Question

Show that the ring M of n x n matrices over Real has no ideals other than 0 and M
Please, help

Answer 1

I have a guidance but don't get. Let me post

Answer 2

Let's suppose that \(I\neq0\) is an ideal other than \(M\).

Answer 3

Let do the smallest case first as guidance said to easily follow, please

Answer 4

Now let's consider the basis \(\{E_{i,j}\}\) as given in the hint ---this is the cannonical basis of \(\mathbb{R}^{n\times n}\). Which means that given any matrix A we can find \(\alpha_{i,j}\) such that
\[ A=\sum_{i,j}\alpha_{i,j}E_{i,j}. \]
Actually \(\alpha_{i,j}\) are the entries of A. All right this far?

Answer 5

yes

Answer 6

the most prosaic way to think about this is if your matrix has one non zero element, then you can multiply it by something to put a non zero element in any entry
since the ideal swallows up the product under multiplication, the idea must the the entire matrix ring

Answer 7

for example if \(A =\left[\begin{matrix} 1&2\\3&1\end{matrix}\right]\), then
\(A= 1\left[\begin{matrix} 1&0\\0&0\end{matrix}\right]=2\left[\begin{matrix} 0&1\\0&0\end{matrix}\right]+3\left[\begin{matrix} 0&0\\1&0\end{matrix}\right]+1\left[\begin{matrix} 0&0\\0&1\end{matrix}\right]\), right?

Answer 8

the rest is just matrix multiplication notation which is good to struggle through but not fascinating

Answer 9

Now there is a problem: the hint says to prove that
\[\large E_{i,j}E_{k,l}=\delta_{j,k}E_{i,l} \]
This is not true!!

Answer 10

*+ before 2, not =

Answer 11

That is true, it leads to 0 is ideal

Answer 12

\[\large E_{2,2}\cdot E_{3,2}=
\begin{pmatrix} 0 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0
\end{pmatrix}\cdot
\begin{pmatrix}
0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 1 & 0
\end{pmatrix} =0\]

Answer 13

because the basis \(E_{ij}\) is independent, hence \(E_{ij}E_{kl}=0\)

Answer 14

so that either \(\delta_{ij}=0\) or \(E_{kl}=0\), but \(E_{kl}\neq 0\), hence only \(\delta_{ij}=0\)

Answer 15

and \(\delta_{ij}\) are entry of ideal, and it =0--> ideal =0, right?

Answer 16

spose \[
M=
\left[ {\begin{array}{cc}
a & 0 \\
0 & 0 \\
\end{array} } \right]
\] in in the ideal
then since \[

\left[ {\begin{array}{cc}
a& 0 \\
0& 0 \\
\end{array} } \right]\times

\left[ {\begin{array}{cc}
0& \frac{b}{a} \\
0 & 0 \\
\end{array} } \right]

\] is in the ideal, so is
\[

\left[ {\begin{array}{cc}
0 & b \\
0 & 0 \\
\end{array} } \right]
\]

Answer 17

and therefore so is every damned matrix

Answer 18

once you got one non zero entry, you got them all

Answer 19

@misty1212 M is a left ideal but not a right one--> hence M is not an ideal

Answer 20

oh my M is a matrix not an ideal
i think i am not really getting the idea across though

Answer 21

If \(I\subset M\) is an ideal (left and right) and if X is a matrix in \(I\) with some non-zero entries (say \(X_{j,k}=\beta\neq0\)) then
\[\large E_{j,k}X(1/\beta)E_{j,k}=E_{j,k}\in I \]

Answer 22

then?

Answer 23

And using what misty did. u can put 1's in row j and column k.

Answer 24

After awhile, I confused though. hehehe... what is an ideal? and what is the relationship between left/ right ideal with ideal?

Answer 25

\[\large \begin{pmatrix} a & 0\\ 0 & 0
\end{pmatrix}\cdot\begin{pmatrix} 0 & 1/a\\ 0 & 0
\end{pmatrix}=\begin{pmatrix} 0 & 1 \\ 0 & 0
\end{pmatrix} \]
\[\large \begin{pmatrix} 1/a & 0\\ 0 & 0
\end{pmatrix}\cdot\begin{pmatrix} a & 0\\ 0 & 0
\end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 0
\end{pmatrix} \]
\[\large \begin{pmatrix} 0 & 0\\ 1/a & 0
\end{pmatrix}\cdot\begin{pmatrix} a & 0\\ 0 & 0
\end{pmatrix}=\begin{pmatrix} 0 & 0 \\ 1 & 0
\end{pmatrix} \]

Answer 26

all these r in \(I\)

Answer 27

an ideal is both left and right ideal.

Answer 28

and they not need equal, right? just being left and right ideal at the same time,right?

Answer 29

yes

Answer 30

Thank you.

Answer 31

I can't figure out a way to put \(E_{2,2}\) in \(I\), though.

Answer 32

:(

Answer 33

sorry. gotta go.

Answer 34

Thanks for response. I will let my argument here, please check if you have time.