Infinitely many ideals on a product Lie Algebra

Question

I'm trying to prove the exercise 2.8)c) of Erdman and Wildon book on Lie Algebras. It says that if $L_{1}$ is isomorphic to $L_{2}$, and the ground field is infinity, then there are infinitely many ideals on $L_{1} \times L_{2}$.

Well, I supposed that on $L \times L$, the diagonal is an ideal, and my idea was to try to show that for every $\lambda \in \mathbb{K}$ the grounfield, $\{ (x, \lambda x) | x \in L \}$ is an ideal of $L \times L$ and so $\{ (x, \lambda x) | x \in L_{1} \}$ is an ideal of $L_{1} \times L_{2}$.

But after trying to do this, I discovered that I'm not even being able to prove that the diagonal is an ideal.

I also tried the following: If $z \in Z(L_{1})$ then $(z, \lambda \varphi(z) ) \in Z(L_{1} \times L_{2})$ for all $\lambda \in \mathbb{K}$, and the ideal will be the cyclic group generated by $(z, \lambda \varphi(z) )$. But I don't know how to proced if $Z(L_{1}) = {0}$ and all those ideals are the same.

Is there any property of lie algebras with trivial center that I can use here?

Edit: Apparently, I have to learn to read. An additional hyphotesis was that $L_{1}$ had dimension $1$, so it's abelian, so it's center is not trivial and my second attack works.

Answer 1

If $x\in L$ then $(x.L)$ is an ideal of $L\times L$

Answer 2

As regards the edit to your post (the way it was originally phrased is wrong, as you noticed yourself, e.g. $L=$ your favourite simple Lie algebra would give a counterexample):

An easier way to see this in the case that $L \neq 0$ is abelian is to notice that in any abelian $k$-Lie algebra $A$, every $k$-subspace of $A$ is an ideal; and then, that any vector space of dimension $\ge 2$ over an infinite field has infinitely many subspaces, namely, if $(e_i)_i\in I$ is a basis, e.g. the subspaces $k \cdot(e_1 + \lambda e_2)$ are distinct for all $\lambda \in k$. In particular also, in your original situation, the original idea to take any nonzero $x \in L$ and look at the spaces $A_\lambda := \{(x, \lambda x): x \in k$} gives infinitely many mutually distinct subspaces (and a fortiori by abelianness, ideals) of $L \times L$.