Associated Topics || Dr. Math Home || Search Dr. Math

### Factor Rings and Ideals

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

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 : =

Z/pZ

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)

as

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).

Forum/Doctor Math whenever you have any math-related questions.

- Doctor Nitrogen, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Modern Algebra

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search