# Why there is not ring homomorphism between these rings?

1. Apr 2, 2010

1. The problem statement, all variables and given/known data

Proof that there is no ring homomorphism between M2(R) [2x2 matrices with real elements] and R2 (normal 2-dimensional real plane).

2. Relevant equations
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3. The attempt at a solution
I have tried to proof this problem with properties of ring homomorphism (or finding such a property of ring homomorphism that doesn't fill in this situation. I have tried nilpotents, inverse elements, etc.)

This would be easy one if those two rings were integral domains. In general speaking, how to proof that there does not exist any ring homomorphism between two rings when the rings aren't integral domains?

Thanks folks.

2. Apr 2, 2010

### Dick

I'll guess M_2(R) has the usual ring structure, but what is your multiplication definition for R^2? And what properties is your homomorphism supposed to have? There's always the zero homomorphism.

3. Apr 3, 2010

Multiplication in R^2 is defined here: (a, b)*(c, d) = (ac, bd). The homework does not specify properties of homomorphism. The assingment of the homework says also that this can be done without long calculations.

If I got it right, I have understood that there is zero homomorphism f: R->S only if S is zero ring (because of condition f(1R)=1S).

4. Apr 3, 2010

Multiplication in R^2 is defined here: (a, b)*(c, d) = (ac, bd). The homework does not specify properties of homomorphism. The assingment of the homework says also that this can be done without long calculations.

If I got it right, I have understood that there is zero homomorphism f: R->S only if S is zero ring (because of condition f(1R)=1S).

5. Apr 3, 2010

### Dick

You are right. I was thinking of groups. Assuming multiplication in M_2(R) is matrix multiplication, that is not commutative. Multiplication as you've defined it in R^2 is. Wouldn't that pose a problem for constructing a homomorphism?

6. Apr 3, 2010

I did actually think of this, but abandoned because of some brainfart :) If I understood correctly, all I need to do is assume first that there is ring homomomorphism, let it to be f: M2 -> R2. Then f(AB)=f(A)*f(B)=f(B)*f(A)=f(BA), but AB=BA is not universaly true (when A and B are matrices).

Is this enough, or do I need to show something more? Is it trivial that f(AB) is different than f(BA), when f is ring homomorphism. I'm not very good algebra, so I can't see it.

7. Apr 3, 2010

8. Apr 3, 2010

### Dick

No, it's not automatic that f(AB) is different from f(BA). I think you have to use that f(AB)=f(BA) to draw some conclusions about the homomorphism. Take for example A=[[0,1],[0,0]] and B=[[0,0],[0,1]]. Then AB=A and BA=0. That tells you f(A)=0. That's a start. I haven't actually thought this all the way through. See what you can come up with.

9. Apr 4, 2010