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Why there is not ring homomorphism between these rings?

  • Thread starter wadd
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  • #1
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Homework Statement



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

Homework Equations


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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.
 

Answers and Replies

  • #2
Dick
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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
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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
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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
Dick
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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).
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
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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?
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
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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).

You would actually need to provide such an example of matrices A and B such that f(AB) is not the same as f(BA). Only then is your proof complete.
 
  • #8
Dick
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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.
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
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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.
OK. Since f(A)=0 and A is not zero matrix, then Ker(f) has other elements than 0, so our homomorphism can't be injective.

So our f can be only surjective now, and I should proof that Im(f) isn't R2. But I haven't any clue how to do that, again.
 
  • #10
Dick
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OK. Since f(A)=0 and A is not zero matrix, then Ker(f) has other elements than 0, so our homomorphism can't be injective.

So our f can be only surjective now, and I should proof that Im(f) isn't R2. But I haven't any clue how to do that, again.
I would try to show f(x)=0 for any x in M2(R). You've already got f([[0,1],[0,0]])=0. Find some more matrices that map to 0.
 

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