## Ideal inverses

Hi Sir.

Should be a quick question about fractional ideals.

How does one work out the inverse of an ideal, when the ideal is generated by multiple elements (esp. 2 elements). i.e. I am looking to work it out intuitively, instead of a fool-proof method for lengthy ideals.

Thanks.

(you have an example in ufd.pdf where you quoted the inverse of

$(2,\alpha)$

to be

$(\alpha^{-1})$

As you’ve guessed, there is a foolproof method using integral basis for ideals. But perhaps it’s not worth going into the general algorithm now.

In the example you mention, if you read the previous paragraphs carefully, you’ll see that actually, $(2,\alpha)=(\alpha)$. That’s why $(2,\alpha)^{-1}=(\alpha^{-1})$. But for fractional ideals that are not principal, it can be trickier to find inverses easily. Here is my suggestion for now: When we factorize elements, we find relations of the form

$(a)=P_1P_2\cdots P_k$

But then, we see that

$P_1^{-1}=P_2\cdots P_k (a^{-1})$

For example, in $Z[\sqrt{-10}]$, we find

$(11)=P_1 P_2$

where

$P_1=(11,\sqrt{-10}+1), \ \ P_2=(11, \sqrt{-10}-1)$

(I think the $P_i$ are non-principal, but I’m too lazy to check at the moment. In any case, they are not obviously principal, which is all that matters for the present demonstration.)

Therefore,

$P_1^{-1}=P_2(11^{-1})=(1, \frac{(\sqrt{-10}-1)}{11})$

The idea here is the same one that recurs many times in the theory. To get information about an ideal, try to make it occur in the factorization of some principal ideal.