a non-empty open subset.  The intersection
$U$ of all the $U_i$ is thus also a non-empty open subset.  
If $k$ is infinite, we may then find a $k$-point of $U$, and hence find a hyperplane
which does not contain any of the $X_i$, as required.

If $k$ is finite, the problem is that $U$ may not contain any $k$-point. (E.g. $X$ might be the union of all the rational hyper lanes!)  It will contain a $k'$-point for some finite extension $k'$ of $k$, and if $\ell$
is the equation (over $k'$) of the corresponding hyperplane, then the product of
all the conjugates (by elements of Gal$(k'/k)$) of $\ell$ will be a higher degree hypersurface which does not contain any of the components of $X$.

For general $A$ you have to a little more careful with this geometric approach,
since after all if Spec $A$ is not irreducible, then neither $\mathbb P^n$
nor $(\mathbb P^n)^*$ will be irreducible either.
One could cut down to the irreducible components of Spec $A$ (after all, the image
of each component of $X$ lies (at least) one of the components of Spec $A$), and then, assuming $A$ is a domain and working first over the generic point of $A$ (where the case of a field applies) one could proceed by Noetherian induction
on $A$ (essentially, induction on dimension) to prove the result.  
I like this geometric way of thinking, but in terms of writing out the details,
its probably simpler at this point to just use the algebraic argument suggested by cant_log.  
