# Thread: Abstract Alegbra - Ideal Rings

1. ## Abstract Alegbra - Ideal Rings

In each case, determine whether the set is an ideal in the ring Z × Z. If it is, prove it; if it isn’t, give a speciﬁc counterexample to one of the axioms.

(a) 8Z × 11Z. (This sets consists of pairs where the ﬁrst element is a multiple of 8 and the second is a multiple of 11.)

(b) <(8,11)> (This is not the same set as the one in (a)!)

This is what I have:

(a)

Ring: Z x Z
Subset: 8Z x 11Z

S = 8Z x 11Z = { (8a, 11b) | a,b ∈ Z }

(8a + 11b) + (8c + 11d) = [ 8(a + c) + 11(b + d)]
∈ S

(0, 0) = [ 8*(0) + 11*(0) ]
∈ S

-(8a, 11b) = [ 8(-a), 11(-b) ]
∈ S

(8a, 11b)
∈ 8Z x 11Z and (x, y) ∈ Z x Z

then, (8a, 11b) * (x, y) = (8(ax), 11(by))
∈ 8Z x 11z

so, this shows that S is an ideal

However, I do not know how to do part b

2. ## Re: Abstract Alegbra - Ideal Rings

Your notation $(8a+11b)\in S$ implies $S = 8\mathbb{Z} \bigoplus 11\mathbb{Z}$ instead of $8\mathbb{Z}\times 11\mathbb{Z}$. Now, these two rings are isomorphic, so I can see how one might be tempted to interchange the notation. Otherwise, everything you wrote for part (a) looks good.

For part (b), $\langle(8,11)\rangle = \{(8a,11a)\in \mathbb{Z}\times\mathbb{Z} \mid a\in\mathbb{Z}\}$

3. ## Re: Abstract Alegbra - Ideal Rings

Oh, on my paper I didn't even have (8a + 11b) + (8c + 11d). I had it as: (8a, 11b) + (8c, 11d) which is what I meant to type the first time. I don't know why I kept putting "+" in the middle of the (). That was just a typo.

For part (b) would this be correct?

Ring: Z x Z
Subset: <(8, 11)>

S = <(8, 11)> = { (8a, 11a) ∈ Z x Z | a ∈ Z }

(8a, 11a) + (8b, 11b) = (8(a + b), 11(a + b)) ∈ S

(0,0) = (8*(0), 11*(0)) ∈ S

-(8a, 11a) = (8(-a), 11(-a)) ∈ S

(8a, 11a) ∈ <(8, 11)> and (b, b) ∈ Z x Z

then, (8a, 11a)*(x, y) = (8(ba), 11(ba)) ∈ <(8, 11)>

so it is an ideal

4. ## Re: Abstract Alegbra - Ideal Rings

Originally Posted by AwesomeHedgehog
Oh, on my paper I didn't even have (8a + 11b) + (8c + 11d). I had it as: (8a, 11b) + (8c, 11d) which is what I meant to type the first time. I don't know why I kept putting "+" in the middle of the (). That was just a typo.

For part (b) would this be correct?

Ring: Z x Z
Subset: <(8, 11)>

S = <(8, 11)> = { (8a, 11a) ∈ Z x Z | a ∈ Z }

(8a, 11a) + (8b, 11b) = (8(a + b), 11(a + b)) ∈ S

(0,0) = (8*(0), 11*(0)) ∈ S

-(8a, 11a) = (8(-a), 11(-a)) ∈ S

(8a, 11a) ∈ <(8, 11)> and (b, b) ∈ Z x Z

then, (8a, 11a)*(x, y) = (8(ba), 11(ba)) ∈ <(8, 11)>

so it is an ideal
You are correct right up until the last part:

If $(8a,11a) \in S, (x,y) \in \mathbb{Z} \times \mathbb{Z}$, then $(8a,11a)\cdot (x,y) = (8ax,11ay)$. Since $(8ax,11ay) \in S$ if and only if $x=y$, $S$ is not an ideal.

5. ## Re: Abstract Alegbra - Ideal Rings

One has to be a bit careful, here, because one often uses the notation <(8,11)> for the ideal generated by the element (8,11).

It should be clear that in this ideal, we have (2,1)*(8,11) = (16,11) also in the ideal, so that (16,11) - (8,11) = (8,0) is in the ideal as well.

Also, (1,2) = (8,22) is in this ideal so that: (8,22) - (8,11) = (0,11) is in the ideal generated by (8,11). From these two facts, we may deduce that the ideal generated by (8,11) contains $8\Bbb Z \times 11\Bbb Z$.

On the other hand, since $8\Bbb Z \times 11\Bbb Z$ certainly has (8,11) as an element, and is an ideal by part (a) of your exercise, $8\Bbb Z \times 11\Bbb Z$ contains the ideal generated by (8,11).

In other words the IDEAL generated by (8,11) is $8\Bbb Z \times 11\Bbb Z$, which is considerably LARGER that the sub-ring generated by (8,11).