Abstract algebra-> Let R be a ring and let M2(R) be the set of 2 x 2 matrices with

[cooljosh2k2]

Abstract algebra--> Let R be a ring and let M2(R) be the set of 2 x 2 matrices with

Homework Statement

Let R be a ring and let M2(R) be the set of 2 x 2 matrices with entries in R.
De fine a function f by:

f(r) = (r 0) <----matrix
...(0 r)

for any r ∈ R

(a) Show that f is a homomorphism.
(b) Find ker(f).

The Attempt at a Solution

I just want to know if this is right:

a) For any r, s in R,
f(r) + f(s) = (r 0)..+..(s 0)
....(0 r)...(0 s) =

((r+s) 0)
(0 (r+s)) = f(r + s).

Therefore, f is a group homomophism.

b) ker f = {r ∈ R: f(r) = zero matrix}.

This is only possible if r=0.

Therefore, ker f = {0}
Hence, ker f = {0}

 

[mighty2000]

.

Hello,

Your solution is correct! Good job. Here is another way to approach part (b):

Since f is a homomorphism, we know that ker(f) is a subgroup of R. Since R is a ring, it is also an abelian group under addition. Therefore, we can use the fundamental theorem of homomorphisms which states that the kernel of a homomorphism is a normal subgroup of the domain. Since ker(f) is a subgroup of R and R is abelian, ker(f) must also be a normal subgroup of R.

Now, let's consider an element (a b) in M2(R) where a and b are in R. This element is in the kernel of f if and only if f(a b) = (0 0). This happens if and only if (a 0) = (0 b). This is only possible if a = b = 0. Therefore, the kernel of f is the set of all (0 0) matrices, which is the zero matrix.

Hence, ker(f) = {(0 0)} = {0}.

Keep up the good work!