# Thread: algebra - finding ideals

1. ## algebra - finding ideals

Just wondering if anyone knows of a quick method for finding ideals from a multiplication table?

e.g say you are looking at Z/6Z, which has ideals {0,2,4} and {0,3} is there any way you can look at the table and just spot immediately what they are?

If anyone can share their wisdom it would be appreciated!

2. Hello,

If you're talking about groups with + not *, I think I can help...

The cardinal of the ideal divides the cardinal of the original set, here it's Z/6Z
Hence, the ideals have possible cardinals : 1,2,3,6.

This also helps in finding the elements of the ideals : starting from 0, the following element of the ideal will be $\frac{n}{\text{cardinal of the ideal}}$, if we work in Z/nZ.

For example, the ideal of Z/6Z which has 3 elements will be the set of elements x such as :
$\{x=\frac 63 k \ \big/ \ k \in \{0;1;2\} \}$

This means that the ideal of Z/6Z with 3 elements is : $\{0;2;4\}$

I hope this is clear enough

3. Originally Posted by hunkydory19
Just wondering if anyone knows of a quick method for finding ideals from a multiplication table?

e.g say you are looking at Z/6Z, which has ideals {0,2,4} and {0,3} is there any way you can look at the table and just spot immediately what they are?
When dealing with $\mathbb{Z}_n$ all subgroups (of the addition group) are ideals (not true in general but for modulo rings it works). Futhermore, all subgroups of $\mathbb{Z}_n$ have the form $k\mathbb{Z}_n = \{ kx| x\in \mathbb{Z}_n\}$ where $k$ is a divisor of $n$.

Thus, in $\mathbb{Z}_6$, your example, the subgroups (which will turn to be ideals are:
$1\mathbb{Z}_6 = \mathbb{Z}_6$
$2\mathbb{Z}_6 = \{ [0],[2],[4]\}$
$3\mathbb{Z}_6 = \{ [0],[3]\}$
$6\mathbb{Z}_6 =\{ [0]\}$.