Show that the ring M of n x n matrices over $\mathbb R$ has no ideals other than 0 an M. Conclude that any ring homomorphism $\phi:M\rightarrow S$ is either identically zero or is injective.
Please, help

Question

Show that the ring M of n x n matrices over $\mathbb R$ has no ideals other than 0 an M. Conclude that any ring homomorphism $\phi:M\rightarrow S$ is either identically zero or is injective.
Please, help

loser66 · Answer

@misty1212

anonymous · Answer

the second statement is a direct consequence of the first
i am not sure what you get to use the prove the first statement, but the following is true: 
if $R$ is a ring and $I$ an ideal in $R$ then  $M_n(I)$ is an ideal in $M_n(R)$

anonymous · Answer

conversely if $J$ is an ideal in $M_n(R)$ then there is some ideal $I$ in $R$ with $J=M_n(I)$

anonymous · Answer

that fact that $\mathbb{R}$is a field means the only ideal in $\mathbb{R}$ are $\mathbb{R}$ and $0$

loser66 · Answer

Thank you. Need time to digest it. :)

anonymous · Answer

yw
btw you probably do not get to use the statement i wrote, it is often written as "if $F$ is simple then $M_n(F)$ is simple

loser66 · Answer

The terminology "simple" is used for group, right? apply it to ring??

loser66 · Answer

One more question: in book it says 
Set of unit $E_{ij}$ is a basis for the linear space of matrices. 
$E_{ij}E_{kl}=\delta_{jk}E_{il}$
I don't understand it. Please, explain
I know $E_{ij}$ is a matrix with 1 in the (i,j) position and 0 elsewhere. But don't ge the product above

loser66 · Answer

Let say, in 2x2 matrices,  $E_{1,1}=\left[\begin{matrix}1&0\0&0\end{matrix}\right]$ is one of the basis.
And $E_{2,1}=\left[\begin{matrix}0&0\1&0\end{matrix}\right]$
 $E_{1,1}E_{2,1}=0$ , so, what is $\delta_{jk}$ ? What is $E_{il}$?

anonymous · Answer

i guess this proof is going to go as follows
if there is a non zero element in any matrix in the ideal, then the ideal contains all matrices

anonymous · Answer

so pick some matrix in the ideal that is not identically zero 
then since that matrix times an other matrix must also be in the ideal, so that you can put an entry anywhere you like in the product of two matrices, where one has one non zero entry

anonymous · Answer

*show, not "so"

loser66 · Answer

Down the case to 2x2 only, show me,please

anonymous · Answer

$$\delta_{ij}$$ is the dirichlet delta, a fancy way of saying $1$ if $i=j, 0$ otherwise

anonymous · Answer

suppose you are working in 2 by 2 matrices, and suppose you have an ideal that is not identically zero
that means there is a matrix in the ideal that contains a non zero element 
call it what you like

loser66 · Answer

oh, got you, so $\delta_{jk}= \delta_{1,2}=0$ and $E_{il}=E_{1,1}$ right?

anonymous · Answer

say $$\left[\begin{matrix}a&0\0&0\end{matrix}\right]$$ is in the ideal
then since the product of any matrix and this matrix is in the ideal, show that the ideal contains all matrices

anonymous · Answer

i.e you can put any number anywhere you like 
and don't forget it is closed under addition, so you really just need to show you can put an entry in any position

loser66 · Answer

In matrix, we don't have commutative property, so no matter what happen, matrix can't have any ideal which defined by left ideal = right ideal

anonymous · Answer

you are supposed to sh ow it can't have any ideal at all!

loser66 · Answer

oh, got it. hehehe...

loser66 · Answer

Thank you so much.

anonymous · Answer

yw