= Solution
{c}
Choose a <basis> $x,y$ of the two-dimensional <abelian Lie algebra> $\mathfrak a$, and lifts with the same names. Every <alternating bilinear form> is determined by $c=\alpha(x,y)$, and every such form is a <Lie algebra two-cocycle> because every <Lie bracket> in $\mathfrak a$ is zero. Thus every <central extension of a Lie algebra> has a <basis> $x,y,z$ with
$$
[x,y]=cz,\qquad [x,z]=[y,z]=0.
$$
If $c=0$, all <Lie brackets> vanish, so this is the <direct sum of Lie algebras> $\mathfrak a\oplus kz\cong k^3$. If $c\ne0$, replace the central <basis vector> by $z'=cz$. Then $[x,y]=z'$, which is the <Heisenberg Lie algebra> $\mathfrak h_3(k)$. This change of <basis> shows that every nonzero $c$ gives the same <Lie algebra isomorphism> class.
These two <Lie algebras> cannot be isomorphic: their <derived algebras> have <dimensions> zero and one, respectively, and a <Lie algebra isomorphism> preserves the <derived algebra>. Moreover, the <Heisenberg Lie algebra> cannot be a <direct sum of Lie algebras> $kz\oplus\mathfrak a$, since that <direct sum of Lie algebras> would have zero <Lie bracket>. Therefore
$$
\boxed{\widetilde{\mathfrak a}\cong k^3\quad\text{or}\quad\mathfrak h_3(k).}
$$
Here “up to isomorphism” means the <Lie algebra isomorphism> class of the middle <Lie algebra>, allowing a change of the identifications at the ends of the <short exact sequence>. If instead <equivalence of central extensions of a Lie algebra> must induce the identity on both $kz$ and $\mathfrak a$, the parameter $c$ cannot change: the formula in part (a) has $t([x,y])=0$. That stricter notion has one class for each $c\in k$, rather than two. This distinction explains the convention required by the question.
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