Augmentation Ideal of Universal Enveloping Algebra

Question

I am confused with the statement that "consider the augmentation map $\epsilon_L :\mathfrak{U}(L) \rightarrow \mathbb{F}$ which is the unique algebra homomorphism induced by $\epsilon_{L}(x)=0 , \forall x \in L$ (I hope $L$ here is seen as the copy $T^1$ in $\mathfrak{U}(L)$) where $\mathfrak{U}(L)$ is the universal enveloping algebra of the Lie algebra $L$ and the kernel of $\epsilon_{L}$ is said to be the augmentation ideal. My confusion is that if $\epsilon_{L}(x)=0 , \forall x \in L$ then does not that imply $ker(\epsilon_{L})$ is just $\mathfrak{U}(L)/\mathbb{F}$?

Answer 1

You wanted to write this $\mathfrak U(L)\backslash\mathbb F$, right? Since it does not make sense to write $\mathfrak U(L)/\mathbb F$, because $\mathbb F$ not is an ideal of $\mathfrak U(L)$. The correct is $\ker \epsilon_L=\mathfrak U(L)\backslash\mathbb F^*$, this is direct by definition of $\epsilon_L$. Moreover, you can show that $\mathfrak U(L)=\mathbb F\oplus\ker\epsilon_L$ if you consider $\mathbb F$ is the subalgbera of $\mathfrak U(L)$ generated by $1.$