In a finitely generated $k$-algebra, the nilradical is $0$ iff the Jacobson radical is $0$. I was solving an exercise in Vakil's notes Foundations of Algebraic Geometry 3.6.K, and eventually proved the following statement:

Let $\mathscr{A}$ be a finitely generated $k$-algebra, where $k$ is any field. Then $\mathscr{N}(\mathscr{A}) = 0\iff \mathscr{R}(\mathscr{A}) = 0$. (All $k$-algebras are commutative here.)

Here is my proof for the hard direction:

Let $0\ne f\in \mathscr{A}$, then $D(f) = \{P \in Spec \mathscr{A}: f\not\in P\}\ne \varnothing$ since $\mathscr{N}(\mathscr{A})= 0$. By Hilbert's Nullstellensatz, if $\mathscr{A}$ is a finitely generated $k$-algebra, any non-empty basic open set $D(f)$ contains a closed point $M$. We conclude that $f\not\in M$.

Does this look correct? Can someone please point to me where this is covered in Eisenbud's Commutative Algebra? And is this statement equivalent to Nullstellensatz?
 A: This statement is much easier than the Nullstellensatz, and admits a straightforward direct proof.
The key point is that a finite dim'l algebra over $k$ is a domain iff it is a field.  
This shows that in a finite dim'l algebra over a field, an ideal is prime iff it is maximal.
This in turn shows that for finite dim'l algebras over a field, the nilradical and Jacobson radical coincide.
The Nullstellensatz is much deeper: it involves showing that a finite type algebra over $k$ that is a field is in fact finite dim'l over $k$.  In your setting of finite dim'l algs. over a field, finite dimensionality is automatic; you don't need to apply the Nullstellensatz to get it.
Added: This answered an earlier version of the question, which had a typo.  See the comments for a discussion and clarification in light of the revised question.
A: With the correction I think the proof is right, if a little slim on details. If it's just for your own use then I think it's great!
In Eisenbud look at the discussion of Jacobson rings in Section 4.5. If you want another presentation of the material there, there is a lot of overlap with this note of Emerton's.
