# Thread: Pre-image of an ideal

1. ## Pre-image of an ideal

Essentially, I believe my question boils down to this: does the correspondence theorem at least partially hold for Lie algebras (or, indeed, any algebra)?

$\displaystyle K, L$ Lie algebras, $\displaystyle K \leq L$. Let $\displaystyle \phi$ be some homomorphism of Lie algebras and let $\displaystyle K\phi \neq 0$ be an ideal of $\displaystyle L\phi$, $\displaystyle K\phi \unlhd L\phi$. Then is it true that $\displaystyle K \unlhd L$?

This seems to be used in a proof in my lecture notes, but I can't work out why it holds (if, indeed, it does)!

2. Originally Posted by Swlabr
Essentially, I believe my question boils down to this: does the correspondence theorem at least partially hold for Lie algebras (or, indeed, any algebra)?

$\displaystyle K, L$ Lie algebras, $\displaystyle K \leq L$. Let $\displaystyle \phi$ be some homomorphism of Lie algebras and let $\displaystyle K\phi$ be an ideal of $\displaystyle L\phi$, $\displaystyle K\phi \unlhd L\phi$. Then is it true that $\displaystyle K \unlhd L$?

This seems to be used in a proof in my lecture notes, but I can't work out why it holds (if, indeed, it does)!

of course not. a trivial counter-example is when $\displaystyle \phi = 0$ and K any subalgebra of L. there must be some more assumptions that you haven't given us!

for example it's true if $\displaystyle \phi$ is injective.

3. Originally Posted by NonCommAlg
of course not. a trivial counter-example is when $\displaystyle \phi = 0$ and K any subalgebra of L. there must be some more assumptions that you haven't given us!

for example it'd be true if $\displaystyle \phi$ was injective.

How about I say "$\displaystyle \phi$ a non - trivial homomorphism"?...

I'll go and have a rummage through my notes to see if I can find it again.

EDIT: It is stipulated that $\displaystyle K \phi \neq 0$. I'll edit that in my first post - thanks for pointing that out.

4. Originally Posted by Swlabr
How about I say "$\displaystyle \phi$ a non - trivial homomorphism"?...

I'll go and have a rummage through my notes to see if I can find it again.
in general, as i said, if $\displaystyle \phi$ is injective, then the result is true in any algebra (very easy to see!).

5. Originally Posted by NonCommAlg
in general, as i said, if $\displaystyle \phi$ is injective, then the result is true in any algebra (very easy to see!).
Can you please expand you "very easy to see"? Thanks.

6. Originally Posted by Swlabr

Can you please expand you "very easy to see"? Thanks.
we can actually prove a more general statement: if $\displaystyle \ker \phi \subseteq K,$ and $\displaystyle \phi(K)$ is an ideal of $\displaystyle \phi(L),$ then $\displaystyle K$ will be an ideal of $\displaystyle L.$ to prove this, let $\displaystyle r \in L, \ a \in K.$ then:

$\displaystyle \phi(ra)=\phi(r) \phi(a) \in \phi(L) \phi(K) \subseteq \phi(K),$ since $\displaystyle \phi(K)$ is an ideal of $\displaystyle \phi(L).$ thus $\displaystyle \phi(ra)=\phi(b),$ for some $\displaystyle b \in K$ and so $\displaystyle ra-b \in \ker \phi \subseteq K.$ hence $\displaystyle ra \in K.$ Q.E.D.

7. Originally Posted by NonCommAlg
we can actually prove a more general statement: if $\displaystyle \ker \phi \subseteq K,$ and $\displaystyle \phi(K)$ is an ideal of $\displaystyle \phi(L),$ then $\displaystyle K$ will be an ideal of $\displaystyle L.$ to prove this, let $\displaystyle r \in L, \ a \in K.$ then:

$\displaystyle \phi(ra)=\phi(r) \phi(a) \in \phi(L) \phi(K) \subseteq \phi(K),$ since $\displaystyle \phi(K)$ is an ideal of $\displaystyle \phi(L).$ thus $\displaystyle \phi(ra)=\phi(b),$ for some $\displaystyle b \in K$ and so $\displaystyle ra-b \in \ker \phi \subseteq K.$ hence $\displaystyle ra \in K.$ Q.E.D.

How does the $\displaystyle (r\phi) (a\phi) = b \phi \Rightarrow (r\phi) (a\phi) - b \phi \in ker \phi$ transfer over to the Lie bracket? Is it just the same because it's a vector space:

$\displaystyle [r\phi,a\phi] = b \phi \Rightarrow [r\phi, a\phi] - b \phi \in ker \phi \leq K$ and so on.

Also, what if $\displaystyle ker \phi \not\leq K$?

8. Originally Posted by Swlabr
How does the $\displaystyle (r\phi) (a\phi) = b \phi \Rightarrow (r\phi) (a\phi) - b \phi \in ker \phi$ transfer over to the Lie bracket? Is it just the same because it's a vector space:

$\displaystyle [r\phi,a\phi] = b \phi \Rightarrow [r\phi, a\phi] - b \phi \in ker \phi \leq K$ and so on.
what i proved is true in all algebras. in Lie algebras we'll have $\displaystyle \phi(b)=[\phi(r), \phi(a)]=\phi([r,a]).$ thus $\displaystyle \phi([r,a]-b)=0,$ i.e. $\displaystyle [r,a] - b \in \ker \phi \subseteq K$ and so $\displaystyle [r,a] \in K.$

Also, what if $\displaystyle ker \phi \not\leq K$?
the result won't necessarily be true anymore.

