R/kZ under multiplication

1. Nov 25, 2013

Mandelbroth

I'm trying to figure out if $\mathbb{R}/k\mathbb{Z}$ would form a field if we allowed for an "induced" multiplication from $\mathbb{R}$, with $r\in\mathbb{R}/k\mathbb{Z}$ identified with $r+k\in\mathbb{R}/k\mathbb{Z}$. So, if we wanted to multiply $3\in\mathbb{R}/4\mathbb{Z}$ by $\frac{1}{2}\in\mathbb{R}/4\mathbb{Z}$, we'd get 3/2, but if we wanted to multiply $3\in\mathbb{R}/4\mathbb{Z}$ and $2\in\mathbb{R}/4\mathbb{Z}$, we'd get 6-4=2. I'm not convinced that this is a field, despite what my professor told me in a really off-topic discussion from polar coordinates.

Clearly, it forms an abelian group under addition. It also has a commutative, associative, and distributive multiplication law with an identity. Now I just need to show that it has an inverse for every nonzero element. I tried to prove this by proving that there isn't a nontrivial proper ideal in $\mathbb{R}/k\mathbb{Z}$, but that isn't getting me anywhere.

Rather than help me with just this specific problem, though, I'd also like to know if there is a general algorithm/procedure/etc. for showing that a commutative ring is a field.

Any help is much appreciated. Thank you.

2. Nov 25, 2013

jgens

Your multiplication has some problems here. Notice that 0+4Z = (0+4Z)(1/2+4Z) = (4+4Z)(1/2+4Z) = 2+4Z using your rule.

3. Nov 26, 2013

Mandelbroth

Aha! There we go. Thank you.

Do you have any idea how I could notice things like that in the future?

4. Nov 26, 2013

Office_Shredder

Staff Emeritus
Anytime you are quotienting out by a set (in this case kZ) the first question should be "are my operations still well-defined?" And until you prove that they are there's no point in proceeding. And by 'well-defined' we always mean what happens if I pick different representatives of the same co-set, so if you tried some examples you would quickly find the problem here.

5. Nov 26, 2013

jgens

Sure. If you have a ring R and a subgroup I then generally R/I inherits multiplication from R only when I is an ideal. So basically whenever you quotient out by something that is not an ideal, you should not expect to end up with a ring.

6. Nov 26, 2013

Mandelbroth

I probably should have thought of that. I just wasn't thinking of $\mathbb{R}$ alone in terms of rings.

Thank you. You've been very helpful and patient.