[proofplan] We verify the Lie algebra axioms directly. Bilinearity of the commutator follows from bilinearity of the associative algebra multiplication. Alternation is immediate from the definition, and the Jacobi identity follows by expanding the three commutators and cancelling all degree-three monomials using associativity. [/proofplan]

[step:Verify bilinearity from bilinearity of multiplication] Let $x_1,x_2,y \in A$ and let $\lambda \in F$. Since multiplication in the associative $F$-algebra $A$ is $F$-bilinear, we have \begin{align*} [x_1 + \lambda x_2,y] &= (x_1 + \lambda x_2)y - y(x_1 + \lambda x_2) \\ &= x_1y + \lambda x_2y - yx_1 - \lambda yx_2 \\ &= (x_1y - yx_1) + \lambda(x_2y - yx_2) \\ &= [x_1,y] + \lambda[x_2,y]. \end{align*} Thus the bracket is $F$-linear in its first argument. Let $x,y_1,y_2 \in A$ and let $\lambda \in F$. Again using $F$-bilinearity of multiplication, \begin{align*} [x,y_1 + \lambda y_2] &= x(y_1 + \lambda y_2) - (y_1 + \lambda y_2)x \\ &= xy_1 + \lambda xy_2 - y_1x - \lambda y_2x \\ &= (xy_1 - y_1x) + \lambda(xy_2 - y_2x) \\ &= [x,y_1] + \lambda[x,y_2]. \end{align*} Therefore $[\cdot,\cdot]: A \times A \to A$ is $F$-bilinear. [/step]

[step:Prove alternation by evaluating the bracket on equal arguments] Let $x \in A$. By the definition of the commutator bracket, \begin{align*} [x,x] = xx - xx = 0. \end{align*} Hence the bracket is alternating. [/step]

[step:Expand the Jacobi expression and cancel every associative monomial]Let $x,y,z \in A$. We compute the Jacobi expression \begin{align*} J(x,y,z) := [x,[y,z]] + [y,[z,x]] + [z,[x,y]]. \end{align*} First, \begin{align*} [x,[y,z]] &= x(yz - zy) - (yz - zy)x \\ &= x(yz) - x(zy) - (yz)x + (zy)x. \end{align*} Using associativity in $A$, this becomes \begin{align*} [x,[y,z]] &= xyz - xzy - yzx + zyx. \end{align*} The other two cyclic terms are \begin{align*} [y,[z,x]] &= y(zx - xz) - (zx - xz)y \\ &= yzx - yxz - zxy + xzy, \end{align*} and \begin{align*} [z,[x,y]] &= z(xy - yx) - (xy - yx)z \\ &= zxy - zyx - xyz + yxz. \end{align*} Adding the three displayed identities gives \begin{align*} J(x,y,z) &= (xyz - xzy - yzx + zyx) \\ &\quad + (yzx - yxz - zxy + xzy) \\ &\quad + (zxy - zyx - xyz + yxz) \\ &= 0. \end{align*} Each monomial $xyz$, $xzy$, $yzx$, $yxz$, $zxy$, and $zyx$ occurs once with coefficient $1$ and once with coefficient $-1$. Therefore \begin{align*} [x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0 \end{align*} for all $x,y,z \in A$.[/step]

[guided]Let $x,y,z \in A$. The Jacobi identity is the statement that the map \begin{align*} J: A \times A \times A &\to A \\ (x,y,z) &\mapsto [x,[y,z]] + [y,[z,x]] + [z,[x,y]] \end{align*} is identically zero. We prove this by expanding every bracket until only products of three elements remain. First expand the inner bracket $[y,z] = yz - zy$ and then the outer commutator with $x$: \begin{align*} [x,[y,z]] &= x[y,z] - [y,z]x \\ &= x(yz - zy) - (yz - zy)x \\ &= x(yz) - x(zy) - (yz)x + (zy)x. \end{align*} Associativity is the algebraic hypothesis used here: it allows us to remove parentheses from products of three elements. Thus \begin{align*} [x,[y,z]] &= xyz - xzy - yzx + zyx. \end{align*} Now expand the second cyclic term. Since $[z,x] = zx - xz$, \begin{align*} [y,[z,x]] &= y[z,x] - [z,x]y \\ &= y(zx - xz) - (zx - xz)y \\ &= y(zx) - y(xz) - (zx)y + (xz)y \\ &= yzx - yxz - zxy + xzy. \end{align*} Again, the last equality uses associativity of multiplication in $A$. Finally, since $[x,y] = xy - yx$, \begin{align*} [z,[x,y]] &= z[x,y] - [x,y]z \\ &= z(xy - yx) - (xy - yx)z \\ &= z(xy) - z(yx) - (xy)z + (yx)z \\ &= zxy - zyx - xyz + yxz. \end{align*} We now add the three expanded expressions: \begin{align*} J(x,y,z) &= [x,[y,z]] + [y,[z,x]] + [z,[x,y]] \\ &= (xyz - xzy - yzx + zyx) \\ &\quad + (yzx - yxz - zxy + xzy) \\ &\quad + (zxy - zyx - xyz + yxz). \end{align*} The terms cancel in opposite-sign pairs: \begin{align*} xyz - xyz &= 0, & -xzy + xzy &= 0, \\ -yzx + yzx &= 0, & -yxz + yxz &= 0, \\ -zxy + zxy &= 0, & zyx - zyx &= 0. \end{align*} Therefore \begin{align*} J(x,y,z) = 0. \end{align*} Since $x,y,z \in A$ were arbitrary, the Jacobi identity holds for all elements of $A$.[/guided]

[step:Conclude that the commutator bracket defines a Lie algebra structure] The bracket $[\cdot,\cdot]: A \times A \to A$ is $F$-bilinear, alternating, and satisfies the Jacobi identity. These are precisely the Lie algebra axioms for the [vector space](/page/Vector%20Space) $A$ over $F$ equipped with the bracket $[\cdot,\cdot]$. Hence $A$ is a Lie algebra over $F$ under the commutator bracket. [/step]