structure of ideals in lie algebra Definition 1: An ideal $P$ of $L$ is called prime if $[H, K] \subseteq P$ with $H, K$ ideals of $L$ implies $H \subseteq P$ or $K \subseteq P$
THEOREM 2: Let $P$ be an ideal of $L .$ Then the following conditions are equivalent:
i) $P$ is prime.
ii) If $[a, H] \subseteq P$ for $a \in L$ and an ideal $H$ of $L,$ then either $a \in P$ or $H \subseteq P$
iii) If $\left[a,<b^{L}>\right] \subseteq P$ for $a, b \in L,$ then either $a \in P$ or $b \in P$

Update 1: For any element $b$ of $L$, $<b^L>$ is the smallest ideal of $L$ containing $b$.

Definition 3: an ideal $Q$ of $L$ to be semiprime if the following condition is satisfied: If $H^{2} \subseteq Q$ for an ideal $H$ of $L,$ then $H \subseteq Q .$
LEMMA 4: An ideal Q of L is semi-prime if and only if $\operatorname{Rad}_{\mathfrak{S}}(L / Q)=(0)$
As in commutative rings 1 we define the irreducibility of ideals as follows:
An ideal $N$ of $L$ is said to be irreducible if $N=H \cap K$ with $H, K$ ideals of $L$ implies $N=H$ or $N=K$
LEMMA 5:
(1) Any prime ideal is semi-prime.
(2) Any prime ideal is irreducible.
(3) Any maximal ideal is irreducible.

Example 6: Let $L$ be a 2-dimensional non-abelian Lie algebra, that is, $L=(x, y)$
with $[x, y]=x.$

Then the ideals of $L$ are $(0),(x)$ and $L$.
$(0)$ is irreducible but neither prime nor semi-prime, for $(x)^{2}=(0) .$ Apparently (0) is not maximal.
$(x)$ is maximal but neither prime nor semi-prime, because $L^{2}=(x) . \quad$
By definition $L$ is prime but not maximal.
..

Example 7: Let $S_{1}, S_{2}$ and $S_{3}$ be finite-dimensional simple Lie algebras.

Let $L=$ $S_{1} \oplus S_{2} \oplus S_{3} .$ Then the ideals containing $S_{1}$ properly are $S_{1} \oplus S_{2}, S_{1} \oplus S_{3}$ and $L$ Therefore $S_{1}$ is semi-prime.
since
$$
\left[S_{1} \oplus S_{2}, S_{1} \oplus S_{3}\right] \subseteq S_{1}=\left(S_{1} \oplus S_{2}\right) \cap\left(S_{1} \oplus S_{3}\right)
$$
$S_{1}$ is neither prime nor irreducible. $S_{1}$ is obviously not maximal.

My Questions:-
In example 6,
Q1 What is the meaning of $L=(x,y)$. Is it a 2-dimensional vector space, I need help to illustrate this?
Q2 What is the meaning of $[x,y]=x$? If this is the case we know also that $[x,y]=x=-[y,x]=-y$ thus $x=-y$??
Q3 why the author state that $L$ itself is prime, however that always books state that the ideal is proper?
In example 7,
Q1 Why $S_1$ is semi-prime??

Update 2 : The source of this question is On prime ideals in Lie algebra
I would really appreciate your help .
 A: Q1: Yes, it's supposed to be a two-dimensional vector space spanned by two basis elements called $x,y$, i.e. $L=\{ax+by: a,b \in K\}$ (where $K$ is whatever ground field we've fixed).
Q2: It means what it says, that we define the Lie bracket via declaring that for the two said basis elements, we have $[x,y]:=x$. I claim that there is only one Lie bracket on $L$ which satisfies this. Namely, for it to be a Lie bracket, we must declare $[y,x]=-[x,y]=-x$ as well as $[x,x]=0=[y,y]$. (Note that $x,y$ are not some variables, but two fixed basis elements. Maybe it would have been better to call them $e_1$ and $e_2$.) Then by bilinearity, necessarily $[ax+by,cx+dy] = (ad-bc)\cdot x$ for all $a,b,c,d \in K$ which describes the Lie bracket on arbitrary elements of the vector space $L$. (Technically, this only proves uniqueness, I leave it to you to check that this unique candidate really satisfies all axioms of a Lie bracket.)
Q3: Well the author of this specific source you're using does not exclude the full Lie algebra in his definition of prime ideals, meaning that for him $L$ is always prime. Other authors might exclude that case, then of course $L$ is not prime. But he is consistent in the paper (note that e.g. in proposition 4 he explicitly excludes the case $P=L$ which would be redundant if it were excluded by definition).
Q1 to example 7: Well, check that it satisfies the definition for all possible $H$. It suffices to check those $H$ which are not contained in $S_1$. The author seems to assume it suffices to check only the three which properly contain $H$, which he sort of does, although formally we should also check the $S_2$ and $S_3$ and $0$, but they are even easier to check.
Added in response to comment: For Lie algebras $L_1, L_2$, the notation $L_1 \oplus L_2$ standardly means the direct sum of vector spaces turned into a Lie algebra by the Lie bracket for which $[l_1+l_2, m_1+m_2] := [l_1,m_1]+[l_2,m_2]$ for $l_i, m_i \in L_i$ (meaning that for all $a_i \in L_i$, we have $[a_1, a_2]=0$, "the direct summands are orthogonal to each other"). Further, for each simple Lie algebra $S$, one has $[S,S]=S$. Consequently e.g.
$$[S_1+S_2, S_1+S_2] \stackrel{\text{bilinearity}}= [S_1, S_1]+\underbrace{[S_1, S_2]}_{0}+\underbrace{[S_2, S_1]}_{0}+[S_2, S_2] = S_1+S_2.$$
