# An Ideal in Z[x]

Homework Statement
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?

matt grime
Homework Helper
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)?

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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?

matt grime
Homework Helper
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.

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.

matt grime
Homework Helper
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?

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.

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?

Hurkyl
Staff Emeritus
Gold Member
... = f(x)g(x) + I. If this product equals I, then f(x)g(x) = 0.
Why?

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?

Dick
Homework Helper
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.

matt grime
Homework Helper
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.)

Dick
Homework Helper
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.)

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

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

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.

matt grime
Homework Helper
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.

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?

matt grime
Homework Helper
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.

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.

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

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.

matt grime
Homework Helper
I is by definition an ideal.

And why have you gone back to this? You were attempting to argue something about zero divisors. It is elementary to show that the ring Z[x]/I is a domain - if it's not a domain I kind find two polynomials ax+b and cx+d such that (ax+b)(cx+d) is a multiple of x^2+1, but that's impossible as x^2+1 is irreducible over Z.

You know what, you're right. My mind is really slow today. What I wanted to show is that if (f(x) + I)(g(x) + I) = I, then either f(x) + I = I or g(x) + I = I.

(f(x) + I)(g(x) + I) = I led to f(x)g(x) in I and this is where I got stuck and confused

Then you wrote: "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."

I understand this clearly now. Thanks a lot again.