For a Lie algebra $L$ of dimension $n$ over a field ${\mathbb F}$ we denote by $\beta(L)$ the maximal dimension of abelian ideals of $L$. In general, $\beta(L)$ is not preserved under extensions of the ground field (see e.g. Example 2.7 in <http://homepage.univie.ac.at/dietrich.burde/papers/burde_39_max_ab.pdf>). 
Do you know any example in which $\beta(L)<\beta(L\otimes\_{\mathbb F} \bar{{\mathbb F}})=n-1$, where $\bar{\mathbb F}$ is the algebraic closure of ${\mathbb F}$? 
(In other words, is it possible that $L\otimes\_{\mathbb F} \bar{{\mathbb F}}$ contains an abelian ideal of codimension 1 and $L$ has no abelian ideal of codimension 1?)


I am mainly interested in the case where $L$ is a restricted Lie algebra over a field of characteristic $p>0$. 



Maximal dimension of abelian ideals of a Lie algebra and extensions of the ground field
There's no such example.


Since this is convenient, I denote by $L$ the Lie algebra over the algebraic closure. Let $A$ be a codimension 1 abelian ideal and let us show that some (possibly other) abelian ideal $A'$ is defined over the ground field, i.e. is a hyperplane that can be defined by a linear equation with coefficients in $K$.


Since $L/A$ is abelian, we have $[L,L]$ contained in $A$. In particular, $A$ is contained in the centralizer of $[L,L]$. In case $A$ is equal to the centralizer of $[L,L]$, this is defined over the ground field and thus we are done. So I now assume that the centralizer of $[L,L]$ is all of $L$ (so $L$ is nilpotent of step 2).


The case when $L$ is abelian is trivial. If the derived subalgebra of $L$ is 1-dimensional, then the Lie algebra law can be viewed as an alternate form. Since $A$ is a codimension 1 isotropic subspace for this form, it is easy to check that the kernel of this alternate form has codimension 2 (and is defined over the ground field) and contains $[L,L]$ because $L$ is nilpotent of step 2. Every hyperplane $A'$ containing this kernel is an abelian ideal; we can pick it to be defined over the ground field.


If the derived subalgebra of $L$ is at least 2-dimensional, there exist two linear forms $f\_1,f\_2$ on $L$ such that the alternate bilinear forms $(x,y)\mapsto b\_i(x,y)=f\_i([x,y])$, $i=1,2$ are not proportional. They can be chosen to be defined over the ground field. Let $K\_i$ be the kernel of $b\_i$. Then $K\_i$ is contained in $A$ (otherwise $b\_i$ would be zero). Besides, $K\_1$ and $K\_2$ have codimension 2 (because $A$ is an isotropic subspace for $b\_i$) and are not equal, because otherwise $b\_1$ and $b\_2$ would be alternate forms on the plane $L/K\_1$ and would thus be proportional as the set of antisymmetric matrices of size 2 is 1-dimensional. So the codimension of $K\_1+K\_2$ is at most 1. Since it's contained in $A$, we deduce that $A=K\_1+K\_2$. So $A$ is defined over the ground field and the proof is complete.


