If $L$ is a solvable Lie algebra over an algebraically closed field of characteristic zero, then the two invariants $$ \alpha(L) = \max \{\dim (\mathfrak{a}) \mid \mathfrak{a} \text{ is an abelian subalgebra of }L\},\\ \beta(L) = \max \{\dim (\mathfrak{b}) \mid \mathfrak{b} \text{ is an abelian ideal of }L\}. $$ coincide, see Proposition $2.6$ here. For an ideal, the question is easier to decide than for a subalgebra. However in general, to decide whether or not there is an abelian subalgebra of given dimension in $L$ the algorithms mentioned use Gröbner bases in one or the other way.
On the other hand, for semisimple Lie algebras the maximal dimension of an abelian subalgebra is known, see the article "Abelian ideals in a Borel subalgebra of a complex simple Lie algebra" by R. Suter (2004) and the paper cited above.

If $L$ is a solvable Lie algebra over an algebraically closed field of characteristic zero, then the two invariants $$ \alpha(L) = \max \{\dim (\mathfrak{a}) \mid \mathfrak{a} \text{ is an abelian subalgebra of }L\},\\ \beta(L) = \max \{\dim (\mathfrak{b}) \mid \mathfrak{b} \text{ is an abelian ideal of }L\}. $$ coincide, see Proposition $2.6$ here. For an ideal, the question is easier to decide than for a subalgebra. However in general, to decide whether or not there is an abelian subalgebra of given dimension in $L$ the algorithms mentioned use Gröbner bases in one or the other way.
On the other hand, for semisimple Lie algebras the maximal dimension of an abelian subalgebra is known, see the article "Abelian ideals in a Borel subalgebra of a complex simple Lie algebra" by R. Suter (2004) and the paper cited above. There it is also proved that the maximal dimension of an abelian ideal in the standard Borel subalgebra $B$ of a simple Lie algebra $L$ coincides with the maximal dimension of a commutative subalgebra of $L$.

If $L$ is a solvable Lie algebra over an algebraically closed field of characteristic zero, then an abelian subalgebra of maximal dimension in $L$ has the same dimension as an abelian ideal of maximal dimension in $L$, see Proposition $2.6$ here. For an ideal, the question is easier to decide than for a subalgebra. However in general, to decide whether or not there is an abelian subalgebra of given dimension in $L$ the algorithms mentioned use Gröbner bases in one or the other way.
On the other hand, for semisimple Lie algebras the maximal dimension of an abelian subalgebra is known, see the article "Abelian ideals in a Borel subalgebra of a complex simple Lie algebra" by R. Suter (2004) and the paper cited above.

If $L$ is a solvable Lie algebra over an algebraically closed field of characteristic zero, then the two invariants $$ \alpha(L) = \max \{\dim (\mathfrak{a}) \mid \mathfrak{a} \text{ is an abelian subalgebra of }L\},\\ \beta(L) = \max \{\dim (\mathfrak{b}) \mid \mathfrak{b} \text{ is an abelian ideal of }L\}. $$ coincide, see Proposition $2.6$ here. For an ideal, the question is easier to decide than for a subalgebra. However in general, to decide whether or not there is an abelian subalgebra of given dimension in $L$ the algorithms mentioned use Gröbner bases in one or the other way.
On the other hand, for semisimple Lie algebras the maximal dimension of an abelian subalgebra is known, see the article "Abelian ideals in a Borel subalgebra of a complex simple Lie algebra" by R. Suter (2004) and the paper cited above.