Annihilator of a Quotient Ring

[Sudharaka]

Hi everyone, :)

I think I need to refresh my memory about annihilators and quotient rings. Hope you can help me with the following example.

I want to find the annihilator of $a'$ and $b'$ of the quotient ring $R=\mathbb{Z}/(a'b')$ where $a',\,b'>1$. So if I go by the definition, $ann(a')=\{r\in \mathbb{R}\mid a'r=0\}=\{a' \mathbb{Z}+b' \mathbb{Z}+(a'\,b')\in \mathbb{R}\mid a'(a' \mathbb{Z}+b'\mathbb{Z})=0\}$Am I correct unto this point?

 

[Sudharaka]

Hi everyone, :)

I think I need to refresh my memory about annihilators and quotient rings. Hope you can help me with the following example.

I want to find the annihilator of $a'$ and $b'$ of the quotient ring $R=\mathbb{Z}/(a'b')$ where $a',\,b'>1$. So if I go by the definition, $ann(a')=\{r\in \mathbb{R}\mid a'r=0\}=\{a' \mathbb{Z}+b' \mathbb{Z}+(a'\,b')\in \mathbb{R}\mid a'(a' \mathbb{Z}+b'\mathbb{Z})=0\}$Am I correct unto this point?

Hi everyone, :)

Since I was in a hurry to find the answer to this question I posted this on StackExchange and I got an answer to this there.

>>Link to Thread<<

 

[Deveno]

Your first post is rather jumbled, and the notation is weird.

Suppose $R = \Bbb Z/(ab)$. I think by:

$a'$ you must mean $a' = a + (ab) \in \Bbb Z/(ab)$.

If we want to know what:

$\text{ann}(a') = \{k + (ab) \in \Bbb Z/(ab): (k + (ab))(a + (ab)) = 0 + (ab)\}$

is, this is the same thing as asking:

For what $k \in \Bbb Z$ do we have:

$ka \equiv 0\text{ (mod }ab)$?

The first thing to notice about this, is that if gcd(k,ab) = 1, this will never happen.

On the other hand, it is obvious that if $k \equiv tb\text{ (mod }ab)$ it will ALWAYS happen.

Now, if:

$(k + (ab))(a + (ab)) = ka + (ab) = (ab)$

it is necessarily the case that $ka \in (ab)$. This means that:

$ka = tab$ for some integer $t$. The integers are cancellative, so we can deduce from this that:

$k = tb$, which is the same thing as saying $k \in (b)$.

Thus $k' \in \text{ann}(a') \iff k \in (b)$, that is:

$\text{ann}(a') = (b') =$ the ideal generated by $b + (ab)$ in the quotient.

Let's look at a specific example, to see how this actually goes down:

Let $a = 4, b = 6$, and we shall work mod 24. As we saw above, we need not consider the congruence class (cosets) of:

1,5,7,11,13,17,19,23 which only leaves 16 cosets to check.

It is immediate that 0,6,12,18 all annihilate 4 (or rather the coset of 4), so we need only show that:

2,3,4,8,9,10,14,15,16,20,21,22 do NOT annihilate 4. This can be done explicitly:

2*4 = 8
3*4 = 12
8*4 = 8
9*4 = 12
10*4 = 16
14*4 = 8
15*4 = 12
16*4 = 16
20*4 = 8
21*4 = 12
22*4 = 16

so it is indeed the case that 6 generates the annihilator.