# Non-commutative ring

• Mar 21st 2009, 02:52 AM
ZetaX
Non-commutative ring
I have the non-commutative ring R which is

$\displaystyle \left[ \begin{array}{ c c } \mathbb Z & \mathbb Q \\ 0 & \mathbb Q \end{array} \right]$

How can I show that every right ideal in the ring R is projective and finitely generated. And how can I use this to show that the right global dimension of R is 1.
• Mar 21st 2009, 08:18 AM
HallsofIvy
Perhaps it would help if you explained what
$\displaystyle \left[\begin{array}{cc}\mathbb{R} & \mathbb{Q} \\ 0 & \mathbb{Q}\end{array}\right]$ means! What does it mean for a matrix to have sets of numbers as entries?
• Mar 21st 2009, 08:55 AM
ZetaX
it mean that you have rational and integers instead of \mathbb Q and mathbb Z.
and you add them as you add integers and same goes for multiplication.
• Mar 22nd 2009, 12:17 AM
NonCommAlg
you need to be reminded that creating multiple accounts is against this forum's rules! so, as a very mild punishment, you'll only get a partial help on this question!

start with observing that the right ideals of your ring, which i'll call it $\displaystyle R$ are exactly in this form: $\displaystyle I_1=\begin{pmatrix}n \mathbb{Z} & \mathbb{Q} \\ 0 & \mathbb{Q} \end{pmatrix}, \ \ I_2 = \begin{pmatrix}n \mathbb{Z} & \mathbb{Q} \\ 0 & 0 \end{pmatrix}, \ \ I_V=\begin{pmatrix}0 & x \\ 0 & y \end{pmatrix},$ where $\displaystyle n \geq 1$ is any integer and

$\displaystyle V=\{\begin{pmatrix}x \\ y \end{pmatrix}: \ \ \begin{pmatrix}0 & x \\ 0 & y \end{pmatrix} \in I_V \}$ is any $\displaystyle \mathbb{Q}$ vector subspace of $\displaystyle \mathbb{Q} \oplus \mathbb{Q}$ and therefore it's either $\displaystyle 0, \ \mathbb{Q} \oplus \mathbb{Q}$ or isomorphic to $\displaystyle \mathbb{Q}.$ also clearly $\displaystyle I_1=nR \cong R, \ I_2=ne_{11}R,$ and $\displaystyle I_1=I_2 \oplus e_{22}R,$

which proves that $\displaystyle I_1,I_2$ are finitely generated projective submodules of $\displaystyle R.$ the case $\displaystyle I_V$ is left for you.

hence $\displaystyle R$ is a right hereditary ring and therefore every submodule of a free R module has to be projective. so $\displaystyle \text{r.gl.dim}(R) \leq 1.$ but $\displaystyle R$ is not semisimple (why?) and thus $\displaystyle \text{r.gl.dim}(R) \neq 0,$

which forces $\displaystyle \text{r.gl.dim}(R)$ to be 1.
• Mar 22nd 2009, 12:38 AM
ZetaX
I think you should give me full help, because I dont have any other account on this forum.

What about left ideals? How can I show that there are left ideals in R which are not projective and finitely generated, and how can I conclude that the left global dimension of R is >1.
• Mar 22nd 2009, 12:43 AM
mr fantastic
Quote:

Originally Posted by ZetaX
I think you should give me full help, because I dont have any other account on this forum.

I think you should spend more than just 20 minutes thinking about what NonCommAlg has posted. What he has posted should give you enough to make progress. When/if you get stuck, you can then say specifically what you can't do.

NB: NonCommAlg has given full help - he was making a joke with that comment. Unfortunately, what you seem to think is full help is getting the complete solution without having to apply any effort. What we think is full help is giving the student who is willing to think a good push in the right direction. Try to be a bit be more grateful - you are clearly in much better shape to answer the question than you were when you first posted it.
• Mar 22nd 2009, 01:00 AM
SimonM
Then why did peteryellow post the same question on a different forum?

MathLinks :: View topic - non-commutative ring
• Mar 22nd 2009, 01:20 AM
mr fantastic
Quote:

Originally Posted by SimonM
Then why did peteryellow post the same question on a different forum?

MathLinks :: View topic - non-commutative ring

There are clearly some off-topic things that need to be sorted out here. In the meantime, please keep all posts in this thread on-topic.
• Mar 22nd 2009, 01:35 AM
ZetaX
Thanks, non-commutative algebra for your help.

I have some questions.

What is e_11 and e_22?
I can see that I_1 and I_2 are finitely generated but how do you get that they are projective?
How do you know that R not semisimple implies that r.gl.dimR is different from 0.
And how can I see that R is not semisimple.

I know that R is right noetherian. Right? Can I use this to say something about that the ring R is not semisimple?
• Mar 22nd 2009, 03:12 AM
NonCommAlg
Quote:

Originally Posted by ZetaX
Thanks, non-commutative algebra for your help.

What is e_11 and e_22?

$\displaystyle e_{11}=\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix}, \ \ e_{22}=\begin{pmatrix}0 & 0 \\ 0 & 1 \end{pmatrix}.$

Quote:

I can see that I_1 and I_2 are finitely generated but how do you get that they are projective?
as i mentioned $\displaystyle I_1 \cong R$ and so it's free and we know every free R-module is projective. also it's in my solution that $\displaystyle I_2$ is a direct summand of $\displaystyle I_1,$ and we know that every direct summand of a

projective module is projective.

Quote:

How do you know that R not semisimple implies that r.gl.dimR is different from 0.
this is a very basic fact that a ring R is semisimple iff every (right) R-module is projective. thus a ring R is semisimple iff $\displaystyle \text{pd.dim}(M)=0,$ for any R-module M, which is equivalent to say that

$\displaystyle \text{r.gl.dim}(R)=0$ because $\displaystyle \text{r.gl.dim}(R)=\text{sup} \{\text{pd}(M): \ \ M \ \text{is a right R module} \}.$ (similary for left global dimension)

Quote:

And how can I see that R is not semisimple.
recall that the Jacobson radical of a semisimple ring is zero. but in our ring $\displaystyle e_{12}=\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}$ is in the Jacobson radical of R. the reason is that for any $\displaystyle z \in R,$ the element $\displaystyle 1-e_{12}z$ is easily

seen to be invertible in $\displaystyle R.$

Note: i won't answer any further questions about this problem anymore. your questions show that you even don't have the minimum background required for problems at this level.

now it's time to get some sleep! (Sleepy)