# Math Help - Rings problem

1. ## Rings problem

Hi guys, not sure if this is the correct forum for Rings but I'll try it here anyways

I'm working on exam questions and I am very lost. Any help or pointers on the following would be really appreciated as I'm running out of revision time!

Q1
Let R be the subring of C defined by R={a+2bi e Z[i]:a,b e Z}.
Show that the set I={2m+2ni e R: m,n e Z} is an ideal of R.

Show that I is not a principal ideal of R

Q2
Let f(x)=x^3+x+2 e Z_5[x] and let I=<f(x)>. How many distinct cosets of I are there?

In q1 I think you can look at the elements 2 and 2i in I to show that it is not a principal ideal?

Thanks!

2. ## First part of Q1

Ok, I think I've proved that I is an ideal of R as follows:

You need to show that p+q e I, 0 e I, -p e I and rp e I where r e R and p e I

The first three are easy enough, the fourth I did the following(I think its right!)
Let r=m+2ni e R
Then rp= (m+2ni)(2a+2bi)=2ma+2mbi+4nai-4nb
=2(ma-2nb)+2(mb+2na)i which is clearly an element of I

I still don't know how to show that I is not a principle Ideal, probably because I don't know how to get the Principal Ideal!
Anyone any ideas?

3. Originally Posted by musicmental85

Q1
Let R be the subring of C defined by R={a+2bi e Z[i]:a,b e Z}.
Show that the set I={2m+2ni e R: m,n e Z} is an ideal of R.

Show that I is not a principal ideal of R
What does principal mean? It means that $I = \left < a \right>$ for some $a\in I$ (this is the set $\{ ra|r\in R\}$).
Try to argue by assumption that there is a $2a+2bi$ such that $\left< 2a+2bi \right> = I$ and reach a contradiction.

4. I know the dfinition for a principal Ideal. I think the problem is that I don't really understand what <a>=I actually means so Assuming <2a+2bi>=I isn't ringing any bells as to where I can go from there?
Do I need to look at each element as part of I? like 2=2(1)+2(0)i which is an element of I

5. Originally Posted by musicmental85
Q2
Let f(x)=x^3+x+2 e Z_5[x] and let I=<f(x)>. How many distinct cosets of I are there?
The division algorithm says that if p(x) is any polynomial in Z_5[x] then we can divide p(x) by f(x) and get a remainder which must be of degree at most 2. So $p(x) = q(x)f(x) + r(x) \in I+r(x)$. This says that p(x) is in the coset I+r(x), and there is one coset for each polynomial r(x) of degree at most 2. All you have to do is to count how many such polynomials there are.

6. For your second question maybe this will help.

7. That link doesn't open but I think that I has precisely 9 distinct cosets in Z_i? I'm rushing so I can't state the proof yet but this would imply that I has 9 distinct cosets in Z_5 also?

8. Originally Posted by musicmental85
That link doesn't open but I think that I has precisely 9 distinct cosets in Z_i? I'm rushing so I can't state the proof yet but this would imply that I has 9 distinct cosets in Z_5 also?
Try it now.

9. Thanks! That did help