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Homomorphic Image of ideal in Lie algebras

I try to prove the following theorem which given without proof here On Prime Ideals of Lie Algebras

Theorem: Let $L$ and $L^{prime}$ be Lie algebras and let $f: L rightarrow L^{prime}$ be a surjective homomorphism. Then an ideal $P$ of $L$ containing $mathrm{Ker} f$ is prime if and only if $f(P)$ is prime in $L^{prime}$.

My Proof: Let $P$ be a prime ideal of $L$, let $H$ and $K$ are two ideals of $L^{prime}$. Suppose that $[H,K] subseteq f(P)$, then $f^{-1}([H,K]) subseteq f^{-1}f(P)$, hence $Big [f^{-1}(H),f^{-1}(K)Big] subseteq P$.
But $P$ is prime, then $f^{-1}(H) subseteq P$ or $f^{-1}(K) subseteq P$. Since $f$ is surjective homomorphism, thus
$H subseteq f(P) textit{ or } K subseteq f(P)$. Therefore $f(P)$ is prime.

Is this proof true?

Is this step $f^{-1}([H,K])= Big [f^{-1}(H),f^{-1}(K)Big]$ true?

Is this step $f^{-1}f(P)=P$ true?

Is this theorem is a trivial conclusion of Isomorphism Theorems of Lie algebra and we do not need this proof?

Mathematics Asked by Hamada Al on November 12, 2021

1 Answers

One Answer

You have correctly identified the weak points of your argument:

(1) Is this step $f^{-1}([H,K])= Big [f^{-1}(H),f^{-1}(K)Big]$ true?

No, in general this is not true. If the Lie bracket on $L$ was zero, then the right hand side would be $0$, but the left hand side would contain ker$(f)$.

However for your purpose all you need is: $$f^{-1}([H,K])supseteq Big [f^{-1}(H),f^{-1}(K)Big],$$ which is true. To see this just apply $f$ to an element of the right hand side, and clearly you will land in $[H,K]$.

(2) Is this step $f^{-1}f(P)=P$ true?

Yes, but only because of one specific bit of information you were given, which you should mention to justify the statement. Namely ker$(f)subseteq P$. Thus if $xin f^{-1}f(P)$ then $f(x)=f(p)$ for some $p in P$, and $$x=(x-p)+p,$$ with $pin P$ and $x-pin$ker$(f)subseteq P$.

Answered by tkf on November 12, 2021

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