# Ring homomorphisms and ideals

1. Apr 27, 2004

### nocheesie

I'm having a very tough time understanding homomorphisms and ideals, probably because i'm very fuzzy with the concept of rings. I'm stuck on the following problem:
Find all the ideals in the following rings:
1. Z
2. Z[7] (Z subscript 7, equivalence classes of 7 i'm guessing)
3. Z[6]
4. Z[12]
5. Z[36]
6. Q (set of rationals)
7. Z (this one wasn't written as a subscript, but as brackets, and "i" as the complex number i. what does that mean?)

Can someone help me to understand these better? I'd really really appreciate it.

2. Apr 27, 2004

### marcus

what about all the multiples of 3?

or all the multiples of 25, or 100?

I think the set of multiples of some number is an interesting set
and isnt it an ideal
(all an ideal means is a subset closed under +
and such that multiplying an element by anything doesnt evict it from
from the set)

but multiplying an even number by anything you still get an even number
multiplying a multiple-of-100 by anything you still get a multiple-of-100

these sets must be ideals!

3. Apr 27, 2004

### Nexus[Free-DC]

Okay, an ideal I in a ring R is a subset of R so that the product of anything in the ideal with anything in the ring is in the ideal again. For example, the even integers are an ideal in the integers because an even number times any integer is even. Same goes for 3,4,5, etc. Note that R is an ideal of R, and so is {0} because anything times zero is zero.

I'll get you started on a few of the questions:

1. You want to find all sets I so that for any i in I, and r in Z you have ir in I. It shouldn't be too hard to see that these Is are just things like {...,-2,0,2,4,6,...}, {...,-6,-3,0,3,6,...} etc. That is, the ideals in Z are {0},Z,2Z,3Z, etc.

6. There are only two ideals in Q, {0} and Q. To see why, consider any ideal other than {0}. It contains some fraction a/b. But then a/b * b/a must be in the ideal, so 1 is in the ideal. So for any c/d in Q, 1*c/d is in the ideal. That is, the ideal contains all the fractions so the ideal is Q.

7. Okay, this is a notation thing. If you have a ring R, then the ring R[x] is all the polynomials of x with coefficients in R. For example Q[x]
would be all polynomials with rational coefficients.

In this case, Z is all polynomials like 5i^5+10i^2-i+19. But i^2=-1 so if you simplify all the powers of i, all you are left with is complex numbers with integer coefficients. These are called the 'Gaussian' integers.

Hope this helps.

4. Apr 27, 2004

### marcus

suggest you learn to write Z7

to see how anybody at Physicsforums writes something, pretend you are going to quote them and respond------so click on "quote" underneath their post
you will see how they wrote something
then you can just click "back" on your browser and go away without responding if you please----you dont have to go thru with responding if you dont want

in the case of that, if you press "quote" on this post you will see
that Z7 is written
Z[s u b]7[/s u b]

but removing the spaces in s u b
-----------------------------

yeah Z7 means the integers modulo 7
which you can think of as equivalence classes
or you can think of them as the set {0, 1, 2, 3, 4, 5, 6}
where after ever + and every x
you mod out any multiple of seven

I dont think Z7 has any ideals excpt the zero ideal
and the whole Z7

because just try
suppose you had an ideal with 2 in it, it has to be closed under addition
so 2+2 has to be in the ideal as well!
2, 2+2, 2+2+2, then the next is 2+2+2+2 = 1
but if 1 is in the ideal then everything is
because you can get anything by adding one to itself.

go ahead and try anything in Z7
like 3
if 3 is in then 3+3+3 = 2
is in and if 2 is in then everything is

Several people here (eg Matt Grime for one) can tell you a theorem that covers all these cases, notice that
7 is a prime number

but sometimes understanding can come just by being patient enough to experiment

Z6 might be different! because 6 is not a prime number
it might be interesting to find the ideals in the ring
Z6

or Z12
------------------

I want to see what comes of this

later we can look at Z
which I think is what happens when you throw in the imaginary number thesquareroot of -1
throw that in with the integers

Personally I dont think the rational numbers have any nontrivial ideals
they got nothing but the zero ideal {0} and the whole ring itself
they are a "field" meaning for one thing that multiplicative inverses are
included, so any ideal with nonzero q in it also has 1
because you can multiply q by 1/q and it should not evict you from the ideal so that means 1 is in the ideal, and then it is all over because an ideal
with 1 in it is the whole ring!

but Z might be good to look at because it does not have inverses
----------------

however first things first
lets get all the ideal of Z6

5. Apr 27, 2004

### marcus

in the Z6 case
and what happens if you multiply 3 by
anything in Z6?

does it ever evict you from {0,3}?
or is the result of multiplying always still in {0,3}?

you have to check it but I think that set is
an ideal in Z6

and there may be another one which is not just
the zero ideal or the whole Z6 ring

6. Apr 27, 2004

### nocheesie

Wow you guys are soooo helpful!! Soo much better than my book which really doesn't explain things very well and has very poor examples. I can't thank you guys enough!

7. Apr 28, 2004

### matt grime

Z_7 and Q are fields, I can tell you a theorem about that, but the general one would be that if I is an ideal and I contains an invertible (multiplicative element) then it is the whole ring, that helps the by hand experimentaion in Z_6 say.

Z is a principal ideal domain - you'd better look that up, and perhaps euclidean domain too, they have useful properties and characterisations.