The universal enveloping algebra of the trivial lie algebra

Given a commutative ring $k$ and a $k$-lie algebra $g$, I need to prove that the universal enveloping algebra $\mathcal{U}\left(g\right)=k\iff g=\left\{ 0\right\}$.

One direction is very easy: If $g=\left\{ 0\right\}$ then through the construction of $\mathcal{U}\left(g\right)$ we get that it is $k$. However, I can't seem to get the converse, and i'm starting to think that it requires a trick which I just can't seem to find (like finding a specific associative algebra and a homomorphism from $g$ to it).

Any hints or ideas? Thanks!

-
Are you familiar with Poincare-Birkhoff-Witt? – Qiaochu Yuan Nov 21 '11 at 20:56
I'm familiar with the fact that i'm about to learn it, but don't think i can use it yet :) – IBS Nov 21 '11 at 21:00

(a) $U(\mathfrak{g})$ is a quotient of the tensor algebra by the relations $X \otimes Y - Y \otimes X = [X,Y]$. Think about how the ideal generated by these relations can intersect the degree one tensors. (The point is that these relations all involve a quadratic term.) If $\mathfrak g \neq 0$, is it possible for its image (thought of as the image of the degree one tensors) in the enveloping algebra to vanish?
(b) If you want to think in terms of the universal property, consider the adjoint representation of $\mathfrak g$ on itself. If $\mathfrak g$ is not abelian, this is non-trivial, so you get a non-trivial rep. of $\mathfrak g$, from which you should be able to draw a conclusion about the enveloping algebra. On the other hand, if $\mathfrak g$ is abelian, you should be able to compute the enveloping aglebra directly.