Relatively prime ideals/comaximal ideals

Question

Let $R$ be a commutative ring.

a) Prove that if $I$, $J$ are proper ideals of $R$ with $I + J = R$ then for any $a, b \in R$ there exists $x \in R$ such that $x + I = a + I$ and $x + J = b + J$.

b) Show that if $I$, $J$ are ideals of $R$ with the property that for all $a, b \in R$ there exists $x \in R$ such that $x + I = a + I$ and $x + J = b + J$ then $I + J = R$.

c) Give an example of a UFD $R$ with relatively prime elements $r$ and $s$ and elements $a$ and $b$ in $R$ such that there is no $x$ in $R$ with $x + \langle r\rangle = a + \langle r\rangle$ and $x + \langle s\rangle = a + \langle s\rangle$.

The first part of the problem I think I've solved.

Since $I + J = R$ there exists $c \in I$ and $d \in J$ such that $c + d = 1$. Let $x = ad + bc$. Then $x + I = (ad + bc) + I$ = {since $bc$ is an element of $I$} = $ad + I$ = {$d = 1-c$} = $a(1-c) + I$ $= a - ac + I$ = {$ac$ is in $I$ since $c$ is in $I$} = $a + I$. In the same way it's possible to show that $x + J = b + J$.

But I'm having a really hard time with b) and c). If I can show that there exists $c \in I$ and $d \in J$ such that $c + d = 1$ then b) is done. But I can't seem to get there. I would also be really grateful for any suggestions on c). (This is not a homework problem to be handed in, just a problem I've been trying really hard to rap my head around). Thankful for any input!

Answer 1

Hint for 2

Pick an $x$ so that $x+I=0+I$ and $x+J=1+J$. Then use this to show that $1=(1-x)+x \in I+J$.

For 3, my algebra is rusty, but by $1)$ this is not possible if $<r>+<s>=R$, which is the case in an Euclidian domain. So I would look at an UFD which is not Euclidian.

By $1)$ and $2)$ the problem in $3)$ simply reduces to finding an UFD and two prime elements $r,s$ so that $1 \notin <r>+<s>$.

Answer 2

I think I might have found an example that solves c).
Consider Z[x]. That is a UFD but not a Euclidean domain.
Let < r > = < x^2 + 2 > and < s > = < x^3 >. Then there are no polynomials a(x) and b(x) such that (x^2 + 2)a(x) + x^3b(x) = 1 since the constant term of the left hand side is equal two times the constant term of a(x), which we can denote a_0. But the equation 2a_0 = 1 has not solution in Z[x] since the coefficients of the polynomial a(x) have to be integers.

By N.S. remark, since we have shown < r > + < s > is not equal to 1, this solves (I hope!) exercise c).