1. ## 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 $a_2?$

2. ## Re: Schur's lemma

Originally Posted by ymar
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 $0$?

3. ## Re: Schur's lemma

PS: by blocks I meant $a_1$ and $a_2.$

4. ## Re: Schur's lemma

Originally Posted by ymar
PS: by blocks I meant $a_1$ and $a_2.$

5. ## Re: Schur's lemma

Originally Posted by NonCommAlg
Sorry, when I started writing, your question wasn't there and I missed it. It's $*$ in the book.

6. ## Re: Schur's lemma

Originally Posted by ymar
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?

7. ## Re: Schur's lemma

Originally Posted by NonCommAlg
$V$ 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?

8. ## Re: Schur's lemma

Originally Posted by ymar
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!

9. ## Re: Schur's lemma

Originally Posted by NonCommAlg
$V$ 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?

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.

10. ## Re: Schur's lemma

Originally Posted by ymar
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 $\overline{A}.$