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Let $L=Fx+Fy$ be the non-abelian Lie algebra with $[x,y]=x$. Let $(\mathfrak{U}(L),i)$ be the universal enveloping algebra of $L$. By definition, $\mathfrak{U}(L)=T(L)/J$ where $T(L)$ is the tensor algebra over $L$ and $J$ is the ideal of $T(L)$ generated by $u\otimes v-v\otimes u-[u,v]$ for $u,v\in L$ and the map $i:L\rightarrow \mathfrak{U}(L)$ satisfies some universal property.

Q. Prove directly that the map $i:L\rightarrow \mathfrak{U}(L)$ is injective i.e. $L\cap J=0$.

What is $i$ here? Well, it is the following composition: $$L\rightarrow T(L) \rightarrow \frac{T(L)}{J}=\mathfrak{U}(L).$$ So perhaps, the author wants that for two dimensional non-abelian Lie algebra this composition is injective, and directly means perhaps without using PBW theorem. I didn't get any direction towards solving this problem. Any hint?

Beginner
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1 Answers1

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It suffices to find an $L$-module $M$ on which $L$ acts faithfully, that is $l\cdot m=0$ for all $m\in M$ entails $l=0$. A module for the Lie algebra $L$ is the same as a module for the ring $\mathfrak{U}(L)$. If $L$ acts faithfully on $M$, then $i(l)$ acts non-trivially on $M$ for nonzero $l\in L$ and so $i(l)\ne0$.

So now all one has to do is find linearly indepdendent matrices $X$ and $Y$ with $XY-YX=X$. I think that's possible....

Angina Seng
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