9. Originally Posted by NonCommAlg
...the result won't necessarily be true anymore.

Hmm. I suppose I should probably stop asking questions in abstraction and just state the Lemma...

"For every finite dimensional Lie algebra $\displaystyle L$ the quotient algebra $\displaystyle L/rad(L)$ is semisimple."

That is to say, for $\displaystyle rad(L)$ the ideal containing every soluble ideal of $\displaystyle L$ then the quotient contains no proper soluble ideals.

Clearly, if $\displaystyle K/rad(L)$ is soluble then $\displaystyle K$ is soluble in $\displaystyle L$, and so it is sufficient to prove that $\displaystyle K \unlhd L$. Your proof requires $\displaystyle rad(L) \leq K$, and I can't quite see why that is. My thought is that there exists some $\displaystyle K'$ such that $\displaystyle K \equiv K'$ mod $\displaystyle rad(L)$ with $\displaystyle rad(L) \leq K'$ but I'm not sure if that is true. (unless...would the canonical map not give the result? $\displaystyle x \mapsto x+ rad(L)$ and taking $\displaystyle \phi ^{-1}$ to be $\displaystyle x+ rad(L) \mapsto x$? Does the set of coset representative of $\displaystyle K$ form an ideal of $\displaystyle L$?)

Thanks for giving me/this problem so much time, by the way!

10. Originally Posted by Swlabr
Hmm. I suppose I should probably stop asking questions in abstraction and just state the Lemma...

"For every finite dimensional Lie algebra $\displaystyle L$ the quotient algebra $\displaystyle L/rad(L)$ is semisimple."
to prove this lemma you don't need the discussion we had:

an ideal of $\displaystyle L/rad(L)$ is in the form $\displaystyle K/rad(L),$ where $\displaystyle K$ is an ideal of $\displaystyle L$ containing $\displaystyle rad(L).$ suppose $\displaystyle K/rad(L)$ is solvable. then, since $\displaystyle rad(L)$ is solvable, $\displaystyle K$ is solvable too. thus $\displaystyle K \subseteq rad(L)$

and hence $\displaystyle K/rad(L) = 0.$ therefore $\displaystyle rad(L/rad(L))=0,$ i.e. $\displaystyle L/rad(L)$ is semisimple.

11. Originally Posted by NonCommAlg
an ideal of $\displaystyle L/rad(L)$ is in the form $\displaystyle K/rad(L),$ where $\displaystyle K$ is an ideal of $\displaystyle L$ containing $\displaystyle rad(L)$.

It's just that one sentence I'm not sure about though - why must $\displaystyle K$ be an ideal of $\displaystyle L$ containing $\displaystyle rad(L)$ if $\displaystyle K/rad(L) \unlhd L/rad(L)$? The above discussion reduced my query to "why must $\displaystyle rad(L) \leq K$ if $\displaystyle K/rad(L) \unlhd L/rad(L)$?", and I'm still not 100% sure about this. I mean, I know it holds I just can't quite "see" it...

12. Originally Posted by Swlabr

It's just that one sentence I'm not sure about though - why must $\displaystyle K$ be an ideal of $\displaystyle L$ containing $\displaystyle rad(L)$ if $\displaystyle K/rad(L) \unlhd L/rad(L)$? The above discussion reduced my query to "why must $\displaystyle rad(L) \leq K$ if $\displaystyle K/rad(L) \unlhd L/rad(L)$?", and I'm still not 100% sure about this. I mean, I know it holds I just can't quite "see" it...
this is just the "correspondence theorem" that you're usually taught right after introducing quotient (Lie) algebras and you should find it in your notes:

suppose L is an algebra and I is an ideal of L. define the map $\displaystyle f: L \to L/I$ by $\displaystyle f(a)=a+I.$ let $\displaystyle J$ be an ideal of $\displaystyle L/I$ and let $\displaystyle K=f^{-1}(J)=\{a \in L: \ f(a) \in J \}.$

1) $\displaystyle K$ is an ideal of $\displaystyle L$ and $\displaystyle I \subseteq K.$

proof: if $\displaystyle a,b \in K$ and $\displaystyle r \in L,$ then $\displaystyle f(ra+b)=f(r)f(a) + f(b) \in J.$ thus $\displaystyle ra+b \in K$ and so $\displaystyle K$ is an ideal of $\displaystyle L.$ also if $\displaystyle s \in I,$ then $\displaystyle f(s)=s+I=0 \in J.$ thus $\displaystyle s \in K,$ i.e. $\displaystyle I \subseteq K.$

2) $\displaystyle K/I = J.$

proof: if $\displaystyle a \in K,$ then $\displaystyle a+I=f(a) \in J,$ by definition of $\displaystyle K.$ thus $\displaystyle K/I \subseteq J.$ conversely, if $\displaystyle f(r)=r+I \in J,$ then $\displaystyle r \in K$ and hence $\displaystyle r+I \in K/I,$ i.e. $\displaystyle J \subseteq K/I.$

it's also clear that if $\displaystyle I_1 \subseteq I_2$ are ideals of $\displaystyle L,$ then $\displaystyle I_2/I_1$ is an ideal of $\displaystyle L/I_1.$ thus we proved that ideals of $\displaystyle L/I$ are exactly in the form $\displaystyle I_2/I_1,$ where $\displaystyle I_1 \subseteq I_2$ are ideals of $\displaystyle L.$

13. Thanks - we didn't cover that though (although I've come across the theorem in groups).