# 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?

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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