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.(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# R/kZ under multiplication

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