Gcd between Ideals

Question

"If two ideals I and J of a ring R are relatively prime then I+J=R "

How can I determinate if I and J are relatively prime when I and J are not principal?

If they are principal I have to take generators and check if gcd is 1.

I have to proof that if I and J are relatively prime, I^n and J^n are too. It is easy when I and J are principal, but when they aren't?

I also have to proof that if IJ=K^m (K ideal and m>=1) I and J are X^m and Y^m for some ideals X and Y of R, and I totally don't know how to do.

Sorry for my english

Answer 1

The first question is a good one . First the main idea of relatively prime is $ \exists a \in I , b \in J$ such that $ a + b = 1$. Now assume I and J are relatively prime.

Now in $I^n + J^n$ consider $ a^n.b^{n-1} + b^n.a^{n-1} = a^{n-1}b^{n-1}(a + b) = a^{n-1}b^{n-1}$.

Then $I^n,J^n$ has $ a^nb^{n-2} + a^{n-1}b^{n-1} = a^{n-1}b^{n-2}(a + b) = a^{n-1}b^{n-2}$. Thus in this manner we atlast get $a^{n-1}b$ is in the sum. Now then we see that $I^n + J^n$ contains $a^n + a^{n-1}b = a^{n-1}(a + b) = a^{n-1}$. Thus $a^{n-1}, b^{n-1}$ is in the sum .Thus we get atlast then a and b in the sum and hence 1 is there in it.Thus they are relatively prime.

Answer 2

First of all, to prove two ideals are equal you may show they are contained one into the other and viceversa. Of course you have $I+J\subset R$, to prove the opposite you just have to show that $1\in I+J$, so as ideals are closed to scalar multiplication you have $r\cdot 1\in I+J$ for every $r\in R$. Notice that each element of $I+J$ is linear combination of element of $I$ and $J$. for the second question, just try to figure it out trying to show that $1$ belongs to that sum as. For the last point, are you sure it is true for every ring? if you take just $R=\mathbb{Z}$ and $I=J=2\mathbb{Z}$, we have $IJ=I^2$ but we can't prove your claim (for $m=2$).