Do linear independent sets of a central (Lie-)algebra remain linear independent in scalar extensions?

Question

Let K'/K be a field extension, L' a K'-(Lie-)algebra and L a K-(Lie-)algebra, such that $L'\subseteq \K'\otimes_K L$ (by injective embedding). Than we can review $L'=\K'\cdot L$. Consider there is a K-linear independent set $U\subseteq L$, which is NOT K'-linear independent in L'. Can we conclude that L is not central?

Answer 1

Edit: OP has significantly altered the question. Answer to new formulation below the bar.

Your first question has nothing to do with the Lie algebra structure, simplicity etc.: If $K' \vert K$ is any field extension, and $V$ is any vector space over $K$, and if by scalar extension $V_{K'}$ we mean $K' \otimes_K V$, then $\dim_K V = \dim_{K'} V_{K'}$. Namely, for any $K$-basis ${v_i: i \in I}$ of $V$ it is straightforward that ${1\otimes v_i: i \in I}$ is a $K'$-basis of $V_{K'}$. This should be one of the first things that is addressed when discussing scalar extensions in any source.

This means that in the lead-up to your second question, necessarily all $k_i =0$. I do not understand how asking about the centroid is then an "alternative formulation". If centroid means what it usually means in this theory, and if $L$ being central simple means that it's simple and $C(L)=K$, then of course you cannot conclude from anything that $C(L) \neq K$.

---

Answer to new formulation: To answer the question in the title: Linearly independet sets in vector spaces remain linearly independent in scalar extensions.

To expand a bit on the new formulation: First of all, ok, one can speak of a $K'$-Lie algebra $L'$ contained in $K' \otimes_K L$. Then it makes sense to speak of $K' \cdot L$, however as a set that is literally the same as $K' \otimes_K L$. This means however that if you're assuming $L' \color{red}{=} K' \cdot L$ you are just assuming that $L'$ $\color{red}{is}$ the entire scalar extension $K' \otimes_K L$, not just contained therein.

But regardless, again: If $U$ is any $K$-linearly independent set in any $K$-vector space $V$, and $K'\vert K$ is any field extension, then the set $U$ considered as subset of $K' \otimes_K V$ (i.e., formally the set $\lbrace 1 \otimes u: u \in U\rbrace$ which is the image of $U$ under the injection $V \hookrightarrow K'\otimes_K V$)) is linearly independent over $K'$. In a way, that is a special case of what I wrote above. So there is no such set $U$ as you assume to exist, hence one cannot draw any conclusions from that.

---

I have a feeling that you must mean something else, but I do not see what it is. Maybe write down a simple example.

Maybe you mean something like this: consider $V := \mathbb C$ as a two-dimensional $\mathbb R$-vector space. Then the vectors $1$ and $i$ are $\mathbb R$-linearly independent. However, when considered over $\mathbb C$, they are of course not. Further, you might mean this by writing $\mathbb C \cdot V$. Your question might make sense as

comparing the $\mathbb C$- vector space $V$ --- to the $\mathbb R$-vector space $Res_{\mathbb C \vert \mathbb R} V$

So I suggest what you actually want to ask about, for a field extension $K'\vert K$, is

a $K'$-Lie algebra $L'$ --- versus its scalar restriction to $K$, maybe call it $L:= Res_{K'\vert K} L'$

Notice that scalar restriction and scalar extension are not inverse to each other. In particular, in the above, the $\mathbb C$-vector space $V$ is not, and should not be called, the scalar extension of the $\mathbb R$-vector space $Res_{\mathbb C \vert \mathbb R} V$.