# Field as a trivial module over its algebra

Let $$k$$ be a field, $$\mathfrak{g}$$ be a lie algebra over $$k$$ and $$\mathcal{A} = \mathcal{U}(\mathfrak{g})$$ be the universal enveloping algebra of $$\mathfrak{g}$$. When dealing with the homology of $$\mathfrak{g}$$ we see $$k$$ as a trivial $$\mathcal{A}$$-module (see Lie algebra homology and Chevalley-Eilenberg chain complex). Is the action simply $$w \cdot \lambda = 0$$ for all $$w \in \mathcal{A}, \lambda \in k$$? Since the ring $$\mathcal{A}$$ is unital that doesn't seem to be the case, otherwise $$k$$ would be trivial: $$1 = 1 \cdot 1 = 0$$.

Recall that $$\mathcal{A}$$ is generated (as a $$k$$-algebra) by the elements of $$\mathfrak{g}$$. An $$\mathcal{A}$$-module structure on $$k$$ is equivalent data to a ring homomorphism $$\mathcal{A} \to \operatorname{End}_{\mathbb{Z}}(k)$$. The "trivial action" is actually a $$k$$-algebra homomorphism $$\mathcal{A} \to \operatorname{End}_k(k)$$ (note that $$\operatorname{End}_k(k) \subseteq \operatorname{End}_{\mathbb{Z}}(k)$$), determined by sending each element of $$\mathfrak{g}$$ to $$0$$ (it's important to check that this is well-defined, i.e. $$[g_1, g_2] - g_1 g_2 + g_2 g_1 \mapsto 0$$ for all $$g_1, g_2 \in \mathfrak{g}$$). Since this is a $$k$$-algebra homomorphism, we have (for example) $$1 \cdot 1 = \operatorname{id}_k(1) = 1$$, giving us no problems.
Side note: there is no such thing as a field where $$1 = 0$$! Indeed part of the definition of a field is that $$1 \neq 0$$.
The other answer is correct, but I'd like to add the following view: $$\mathcal U(\mathfrak g)$$ has a natural filtration $$\mathcal U(\mathfrak g)_0 \subsetneq \mathcal U(\mathfrak g)_1 \subsetneq \mathcal U(\mathfrak g)_2 \subsetneq ...$$ (inherited from the tensor algebra), where the zero-th step is just the ground field $$k$$.
Now it's clear from the filtration properties that $$\mathcal U(\mathfrak g)_0$$ is a subalgebra, but more importantly for us, it's also a quotient. Namely, the set $$I:= \{0\} \cup \{u\in \mathcal U(\mathfrak g): u \notin \mathcal U(\mathfrak g)_0 \}$$ is an ideal in $$\mathcal U(\mathfrak g)$$. This ideal is the image in $$\mathcal U(\mathfrak g)$$ of the ideal $$\hat I := \color{red}{0}\oplus \mathfrak g \oplus (\mathfrak g \otimes \mathfrak g) \oplus (\mathfrak g \otimes \mathfrak g \otimes \mathfrak g) \oplus ...$$ in the tensor algebra $$T(\mathfrak g) := \color{blue}{k}\oplus \mathfrak g \oplus (\mathfrak g \otimes \mathfrak g) \oplus (\mathfrak g \otimes \mathfrak g \otimes \mathfrak g) \oplus ...$$ We get a projection $$U(\mathfrak g) \twoheadrightarrow U(\mathfrak g)/I \simeq U(\mathfrak g)_0 \simeq k$$ with trivial $$\mathfrak g$$-action.
As an instructive example, consider an abelian Lie algebra $$\mathfrak g$$ with $$k$$-basis $$x_1, .., x_d$$. Then $$U(\mathfrak g) \simeq k[x_1, ..., x_d]$$, the polynomial ring in $$d$$ variables. The above projection is just the map sending such a polynomial to its constant term,
$$\sum c_{\alpha_1, ..., \alpha_d} \,x_1^{\alpha_1} \cdots x_d^{\alpha_d} \;\mapsto \; c_{(0,...,0)}$$ If we let $$U(\mathfrak g)$$ act on $$k$$ via this projection, indeed $$x_i \cdot \lambda=0$$ for all $$x_i \in \mathfrak g, \lambda \in k$$, but still $$1 \cdot \lambda = \lambda$$. (Note in general $$1 \in U(\mathfrak g)_0$$ but $$\mathfrak g \setminus \{0\} =U(\mathfrak g)_1\setminus U(\mathfrak g)_0$$. The unit $$1$$ is not in the Lie algebra, it's in a different filtration step.)