# One-dimensional Lie algebra with non-trivial bracket operation

We can define a Lie algebra letting $\mathbb{R}$ be the vector space and also the field. We can then have $[x,y]=xy-yx=0$ for all $x,y$.

Is there a one-dimensional Lie algebra such that $[x,y]$ is not identically zero?

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No, this won't happen. To see this, let $\mathfrak{g}$ be a one-dimensional Lie algebra over a field $F$. Since $\mathfrak{g}$ is one-dimensional, we can write $\mathfrak{g}=Fz$ for any $z \in \mathfrak{g}-\{0\}$.
Now let $x, y \in \mathfrak{g}$. We will show that $[x,y]=0$. Fix $z \in \mathfrak{g}-\{0\}$ and write $x = az$ and $y = bz$ for some $a, b \in F$. Then $[x,y] = [az,bz] = ab[z,z] = ab*0 = 0 \in \mathfrak{g}$
The second equality follows from the bilinearity of the bracket. The third equality makes use of the alternating axiom of a Lie algebra: $[z,z] = 0 \ \ \forall z \in \mathfrak{g}$.
Thus, any one-dimensional Lie algebra is abelian. That is, the bracket is identically zero. Note, however, that the converse is not true. There are abelian Lie algebras which are not one-dimensional, such as $\mathbb{R}^n$ with the trivial bracket ($n \geq 2$).