Lie algebra and enveloping algebra

I have to prove that a Lie algebra over the field $k$ is trivial if and only if the enveloping algebra $U(L)=k$.

I have an idea of proof: If $L=\{0\}$ we have that the tensor algebra $T^m=\{0\}$ for all $m \neq 0$, so we have $U(L)=k$. We have that always exists an injection of $L$ in the enveloping algebra $U(L)$, which has dimension $1$. So $\dim(L) \le 1$.

How can I finish my proof?

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Trivial here must mean $L = \{0\}$ because if $L \neq \{0\}$ then $U(L)$ is infinite dimensional. –  Jim Feb 22 '13 at 21:36
Jes, triavial means $L=\{0\}$ –  ArthurStuart Feb 22 '13 at 21:37
@all Thanks! Always learning new things here... –  rschwieb Feb 22 '13 at 22:05

To finish note that it is never the case that the identity element $1 \in U(L)$ is contained in $L$ (when we think of $L$ as a linear subspace of $U(L)$) so it cannot be the case that $\dim L = \dim U(L)$ (otherwise $L = U(L)$). Thus $\dim L < 1$. This gives $\dim L = 0$ hence $L = \{0\}$.
The reason $1$ is not in $L \subseteq U(L)$ is because of the way $U(L)$ is constructed. You start with the tensor algebra $T(L)$ where $L$ is in degree $0$. Then you quotient out by the ideal generated by elements of the form $[xy] - x \otimes y + y \otimes x$. Noteice that these elements are sums of monomials of degree $1$ and higher. So the ideal you are quotienting by is contained entirely in $T(L)^+$, the two sided ideal of $T(L)$ generated by homogeneous elements of degree $1$ or greater. So if $x \in L$ the element $1 - x$ does not go to zero in the quotient (it's not contained in $T(L)^+$). So in $U(L)$ the expression $1 = x$ is not true for any $x \in L$.
If you are more specific about what confuses you it will be easier to help. Is it that you don't understand why $\dim L < 1$ finishes your argument? Or is it that you don't understand by $1 \notin L$? –  Jim Feb 22 '13 at 21:44
And why $dim(L) <1$? –  ArthurStuart Feb 22 '13 at 22:05