Definition: 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: 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 $b \in P$
iv) If $\left[<a^{L}>,<b^{L}>\right] \subseteq P$ for $a, b \in L,$ then either $a \in P$ or $b \in P$
Why the definition can not be: An ideal $P$ of $L$ is called prime if $[x, y] \in P$ with $x, y$ elements of $L$ implies $x \in P$ or $y \in P$
If anyone say that $[x,x]=0 \in P$ and $x \notin P$ so there is no prime ideals.. I think we can exclude this case and say :An ideal $P$ of $L$ is called prime if $0\neq [x, y] \in P$ with $x, y$ elements of $L$ implies $x \in P$ or $y \in P$
Any participation would be appreciated.
Suggested definition 1: An ideal P of L is called prime if [x,y]∈P with x,y elements of L implies x∈P or y∈P comment on it form Mr.@TorstenSchoeneberg: since [x,x]=0∈P and x∉P so there is no prime ideals.
Suggested definition 2An ideal P of L is called prime if [x,y]∈P with x,y elements of L implies x∈P or y∈P comment on it form Mr.@TorstenSchoeneberg: With that proposed new definition, (0) would be a prime ideal in any algebra. Because for P:={0}, if x,y∈L with 0≠[x,y]∈P, then ... whatever, as there are no such x,y. Thanks To Mr.@TorstenSchoeneberg