# An Ideal in Z[x]

1. Jul 14, 2008

### e(ho0n3

The problem statement, all variables and given/known data
Prove that the ideal I = $$\langle x^2 + 1 \rangle$$ is prime in Z[x] but not maximal.

The attempt at a solution
I'm having a hard time doing this because Z[x] is not a field. I know that x2 + 1 is irreducible in Z[x] so the proof must hinge on this fact.

Let f(x) and g(x) belong to Z[x] and suppose f(x)g(x) is in I. Then there is some q(x) in Z[x] such that f(x)g(x) = (x2 + 1)q(x). How can I show that either f(x) or g(x) belongs to I? How does the irreducibility of x2 + 1 come into play here?

2. Jul 14, 2008

### matt grime

A field has no ideals (other than 0 and itself) so I don't get your comment about it being harder (unless it was a joke).

What are the definitions of maximal and prime ideal (in terms of quotients)?

Last edited: Jul 14, 2008
3. Jul 14, 2008

### e(ho0n3

It's hard because most of the results I have concerning ideals involve fields.

I know that if R is a commutative ring with unity and A is an ideal of R, then

(1) A is prime iff R/A is an integral domain, and
(2) A is maximal iff R/A is a field.

I can try proving that Z[x]/I is an integral domain to show that I is prime but that's tedious. Is there a slicker way?

4. Jul 14, 2008

### matt grime

It's not tedious, it is simple and takes about one line.

How can most of the results you know about ideals involve fields? As I pointed out fields don't have any non-trivial ideals.

5. Jul 14, 2008

### e(ho0n3

I forgot that Z[x]/I is already a ring since I is an ideal. So all I have to do is show that Z[x]/I contains unity and no zero divisors. I guess this all follows from the fact that Z[x] is an integral domain. Correct?

I have a theorem that states: Let F be a field an let p(x) belong to F[x]. Then $$\langle p(x) \rangle$$ is a maximal ideal of F[x] if and only if p(x) is irreducible in F. Unfortunately I can't use this.

I was also trying to use the division algorithm to show that I is a prime ideal but that only works for fields. The chapter where this problem is from also states a bunch of reducibility results but they all involve fields. This is why I wrote what I wrote.

6. Jul 14, 2008

### matt grime

What, independently of I? You need to prove something because x^2+1 is irreducible, remember. It is trivially true that the quotient isn't a field (just consider the polynomial 2 - it can't have a multiplicative inverse), and equally trivial that it contains a unit.

7. Jul 14, 2008

### e(ho0n3

You're right. I haven't used the fact that xx + 1 is irreducible. Is this used to show that Z[x]/I has no zero divisors?

Consider the product (f(x) + I)(g(x) + I) = f(x)g(x) + I. If this product equals I, then f(x)g(x) = 0. It follows that the coefficients of f(x)g(x), which are just products of integers equals, equal 0. This means one of the integer factors of each of the coefficients is 0 and so one of f(x) or g(x) must be the 0 polynomial. Nowhere did I use the fact that xx + 1 is irreducible. Where did I go wrong?

8. Jul 14, 2008

### Hurkyl

Staff Emeritus
Why?

9. Jul 14, 2008

### e(ho0n3

Hmm...I guess that isn't necessarily true. If f(x)g(x) + I = I, then all I can say is that f(x)g(x) is a member of I and so there is some q(x) in Z[x] such that f(x)g(x) = (xx + 1)q(x). Now what?

10. Jul 14, 2008

### Dick

Let r_f(x) and r_g(x) be the remainders after division of f(x) and g(x) by (x^2+1). If that relation is true then r_f(x)*r_g(x)=0. Now show Z[x] has no zero divisors.

11. Jul 15, 2008

### matt grime

Ugh, all this dividing and so on. Horrible. This is Z[x], after all, so 'division' isn't that nice to contemplate (we should really be thinking about the cosets f(x)+I as used above, anyway).

I much prefer thinking of it as "x^2=-1".

This makes it clear that we can think of Z[x]/(x^2+1) as

{a+bx : a,b in Z and x^2=-1}

and so zero divisors in Z[x]/(x^2+1) would imply that there are two deg 1 polys f(x) and g(x) such that f(x)g(x)=n(x^2+1) for some integer n.

(Of course, you can always tensor with Q and use division algorithms there.)

12. Jul 15, 2008

### Dick

You are right. After rereading it, it looked so ugly I was going to delete it. Then the forum went offline.

13. Jul 15, 2008

### e(ho0n3

Why is it that you can treat xx as -1? I never understood this.

14. Jul 15, 2008

### e(ho0n3

Oh, and since f(x)g(x) = n(x2 + 1), how is it possible for f(x) or g(x) to be in I? I don't understand how I is a prime ideal yet.

15. Jul 15, 2008

### matt grime

You treat x^2 as -1 since you're treating x^2+1 as zero: that's what the notation Z[x]/(x^2+1) means.

I don't understand your last post. The point was that if it is conceivable that two linear polys multiply together to be in I, then it must follow that f(x)g(x)=n(x^2+1). Polys of the form n(x^2+1) are the only polys of degree two in I. But x^2+1 is irreducible, so there you can't write (an integer multiple of) it as a product of two deg 1 polys.

16. Jul 15, 2008

### e(ho0n3

So in other words, f(x) and g(x) can't be linear. Right? But then what other choice do they have? How can I show that f(x) or g(x) is in I and thus prove that I is a prime ideal?

17. Jul 15, 2008

### matt grime

How many ways can I write n(x^2+1) as a product of two polynomials, recalling x^2+1 is irreducible over Z....? This is not hard algebra, this is stuff you were taught in high school.

18. Jul 15, 2008

### e(ho0n3

You could consider n as one polynomial and x2 + 1 as another polynomial.

Now what? Are you suggesting that f(x) must be n (or x2 + 1 ) and g(x) must by x2 + 1 (or n)? But this will contradict the fact that f(x) and g(x) have degree at most 1.

19. Jul 15, 2008

### matt grime

Uh, what? You've lost me as to what it is that you think that you're doing.

20. Jul 15, 2008

### e(ho0n3

I'm trying to show that I is an ideal. I'm using the following definition.

A prime ideal A of a commutative ring R is a proper ideal of R such that a, b in R and ab in A imply a in A or B in B.

So I have to show that either f(x) or g(x) is in I.