Coprime ideals in ring

Question

Given the commutative ring R with unity and $a,b,c \in R$, prove that if the following are coprime principal ideals $$ ⟨a⟩=\{\ ar\ |\ r∈R\ \}\ and\ ⟨b⟩=\{\ br\ |\ r∈R\ \} $$ and $a\ |\ bc$ (a divides bc), then $a\ |\ c$ (a divides c).

I know that I have to show that a and b are "coprime elements".

For example in $ \mathbb{Z}$, the ideals $m \mathbb{Z}$ and $n \mathbb{Z}$ are coprime ideals when $ \mathbb{Z} = m\mathbb{Z}+ n\mathbb{Z} = GCD(m,n) \mathbb{Z}$ and $GCD(m,n) = \pm1$. So I understand the concept, but when the ring is arbitrary, I'm not sure how can I prove the statement.

In the quotient ring $R/\langle a \rangle$, the element $b+\langle a \rangle$ is a unit. Now, use the fact that units are never zero divisors. — Geoffrey Trang, Apr 29 '21 at 18:47

score 2 · Accepted Answer · answered Apr 29 '21 at 18:39

2

It's the same as in the case with $\mathbb Z$. If $I$ and $J$ are coprime ideals of a ring $R$, this means that $I+J = R$. So in particular, you can write $1 = i+j$ for $i \in I, j\in J$. Write out what this means for your ideals and a proof will be readily apparent.

answered Apr 29 '21 at 18:39

D_S

33,891

That was the thing I tried, $ar + bs = 1$, but I'm still stuck on how I get the contradiction. Could you give me a little more details? Thanks in advance! – Paul Apr 29 '21 at 18:44
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Multiply both sides by $c$ – D_S Apr 29 '21 at 18:58

Coprime ideals in ring

1 Answers1