# Ideals of the Gaussian integers

1. Aug 22, 2011

### murmillo

I'm working on an exam that Michael Artin once gave, where one of the questions is basically,

Consider the homomorphism from Z[x] to Z given by x --> i. What does this homomorphism tell you about the ideals of Z?

So far I haven't come up with anything. I know in advance that the ideals of the Gaussian integers are principal, but I don't see how I can prove that just by looking at the given homomorphism. I know that there is a bijective correspondence between ideals of Z[x] containing x^2 + 1 and ideals of Z, but I'm stuck. For example, if I look at the ideal generated by x^2 + 1 and x+1, how do I know whether this is a proper ideal or not? If it is proper, then I know that the ideal generated by i + 1 is a proper ideal of Z. But I don't know how to tell whether or not the ideal generated by x^2 + 1 and x+1 is proper. Do you guys think I'm going about this problem the right way? I've spent like half an hour thinking about it and am not making good progess.

2. Aug 23, 2011

### Landau

Well, it consists of all elements of the form $$r(x^2+1)+s(x+1)$$ with r,s ring elements. So it cannot contain, e.g., the element x.

3. Aug 23, 2011

### Hurkyl

Staff Emeritus
While true, that conclusion is certainly not immediate.

4. Aug 23, 2011

### Hurkyl

Staff Emeritus
Have you tried simplifying?

e.g. the ideal <x-3, 2x+5> is equal to the ideal <x-3, 11>.

I often find questions about ideals are easier to solve by looking at the quotient rings.

5. Aug 23, 2011

### murmillo

Yes, I've tried simplifying, but it might be difficult to do in general. I see that if I multiply x by x+1 and subtract x^2 + 1 I get x-1, then I can add it to x+1 to get 2x, then I can multiply 2 by x+1 and subtract 2x to get 2, but I don't know if it's possible to get 1 in the ideal. I also tried with x+n instead of x+1 but haven't had success. Something tells me I might be going about this problem the wrong way.

6. Aug 23, 2011

### Hurkyl

Staff Emeritus
So <x^2 + 1, x+1> = <x^2 + 1, x+1, x-1, 2>. You said you haven't had much luck finding interesting new elements to add to the list of generators. Can you remove any, to get a simpler generating set?

7. Aug 23, 2011

### murmillo

Well I can get <x^2 + 1, x+1, x-1, 2> = <x^2 + 1, x-1, 2> = <x-1, 2>. This ideal can't be principal because x-1 is irreducible. Here it seems that I can't get x in this ideal, because everything generated by 2 is divisible by 2, and I'd have to multiply x-1 by something divisible by 2 if I'm to get a linear combination that cancels out terms with x's, but that would leave me with something divisible by 2. But I'm not sure what that tells me about the corresponding ideal in Z.