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Math Help - Schur's lemma

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    Schur's lemma

    I have begun reading Herstein's book, "Noncommutative Rings". I like it very much in general, great style. So if someone reading this hasn't seen it yet, I think I can recommend it.

    I'm writing this because I've having difficulty understanding a part of the text about Schur's lemma. I understand the statement and the proof since they're straightforward. What I'm having trouble with is the special case he's talking about later on. He says this:

    Let F be a field and let F_n be the ring of all n\times n matrices over F. We consider F_n as the ring of all linear transformations on the vector space V of n-tuples of elements of F. If A is a subset of F_n let \overline{A} be the subalgebra generated by A over F. Clearly V is a faithful F_n, and so, \overline{A}-module. V is, in addition, both a unitary and irreducible F_n-module.

    We say the set of matrices A is irreducible if V is an irreducible \overline{A}-module. In matrix terms this merely says that there is no invertible matrix S in F_n so that

    S^{-1}aS=\left(\begin{array}{c|c} a_1 & 0 \\ \hline * & a_2 \end{array}\right)

    for all a\in A.
    I understand everything but the last sentence. Are the blocks supposed to be square matrices? Of the same size? What is the quantificator for a_1 and a_2?
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    Re: Schur's lemma

    Quote Originally Posted by ymar View Post
    I have begun reading Herstein's book, "Noncommutative Rings". I like it very much in general, great style. So if someone reading this hasn't seen it yet, I think I can recommend it.

    I'm writing this because I've having difficulty understanding a part of the text about Schur's lemma. I understand the statement and the proof since they're straightforward. What I'm having trouble with is the special case he's talking about later on. He says this:

    I understand everything but the last sentence. Are the blocks supposed to be square matrices? Of the same size? What is the quantificator for a_1 and a_2?
    strange, shouldn't the * be 0?
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    Re: Schur's lemma

    PS: by blocks I meant a_1 and a_2.
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    Re: Schur's lemma

    Quote Originally Posted by ymar View Post
    PS: by blocks I meant a_1 and a_2.
    you didn't answer my question!
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    Re: Schur's lemma

    Quote Originally Posted by NonCommAlg View Post
    you didn't answer my question!
    Sorry, when I started writing, your question wasn't there and I missed it. It's * in the book.
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    Re: Schur's lemma

    Quote Originally Posted by ymar View Post
    Sorry, when I started writing, your question wasn't there and I missed it. It's * in the book.
    V being a reducible \overline{A}-module means V has a non-trivial \overline{A}-submodule V_1. obviously V_1 is also an F-subspace of V and thus there exists an F-subspace V_2 such that V = V_1 \oplus V_2. note that both V_1,V_2 are non-zero. now suppose that B_1, B_2 are some F-basis for V_1, V_2 respectively. then B=B_1 \cup B_2 is an F-basis for V. clearly ab_1 \in V_1 for all a \in A and b_1 \in B_1 because V_1 is an \overline{A}-module. if we look at the elements of A as linear transformations, how will the matrix of an element a \in A, with respect to B, look like?
    Last edited by NonCommAlg; December 16th 2011 at 05:08 PM.
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    Re: Schur's lemma

    Quote Originally Posted by NonCommAlg View Post
    V being a reducible \overline{A}-module means V = V_1 \oplus V_2 for some non-zero \overline{A}-modules V_1, V_2.
    Why is it so? Herstein's definition of irreducible module is this:

    M is said to be an irreducible R-module if MR\neq (0) and if the only submodules of M are (0) and M.

    When you say "reducible", do you mean not irreducible? Do you use the same definition of "irreducible" as Herstein does? I'm asking because I don't see why what you said is equivalent to V either having a trivial scalar multiplication or having a nontrivial proper submodule.

    Of course, if V=V_1\oplus V_2, then V_1 is a nontrivial proper submodule. But I don't understand why the other implication would hold. Say V contains a nontrivial proper submodule V_1. Then it's not true for general modules that V\cong V/V_1\oplus V_1, right?
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    Re: Schur's lemma

    Quote Originally Posted by ymar View Post
    Why is it so? Herstein's definition of irreducible module is this:

    M is said to be an irreducible R-module if MR\neq (0) and if the only submodules of M are (0) and M.

    When you say "reducible", do you mean not irreducible? Do you use the same definition of "irreducible" as Herstein does? I'm asking because I don't see why what you said is equivalent to V either having a trivial scalar multiplication or having a nontrivial proper submodule.

    Of course, if V=V_1\oplus V_2, then V_1 is a nontrivial proper submodule. But I don't understand why the other implication would hold. Say V contains a nontrivial proper submodule V_1. Then it's not true for general modules that V\cong V/V_1\oplus V_1, right?
    ok, you're right. i made a small mistake! i was thinking of "completely reducible" not just "reducible"! it now makes sense why Herstein used * instead of 0 at the bottom left of the matrix. anyway, i fixed the mistake and you should see my previous comment again!
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    Re: Schur's lemma

    Quote Originally Posted by NonCommAlg View Post
    V being a reducible \overline{A}-module means V has a non-trivial \overline{A}-submodule V_1. obviously V_1 is also an F-subspace of V and thus there exists an F-subspace V_2 such that V = V_1 \oplus V_2. note that both V_1,V_2 are non-zero. now suppose that B_1, B_2 are some F-basis for V_1, V_2 respectively. then B=B_1 \cup B_2 is an F-basis for V. clearly ab_1 \in V_1 for all a \in A and b_1 \in B_1 because V_1 is an \overline{A}-module. if we look at the elements of A as linear transformations, how will the matrix of an element a \in A, with respect to B, look like?
    Thank you. I'm sorry I couldn't reply earlier but I was abroad and I had limited access to the net.

    I understand now, except for one little doubt. Does the algebra \overline{A} have to have identity? Herstein's rings in general needn't have it, and I don't know about his algebras. He sometimes assumes that the reader knows what he means by things and I think this is one of these cases, unless I missed something in the text. If \overline{A} has identity, then V_1 is an F-subspace of V, because for f\in F and v\in V_1, we have

    fv=\begin{bmatrix} f & & & \\ & f & & \text{\huge{0}} \\ \text{\huge{0}} & &  \ddots & \\ & & & f \end{bmatrix}v \in V_1,

    because V_1 is an \overline{A}-module.
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    Re: Schur's lemma

    Quote Originally Posted by ymar View Post
    I understand now, except for one little doubt. Does the algebra \overline{A} have to have identity?
    yes because  \overline{A} is the algebra generated by F and A and so \overline{A} contains a copy of F, i.e. \overline{F} = \{aI : \ a \in F \}, where I is the identity matrix. thus I is the identity element of \overline{A}.
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