Herstein's matrix proof

• Nov 15th 2011, 04:36 PM
ThatPinkSock
Herstein's matrix proof
Herstein's abstract algebra: Chapter 4 section 3 number 19

If R = {( a b ) | a, b, c real} and I = {( 0 b ) | b real }
.........{( 0 c ) |.................}..............{( 0 0 ) |...........}

values in between the {} are matrices.

Show that:

(a) R is a ring
(b) I is an ideal of R
(c) R/I ≈ F ⊕ F, where F is the field of real numbers
• Nov 15th 2011, 04:40 PM
Drexel28
Re: Herstein's matrix proof
Quote:

Originally Posted by ThatPinkSock
Herstein's abstract algebra: Chapter 4 section 3 number 19

If R = {( a b ) | a, b, c real} and I = {( 0 b ) | b real }
.........{( 0 c ) |.................}..............{( 0 0 ) |...........}

values in between the {} are matrices.

Show that:

(a) R is a ring
(b) I is an ideal of R
(c) R/I ≈ F ⊕ F, where F is the field of real numbers

What have you tried? I mean, there is no trickery here, it's just doing it. Are you stuck somewhere in particular, because I know you don't expect us to just do it for you.
• Nov 20th 2011, 05:55 PM
ThatPinkSock
Re: Herstein's matrix proof
Quote:

Originally Posted by Drexel28
What have you tried? I mean, there is no trickery here, it's just doing it. Are you stuck somewhere in particular, because I know you don't expect us to just do it for you.

I expected no such thing!

(a) It suffices to show that R is a subring of M₂(R).

Since the difference and products of upper triangular matrices are again upper triangular, this is indeed the case.

For specifically, let
A = (a b)
......(0 c), and

B = (d e)
......(0 f) be in R.

Then, A - B =
(a-d b-e)
(0 c-f) is also in R.

AB =
(.0......cf...) is in R.

b) For A = (0 a; 0 0), B = (0 b; 0 0) in I,
A + B = (0 a+b; 0 0) is in R.

For A = (0 a; 0 0) and I, and C = (b c; 0 d) in R:

AC =
(0 a)(b c)...(0 ac)
(0 0)(0 d).=(0 0) is in R, and

CA =
(b c)(0 a)...(0 ab)
(0 d)(0 0).=(0 0) is in R.

Hence, I is an (two-sided) ideal of R.

(c) Consider the map F : R → F ⊕ F defined by F((a b; 0 c)) = (a, c).
It's easy to see that this is a surjective ring homomorphism,
with ker F = {(0 b; 0 0) : b in F} = I.

So, the result follows from the first isomorphism theorem.

Everything look alright?