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Suppose $I$ and $J$ are ideals of a Lie Algebra L. I know that we have the fact that:

$\frac{I+J}{J} \cong \frac{I}{I\cap J}$

Prove that the ideals of $\frac{L}{I}$ - the quotient algebra of L defined by $x + I$ $x \in L$ are of the form $\frac{K}{I}$ where K is an ideal of L containing I.

user1066113
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  • Try defining a homomorphism $f\colon I\to (I+J)/J$ (there's an obvious candidate). What's its kernel? Is it surjective? – egreg May 26 '13 at 10:05
  • I know that I can do this to prove the fact that I've stated. by using kernal and image of the homomorphism. That's not what I was asking though. – user1066113 May 26 '13 at 10:21
  • Sorry, I didn't notice. However, the first part is not needed for the second one. Just use direct and inverse images by the canonical homomorphism $L\to L/I$. – egreg May 26 '13 at 10:24
  • could you explain in more detail please :) – user1066113 May 26 '13 at 10:30

1 Answers1

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If $f\colon L\to L'$ is a surjective homomorphism of Lie algebras, then, for any ideal $I$ of $L$,

$$ f^\to(K)=\{f(x):x\in K\} $$

is an ideal of $L'$.

For any ideal $K'$ of $L'$, $$ f^\gets(K')=\{x\in L:f(x)\in K'\} $$ is an ideal of $L$ (here surjectivity is not needed), containing $\ker f$.

If an ideal $K$ of $L$ contains $\ker f$, then $f^\gets(f^\to(K))=K$. Moreover, for any ideal $K'$ of $L'$, $f^\to(f^\gets(K'))=K'$. The proofs of these two statements are easy.

Therefore there is a bijection, given by $f^\to$ and $f^\gets$ between the ideals of $L$ containing $\ker f$ and the ideals of $L'$.

Apply this to the canonical homomorphism $L\to L/I$, when $I$ is an ideal of $L$.

egreg
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