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Math Help - Pre-image of an ideal

  1. #1
    MHF Contributor Swlabr's Avatar
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    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)?

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

    Thanks in advance.
    Last edited by Swlabr; May 15th 2009 at 12:11 AM.
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  2. #2
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    Quote Originally Posted by Swlabr View Post
    Essentially, I believe my question boils down to this: does the correspondence theorem at least partially hold for Lie algebras (or, indeed, any algebra)?

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

    Thanks in advance.
    of course not. a trivial counter-example is when \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 \phi is injective.
    Last edited by NonCommAlg; May 15th 2009 at 12:04 AM.
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    of course not. a trivial counter-example is when \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 \phi was injective.

    How about I say " \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 K \phi \neq 0. I'll edit that in my first post - thanks for pointing that out.
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    Quote Originally Posted by Swlabr View Post
    How about I say " \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 \phi is injective, then the result is true in any algebra (very easy to see!).
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    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    in general, as i said, if \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.
    Last edited by Swlabr; May 15th 2009 at 04:04 AM.
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    Quote Originally Posted by Swlabr View Post

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

    \phi(ra)=\phi(r) \phi(a) \in \phi(L) \phi(K) \subseteq \phi(K), since \phi(K) is an ideal of \phi(L). thus \phi(ra)=\phi(b), for some b \in K and so ra-b \in \ker \phi \subseteq K. hence ra \in K. Q.E.D.
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  7. #7
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    we can actually prove a more general statement: if \ker \phi \subseteq K, and \phi(K) is an ideal of \phi(L), then K will be an ideal of L. to prove this, let r \in L, \ a \in K. then:

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

    How does the (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:

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

    Also, what if ker \phi \not\leq K?
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    Quote Originally Posted by Swlabr View Post
    How does the (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:

    [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 \phi(b)=[\phi(r), \phi(a)]=\phi([r,a]). thus \phi([r,a]-b)=0, i.e. [r,a] - b \in \ker \phi \subseteq K and so [r,a] \in K.


    Also, what if ker \phi \not\leq K?
    the result won't necessarily be true anymore.
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  9. #9
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    ...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 L the quotient algebra L/rad(L) is semisimple."

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

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

    Thanks for giving me/this problem so much time, by the way!
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    Quote Originally Posted by Swlabr View Post
    Hmm. I suppose I should probably stop asking questions in abstraction and just state the Lemma...

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

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

    and hence K/rad(L) = 0. therefore rad(L/rad(L))=0, i.e. L/rad(L) is semisimple.
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  11. #11
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    an ideal of L/rad(L) is in the form K/rad(L), where K is an ideal of L containing rad(L).

    It's just that one sentence I'm not sure about though - why must K be an ideal of L containing rad(L) if K/rad(L) \unlhd L/rad(L)? The above discussion reduced my query to "why must rad(L) \leq K if 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...
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  12. #12
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    Quote Originally Posted by Swlabr View Post

    It's just that one sentence I'm not sure about though - why must K be an ideal of L containing rad(L) if K/rad(L) \unlhd L/rad(L)? The above discussion reduced my query to "why must rad(L) \leq K if 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 f: L \to L/I by f(a)=a+I. let J be an ideal of L/I and let K=f^{-1}(J)=\{a \in L: \ f(a) \in J \}.

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

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

    2) K/I = J.

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

    it's also clear that if I_1 \subseteq I_2 are ideals of L, then I_2/I_1 is an ideal of L/I_1. thus we proved that ideals of L/I are exactly in the form I_2/I_1, where I_1 \subseteq I_2 are ideals of L.
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  13. #13
    MHF Contributor Swlabr's Avatar
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    Thanks - we didn't cover that though (although I've come across the theorem in groups).
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