2
votes
2answers
99 views

Ring such that $q^2\mid p^2$ does not imply $q\mid p$?

Let $R$ be a commutative ring with $1$ and suppose $q^2\mid p^2,$ for $p,q \in R$. Unless $R$ is a UFD, I don't believe I can conclude that $q\mid p,$ but I would like to know a concrete ...
1
vote
6answers
646 views

Proof that $\mathbb{Z}$ has no zero divisors

Everyone knows the rules of zero divisors like $$\forall \alpha,\beta\in\mathbb{R}\;:\;\alpha\cdot\beta = 0\Rightarrow\alpha=0\vee \beta=0.$$ But how can I prove it for $\mathbb{Z}$? My first try was ...
6
votes
3answers
403 views

If $R$ is a commutative ring with identity, and $a, b$ are its elements that are divisible by each other, is it true that they must be associates?

Thank you very much! My problem is: If $R$ is a commutative ring with identity, and $a, b$ are its elements that are divisible by each other, is it true that they must be associates? Here, $a$ ...