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Factor Rings and Ideals

Date: 04/22/2003 at 16:16:28
From: Kris
Subject: Factor rings and ideals

I have three questions and no idea where to begin. Please help.

1. Give an example to show that a factor ring of an integral domain 
   may be a field.

2. Let R = {a + b*sqrt2|a, b in Z} and let R prime consist of all 2x2 
   matrices of the form |a 2b|  for all a, b in Z.  
                        |b  a| 
   Show that R and R prime are isomorphic rings.

3. Let R and R prime be rings and let phi:R to R prime be a ring 
   homomorphism such that phi[R] is not equal to {0 prime}. Show that 
   if R has unity 1 and R prime has no 0 divisors, then phi (1) is 
   unity for R prime.

Date: 05/06/2003 at 19:56:12
From: Doctor Nitrogen
Subject: Re: Factor rings and ideals

Hi, Kris:

(1) Look at the factor ring : =


where p is a prime. How many elements does it have? With some effort 
you will be able to convince yourself that this factor ring as defined 
will always be a field and has no 0 divisors. 

(2) You will want to define a homomorphism theta:

   theta: R -------------> R' (your R prime)


   theta( a + b*sqrt(2) ) 

    =  |a   2b|
       |b    a|

with a + b*sqrt(2) in R, a, b in Z, and the 2 X 2 matrix in R'.

Now prove that this homomorphism is well-defined; that is, if p and q 
are both in R you must show

 (A)              theta(p + q) = theta(p) + theta(q),        

 (B)              theta(p*q) = theta(p)*theta(q),

where theta(p), theta(q), are both in R', p, q are in R.

For (A) note that, for c, d in Z:

   theta(a + b*sqrt(2) + c + d*sqrt(2) )

   = theta( (a + c) + (b + d)*sqrt(2) )

   = |(a + c)                   2(b + d)|
     |(b + d)                   (a + c) |

   = |a     2b|        +    |c        2d|
     |b      a|        +    |d         c|

   = theta(a + b*sqrt(2)  +   theta(c + d*sqrt(2) ).

And also, for (B):

   theta( (a + b*sqrt(2))*(c + d*sqrt(2) ) )

   = theta( (ac + 2bd) + (ad + bc)*sqrt(2) )

   = |ac + 2bd                  2(ad + bc)|
     |bc + ad                   2bd + ac  |

   = | a     2b | * | c      2d  |
     | b      a | * | d        c |

   = theta(a + b*sqrt(2))*theta(c + d*sqrt(2)).

To show this map is onto, suppose there is an element in R', say

                      | r         2s |
                      | s          r |

which is NOT the image of at least one element in R. Then there is no 
element in R that can be mapped to that element above, which is in R'. 
But there exist integers r, s in Z such that

                      r + s*sqrt(2)

will be in R. But this element in R must get mapped to R' by theta, a 
contradiction, meaning every element in R' must be the image of at 
least one element in R.

To prove theta is 1-1, try to convince yourself the only element in R 
that gets mapped to the 0 element in R' is

   0 + 0*sqrt(2) = 0,

that  is, that 

   0 = 0 + 0*sqrt(2) -------------> | 0      2(0) |
                                    | 0        0  |

     = 0_2 (the 2 X 2  0 matrix).

Then since the map theta really is well defined, is both 1-1 and 
onto, it must be an isomorphism from R to R'.

(3) You might want to let r be an element of R, and then look at

   phi(r) = phi(1*r) = phi(1)*phi(r).

I hope this helped answer the questions you had concerning your 
mathematics problem. You are welcome to return to The Math 
Forum/Doctor Math whenever you have any math-related questions.

- Doctor Nitrogen, The Math Forum
Associated Topics:
College Modern Algebra

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